Energy Levels and Spectral Series | Kumar Physics Classes

If you are searching for a Physics Tutor and still find Energy Levels and Spectral Series difficult, contact Kumar Sir for one-to-one online Physics classes.
📞 +91-9958461445 | 📧 kumarsirphysics@gmail.com | 🌐 kumarphysicsclasses.com

Chapter: Atoms · NCERT Class 12 Physics

Energy Levels and Spectral Series

Quantised energy levels, electron transitions, photon emission & absorption, spectral series, excitation & ionisation energy — with full theory, diagrams, numericals & PYQs.

⚡ Energy Level Diagram 🔄 Electron Transition 💡 Photon Emission 🔆 Photon Absorption 📐 Frequency Formula 🌊 Wavelength Formula ⬆ Excitation Energy ⚛ Ionisation Energy 🌈 Spectral Series 📊 Important Graphs 🔢 Numericals 📝 PYQs

1 Energy Level Diagram

⚡ Quantised Energy Levels

Bohr showed that the electron in a hydrogen atom can only occupy certain discrete (quantised) energy levels. The electron cannot have an energy value between these levels. This is the cornerstone of atomic physics.

🔵 Ground State

The lowest energy state of an atom where the electron is at n = 1. For hydrogen, the ground state energy is −13.6 eV. The atom is most stable in this state.

🟢 Excited States

When an atom absorbs energy, the electron jumps to a higher orbit (n = 2, 3, 4 …). These states are called excited states. Atoms do not remain excited for long (~10⁻⁸ s).

🔴 Ionisation

When n → ∞, the electron is completely removed from the atom. The energy at this point is 0 eV (reference level). The energy required to reach this state from n = 1 is the ionisation energy.

Fig 1.1 — Hydrogen Atom Energy Level Diagram (NCERT Style) showing Lyman, Balmer, and Paschen series transitions

⚡ n=1: −13.6 eV (Ground State)

🔵 n=2: −3.4 eV

🟡 n=3: −1.51 eV

🟠 n=4: −0.85 eV

🔴 n=5: −0.54 eV

🟣 n=6: −0.38 eV

♾ n=∞: 0 eV (Free Electron)

    Key Formula: Energy of nth level in hydrogen atom: Eₙ = −13.6 / n² eV
    
So E₁ = −13.6 eV, E₂ = −13.6/4 = −3.4 eV, E₃ = −13.6/9 = −1.51 eV, etc.

2 Transition of Electron

⬆ Upward Transition (Absorption)

When an atom absorbs a photon of exactly the right energy, the electron jumps from a lower energy level to a higher one. The photon energy must exactly equal the energy difference: ΔE = E_upper − E_lower.

⬇ Downward Transition (Emission)

When an excited electron returns to a lower energy level, it emits a photon carrying energy equal to the energy difference between the two levels. This is the basis of atomic emission spectra.

🔆 Absorption Process

Atom in lower state + photon (energy = ΔE) → Atom in higher state. An absorption line (dark line) appears in the spectrum at the corresponding wavelength/frequency.

💡 Emission Process

Atom in higher state → Atom in lower state + photon (energy = ΔE). Bright lines appear in the emission spectrum at specific wavelengths corresponding to allowed transitions.

Fig 2.1 — Upward Transition (Absorption)

Fig 2.2 — Downward Transition (Emission)

3 Photon Emission and Absorption

💡 Photon Emission

When an electron transitions from a higher energy state (E₂) to a lower energy state (E₁), it emits a photon. The energy of the emitted photon equals the energy difference between the two states.

        ΔE = E₂ − E₁ = hν
        
where hν is the energy of the emitted photon

Energy is conserved: the energy lost by the electron exactly equals the energy gained by the emitted photon.

🔆 Photon Absorption

When an atom absorbs a photon of exactly the right energy, the electron jumps from a lower state (E₁) to a higher state (E₂). The photon energy must be precisely equal to the energy gap — partial absorption is NOT possible.

        hν = E₂ − E₁
        
Photon energy must exactly match the energy gap

Fig 3.1 — Energy Conservation in Photon Emission and Absorption

Photon Energy

ΔE = hν

ΔE = energy difference (J or eV)
h = Planck's constant = 6.626 × 10⁻³⁴ J·s
ν = frequency of photon (Hz)

Energy – Wavelength

E = hc/λ

c = speed of light = 3 × 10⁸ m/s
λ = wavelength (m)
E = energy of photon (J)

Frequency from Energy

ν = ΔE/h

ν = frequency of emitted/absorbed photon
ΔE = E₂ − E₁
h = 6.626 × 10⁻³⁴ J·s

4 Frequency and Wavelength of Emitted Photon

📐 Derivation of Frequency Formula

Bohr's postulate: Energy of nth level in hydrogen atom: Eₙ = −13.6/n² eV

When electron transitions from orbit nᵢ (higher) to nf (lower), energy of emitted photon:
ΔE = Eₙᵢ − Eₙf = −13.6/nᵢ² − (−13.6/nf²) = 13.6(1/nf² − 1/nᵢ²) eV

Using E = hν: ν = ΔE/h = (13.6 eV × 1.6×10⁻¹⁹ J/eV) × (1/nf² − 1/nᵢ²) / (6.626×10⁻³⁴)

This gives the Rydberg formula: 1/λ = R_H (1/nf² − 1/nᵢ²)
where R_H = 1.097 × 10⁷ m⁻¹ (Rydberg constant)

Frequency Formula

ν = R_H · c · (1/nf² − 1/nᵢ²)

R_H = 1.097 × 10⁷ m⁻¹ | c = 3 × 10⁸ m/s
nf = final (lower) orbit | nᵢ = initial (higher) orbit

Wavelength Formula (Rydberg)

1/λ = R_H (1/nf² − 1/nᵢ²)

λ = wavelength of emitted radiation (m)
R_H = Rydberg constant = 1.097 × 10⁷ m⁻¹
nf < nᵢ (downward transition)

🔢 Solved Examples

Example 4.1

Calculate the wavelength and frequency of the photon emitted when an electron in hydrogen atom transitions from n = 3 to n = 2.

Solution

Using Rydberg formula: 1/λ = R_H (1/nf² − 1/nᵢ²)
nf = 2, nᵢ = 3, R_H = 1.097 × 10⁷ m⁻¹
1/λ = 1.097 × 10⁷ × (1/4 − 1/9) = 1.097 × 10⁷ × (9−4)/36 = 1.097 × 10⁷ × 5/36
1/λ = 1.524 × 10⁶ m⁻¹
λ = 6563 Å = 656.3 nm (Red line — Hα line of Balmer series)
ν = c/λ = 3×10⁸ / 656.3×10⁻⁹ = 4.57 × 10¹⁴ Hz

λ = 656.3 nm | ν = 4.57 × 10¹⁴ Hz

Exam Tip: The n=3 → n=2 transition (Hα line, 656.3 nm) is the longest wavelength in the Balmer series. Memorize it — it appears in almost every exam.

Example 4.2

Find the wavelength of photon emitted in the transition n = 4 to n = 1 (Lyman series).

Solution

nf = 1, nᵢ = 4
1/λ = 1.097×10⁷ × (1/1² − 1/4²) = 1.097×10⁷ × (1 − 1/16) = 1.097×10⁷ × 15/16
1/λ = 1.028×10⁷ m⁻¹
λ = 97.2 nm (UV region — Lyman series)

λ = 97.2 nm (UV)

5 Excitation Energy

📖 Definition

Excitation energy is the minimum energy required to excite an atom from its ground state (n=1) to a particular excited state (n = 2, 3, 4 …).

Excitation Energy Formula

E_exc = Eₙ − E₁

Eₙ = energy of excited state | E₁ = ground state energy (−13.6 eV)

📊 Excitation Energies for Hydrogen

Transition	Eₙ (eV)	Excitation Energy (eV)
n=1 → n=2	−3.4	−3.4 − (−13.6) = 10.2 eV
n=1 → n=3	−1.51	−1.51 − (−13.6) = 12.09 eV
n=1 → n=4	−0.85	−0.85 − (−13.6) = 12.75 eV
n=1 → n=5	−0.54	−0.54 − (−13.6) = 13.06 eV
n=1 → n=∞	0	0 − (−13.6) = 13.6 eV

🔢 Numericals on Excitation Energy

Numerical 5.1

What is the excitation energy required to excite hydrogen from ground state to the second excited state?

Solution

Second excited state = n = 3 (Note: n=2 is first excited state, n=3 is second)
E₃ = −13.6/9 = −1.511 eV
E₁ = −13.6 eV
Excitation Energy = E₃ − E₁ = −1.511 − (−13.6) = 12.09 eV

12.09 eV

Exam Tip: "nth excited state" means n = (n+1)th orbit. Ground state = 1st orbit. First excited state = 2nd orbit. Second excited state = 3rd orbit.

Numerical 5.2

An electron in a hydrogen atom is in the n = 4 state. How much energy is released if the electron falls to n = 2?

Solution

E₄ = −13.6/16 = −0.85 eV
E₂ = −13.6/4 = −3.4 eV
Energy released = E₄ − E₂ = −0.85 − (−3.4) = 2.55 eV

2.55 eV

6 Ionisation Energy

⚛ Definition

Ionisation energy is the minimum energy required to completely remove an electron from the atom (from ground state to n = ∞). For hydrogen:

Ionisation Energy = 0 − (−13.6) = 13.6 eV

Ionisation potential = the potential difference through which an electron must be accelerated to gain energy equal to the ionisation energy.
For hydrogen: V = 13.6 V

📊 Ionisation from Different States

Initial State	Energy (eV)	Ionisation Energy
n = 1 (ground)	−13.6	13.6 eV
n = 2	−3.4	3.4 eV
n = 3	−1.51	1.51 eV
n = 4	−0.85	0.85 eV
n = 5	−0.54	0.54 eV

Formula

E_ion = 0 − Eₙ = |Eₙ| = 13.6/n² eV
For ground state (n=1): E_ion = 13.6 eV

Application

Ionisation energy determines the chemical reactivity of atoms. It is the energy that binds the electron to the nucleus.

Key Value

Hydrogen: 13.6 eV = 2.18 × 10⁻¹⁸ J
Ionisation potential: 13.6 V

Numerical 6.1

The energy of an electron in hydrogen atom is −3.4 eV. What is its kinetic energy and ionisation energy?

Solution

Total energy = −3.4 eV (this is n=2 state)
Kinetic energy = −(Total energy) = +3.4 eV (KE is always positive and equals |E|)
Potential energy = 2 × Total energy = −6.8 eV
Ionisation energy = 0 − (−3.4) = 3.4 eV

KE = 3.4 eV | IE = 3.4 eV

Key Relation: In Bohr's model: KE = |E|, PE = 2E, Total E = −KE

Numerical 6.2

An electron is in the third excited state of hydrogen. What energy must be supplied to ionise this atom?

Solution

Third excited state = n = 4
E₄ = −13.6/16 = −0.85 eV
Energy to ionise = 0 − (−0.85) = 0.85 eV

0.85 eV

7 Spectral Series Comparison

All spectral series of hydrogen arise due to electron transitions to different final orbits. Each series appears in a different region of the electromagnetic spectrum.

Series	Final Orbit (nf)	Initial Orbits (nᵢ)	Region	Wavelength Range	Visibility	Exam Importance
Lyman	n = 1	2, 3, 4, 5 …	Ultraviolet (UV)	91 nm – 122 nm	Invisible (UV)	⭐⭐⭐⭐⭐ (Very High)
Balmer	n = 2	3, 4, 5, 6 …	Visible + Near UV	364 nm – 656 nm	Visible (ROYGBIV)	⭐⭐⭐⭐⭐ (Very High)
Paschen	n = 3	4, 5, 6, 7 …	Infrared (IR)	820 nm – 1875 nm	Invisible (IR)	⭐⭐⭐⭐ (High)
Brackett	n = 4	5, 6, 7, 8 …	Far Infrared	1.46 μm – 4.05 μm	Invisible (Far IR)	⭐⭐⭐ (Medium)
Pfund	n = 5	6, 7, 8, 9 …	Far Infrared	2.28 μm – 7.46 μm	Invisible (Far IR)	⭐⭐ (Low–Medium)
Humphreys	n = 6	7, 8, 9 …	Far Infrared	> 3.28 μm	Invisible	⭐ (Low)

🟣 Lyman Series (UV)

Transitions to n=1. Shortest wavelength series. First line: n=2→1, λ=122 nm. Series limit: 91 nm. Region: UV. Named after Theodore Lyman.

🔵 Balmer Series (Visible)

Transitions to n=2. Only series visible to naked eye. Hα (656 nm, red), Hβ (486 nm, blue-green), Hγ (434 nm, violet). Named after Johann Balmer.

🟠 Paschen Series (IR)

Transitions to n=3. First line: n=4→3, λ=1875 nm. Lies in near-infrared. Named after Friedrich Paschen.

🟢 Brackett Series

Transitions to n=4. Far infrared region. First line: n=5→4. Named after Frederick Brackett.

🔴 Pfund Series

Transitions to n=5. Far infrared. Named after August Herman Pfund. Less commonly tested in CBSE/JEE.

🔢 Rydberg Formula

Applies to ALL series: 1/λ = R_H(1/nf² − 1/nᵢ²)
R_H = 1.097 × 10⁷ m⁻¹

8 Important Graphs

Fig 8.1 — Energy vs n: curve flattens as n→∞, showing convergence to ionisation level

Fig 8.2 — Spectral Series wavelength comparison across the EM spectrum

Fig 8.3 — Energy Level Convergence: levels get closer as n increases, converging at 0 eV

Fig 8.4 — Frequency vs Transition Energy (linear: ν = ΔE/h)

9 Important Formulae

Energy of nth Level

Eₙ = −13.6/n² eV

n = principal quantum number (1, 2, 3 …)
−ve sign: electron is bound to nucleus
E₁ = −13.6 eV (Ground State)

Photon Energy (Emission)

ΔE = E_i − E_f = hν

E_i = higher energy level
E_f = lower energy level
h = 6.626 × 10⁻³⁴ J·s | ν = frequency

Frequency Formula

ν = (E_i − E_f) / h

ν = frequency of photon (Hz)
E in Joules; divide eV by 1.6×10⁻¹⁹ to convert
Or: ν = R_H · c · (1/nf² − 1/ni²)

Rydberg Formula

1/λ = R_H (1/nf² − 1/ni²)

R_H = 1.097 × 10⁷ m⁻¹ (Rydberg constant)
nf = final (lower) orbit
ni = initial (higher) orbit | ni > nf

Excitation Energy

E_exc = Eₙ − E₁

E₁ = −13.6 eV (ground state)
Eₙ = energy of excited state
Always positive value

Ionisation Energy

E_ion = 0 − Eₙ = |Eₙ|

From ground state: 13.6 eV
From nth state: 13.6/n² eV
Ionisation potential (V) = E_ion/e

Photon – Wavelength

E = hc/λ

c = 3 × 10⁸ m/s
h = 6.626 × 10⁻³⁴ J·s
λ in metres, E in Joules

Energy in eV ↔ J

1 eV = 1.6 × 10⁻¹⁹ J

To convert eV → J: multiply by 1.6×10⁻¹⁹
To convert J → eV: divide by 1.6×10⁻¹⁹
Use this for all formula calculations

Number of Spectral Lines

N = n(n−1)/2

n = number of energy levels involved
Maximum lines possible for n levels
e.g. n=4 levels → 6 possible lines

    Constants to memorise: 
    h = 6.626 × 10⁻³⁴ J·s  | 
    c = 3 × 10⁸ m/s  | 
    R_H = 1.097 × 10⁷ m⁻¹  | 
    1 eV = 1.6 × 10⁻¹⁹ J  | 
    e = 1.6 × 10⁻¹⁹ C
  

10 NCERT Examples

NCERT Example 12.1

The radius of the innermost electron orbit of a hydrogen atom is 5.3 × 10⁻¹¹ m. What are the radii of the n = 2 and n = 3 orbits?

Given

r₁ = 5.3 × 10⁻¹¹ m (Bohr radius a₀)

Formula

rₙ = n² × r₁ = n² × a₀

Solution

r₂ = (2)² × 5.3 × 10⁻¹¹ = 4 × 5.3 × 10⁻¹¹ = 2.12 × 10⁻¹⁰ m
r₃ = (3)² × 5.3 × 10⁻¹¹ = 9 × 5.3 × 10⁻¹¹ = 4.77 × 10⁻¹⁰ m

r₂ = 2.12 × 10⁻¹⁰ m | r₃ = 4.77 × 10⁻¹⁰ m

Exam Tip: rₙ ∝ n² — radius increases as square of quantum number. Always start from r₁ = 0.529 Å = 5.29 × 10⁻¹¹ m.

NCERT Example 12.2

The ground state energy of hydrogen atom is −13.6 eV. What are the kinetic and potential energies of the electron in this state?

Given

Total energy E = −13.6 eV

Formula

In Bohr model: KE = −E = |E| and PE = 2E

Solution

KE = −(−13.6) = +13.6 eV
PE = 2 × (−13.6) = −27.2 eV
Verification: Total E = KE + PE = 13.6 + (−27.2) = −13.6 eV ✓

KE = 13.6 eV | PE = −27.2 eV

Key relation: Total Energy = −KE = PE/2. Memorize: E = −KE = PE/2

NCERT Example 12.3

Using the Rydberg formula, calculate the wavelengths of the first four spectral lines in the Lyman series of hydrogen spectrum.

Given

Lyman series: nf = 1, nᵢ = 2, 3, 4, 5; R_H = 1.097 × 10⁷ m⁻¹

Solution

Line 1 (nᵢ=2→1): 1/λ = 1.097×10⁷ × (1 − 1/4) = 1.097×10⁷ × 3/4 → λ = 121.5 nm (Lyman α)
Line 2 (nᵢ=3→1): 1/λ = 1.097×10⁷ × (1 − 1/9) = 1.097×10⁷ × 8/9 → λ = 102.5 nm (Lyman β)
Line 3 (nᵢ=4→1): 1/λ = 1.097×10⁷ × (1 − 1/16) = 1.097×10⁷ × 15/16 → λ = 97.2 nm (Lyman γ)
Line 4 (nᵢ=5→1): 1/λ = 1.097×10⁷ × (1 − 1/25) = 1.097×10⁷ × 24/25 → λ = 94.9 nm (Lyman δ)

121.5 nm | 102.5 nm | 97.2 nm | 94.9 nm

Exam Tip: Lyman α (121.5 nm) is the longest wavelength in the Lyman series. All Lyman lines are in UV — invisible to human eye.

11 Numericals

🟢 Easy Numericals

Easy

Numerical 11.1

Calculate the energy of a photon emitted when an electron in hydrogen transitions from n = 2 to n = 1.

Solution

E₁ = −13.6 eV, E₂ = −3.4 eV
ΔE = E₂ − E₁ = −3.4 − (−13.6) = 10.2 eV

10.2 eV

Easy

Numerical 11.2

Find the frequency of photon emitted in transition n = 3 to n = 1 in hydrogen.

Solution

E₃ = −1.51 eV, E₁ = −13.6 eV
ΔE = −1.51 − (−13.6) = 12.09 eV = 12.09 × 1.6×10⁻¹⁹ = 1.934×10⁻¹⁸ J
ν = ΔE/h = 1.934×10⁻¹⁸ / 6.626×10⁻³⁴ = 2.92 × 10¹⁵ Hz

2.92 × 10¹⁵ Hz

Easy

Numerical 11.3

What is the wavelength of the Hα line (n=3 → n=2) in the Balmer series?

Solution

1/λ = R_H(1/4 − 1/9) = 1.097×10⁷ × 5/36 = 1.524×10⁶ m⁻¹
λ = 1/1.524×10⁶ = 6563 Å = 656.3 nm

656.3 nm (Red)

🟡 Medium Numericals

Medium

Numerical 11.4

A photon of wavelength 97.2 nm is absorbed by a hydrogen atom. Find the initial and final states of the electron and identify the spectral series.

Solution

1/λ = 1/97.2×10⁻⁹ = 1.028×10⁷ m⁻¹
R_H(1/nf² − 1/ni²) = 1.028×10⁷
Try nf=1: 1.097×10⁷(1 − 1/ni²) = 1.028×10⁷ → 1 − 1/ni² = 0.937 → 1/ni² = 0.063 → ni² ≈ 15.9 → ni = 4
Transition: n=1 → n=4 (absorption), Lyman series

n=1 → n=4 | Lyman Series (UV)

Exam Tip: For absorption, the final state is higher than the initial. For emission, it's the reverse.

Medium

Numerical 11.5

How many spectral lines will be emitted when hydrogen atoms, all in the n = 4 state, make transitions to the ground state?

Solution

Possible transitions from n=4:
4→3, 4→2, 4→1, 3→2, 3→1, 2→1
Total = 6 lines
Using formula: N = n(n−1)/2 = 4×3/2 = 6 lines

6 spectral lines

Medium

Numerical 11.6

Calculate the shortest wavelength in the Balmer series of hydrogen spectrum. (R_H = 1.097 × 10⁷ m⁻¹)

Solution

Shortest wavelength → maximum energy → nᵢ = ∞
1/λ_min = R_H(1/4 − 0) = R_H/4 = 1.097×10⁷/4 = 2.743×10⁶ m⁻¹
λ_min = 1/2.743×10⁶ = 3646 Å = 364.6 nm (Series limit)

364.6 nm (Balmer Series Limit)

Medium

Numerical 11.7

An electron in hydrogen is excited from n=1 to n=3. How many different photons can be emitted when it returns to ground state?

Solution

From n=3, possible transitions: 3→2, 3→1, 2→1
N = 3(3−1)/2 = 3 photons possible.
Energies: 3→2: 1.89 eV; 3→1: 12.09 eV; 2→1: 10.2 eV

3 different photons

🔴 Advanced Numericals

Advanced

Numerical 11.8

A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted?

Solution

Ground state energy = −13.6 eV. After absorbing 12.5 eV: Total energy = −13.6 + 12.5 = −1.1 eV
Find n: −13.6/n² = −1.1 → n² = 13.6/1.1 = 12.36 → n ≈ 3.5, so n = 3 is highest accessible (E₃ = −1.51 eV, since −13.6 + 12.5 = −1.1 > −1.51)
Wait: −1.1 > −1.51, so n=3 is accessible (−1.51 < −1.1). n=4 would be −0.85, which is above −1.1. So electron can reach n=3.
Possible emissions: 3→2, 3→1, 2→1
Lyman series: n=2→1 (λ=121.5 nm) and n=3→1 (λ=102.5 nm)
Balmer series: n=3→2 (λ=656.3 nm)

Lyman (121.5 nm, 102.5 nm) + Balmer (656.3 nm)

Exam Trick: Electron beam energy must be ≥ excitation energy to reach that level. Find the highest n reachable, then list all downward transitions.

Advanced

Numerical 11.9

Find the ratio of the longest to shortest wavelength in the Lyman series.

Solution

Longest λ (Lyman α, nᵢ=2→1): 1/λ_max = R_H(1 − 1/4) = 3R_H/4
Shortest λ (series limit, nᵢ=∞→1): 1/λ_min = R_H
Ratio λ_max/λ_min = (1/R_H) / (4/3R_H) = 4/3
λ_max/λ_min = 4/3

λ_max/λ_min = 4/3

Advanced

Numerical 11.10

For what kinetic energy of a neutron will the de Broglie wavelength be equal to 1.40 × 10⁻¹⁰ m? (mass of neutron = 1.675 × 10⁻²⁷ kg)

Solution

λ = h/mv → mv = h/λ = 6.626×10⁻³⁴ / 1.40×10⁻¹⁰ = 4.73×10⁻²⁴ kg·m/s
KE = (mv)²/2m = (4.73×10⁻²⁴)² / (2 × 1.675×10⁻²⁷)
KE = 22.37×10⁻⁴⁸ / 3.35×10⁻²⁷ = 6.67×10⁻²¹ J = 6.67×10⁻²¹ J = 0.042 eV

KE = 6.67 × 10⁻²¹ J ≈ 0.042 eV

12 Previous Year Questions (PYQs)

🔵 CBSE PYQs (25 Questions)

CBSE 2023

CBSE Q1

Using Bohr's postulates, derive the expression for the energy of an electron in the nth orbit of hydrogen atom.

Solution

From Bohr's postulate: L = mvr = nh/2π ... Coulomb force = centripetal force: ke²/r² = mv²/r → ke²/r = mv². Combining: r = n²h²/4π²mke² = n²a₀. KE = mv²/2 = ke²/2r. PE = −ke²/r. Total E = KE + PE = ke²/2r − ke²/r = −ke²/2r. Substituting r: Eₙ = −me⁴k²/(2ℏ²n²) = −13.6/n² eV

Eₙ = −13.6/n² eV

CBSE 2023

CBSE Q2

The energy of the ground state of hydrogen atom is −13.6 eV. Find the energy required to excite it to the first excited state.

Solution

First excited state = n=2; E₂ = −13.6/4 = −3.4 eV. ΔE = E₂ − E₁ = −3.4 − (−13.6) = 10.2 eV

10.2 eV

Exam Tip: First excited state = n=2, not n=1.

CBSE 2022

CBSE Q3

Draw the energy level diagram of hydrogen atom showing at least 5 energy levels. Mark the transitions responsible for Lyman and Balmer series.

Solution

Draw horizontal lines for n=1 (−13.6 eV), n=2 (−3.4 eV), n=3 (−1.51 eV), n=4 (−0.85 eV), n=5 (−0.54 eV). Lyman: arrows from n≥2 to n=1 (UV). Balmer: arrows from n≥3 to n=2 (visible). See Section 1 diagram above.

See energy level diagram (Section 1)

CBSE 2022

CBSE Q4

Calculate the wavelength of the first line of the Balmer series (n=3→2) given R = 1.097 × 10⁷ m⁻¹.

Solution

1/λ = R(1/4 − 1/9) = 1.097×10⁷ × 5/36 = 1.524×10⁶ m⁻¹ → λ = 656.3 nm

656.3 nm (Hα line, Red)

CBSE 2021

CBSE Q5

Name the series of hydrogen spectrum which lies in the (i) UV region (ii) Visible region (iii) Infrared region.

Solution

(i) Lyman series — UV region (nf=1)
(ii) Balmer series — Visible region (nf=2)
(iii) Paschen, Brackett, Pfund series — Infrared region (nf=3,4,5)

Lyman (UV) | Balmer (Visible) | Paschen/Brackett/Pfund (IR)

CBSE 2021

CBSE Q6

What is the maximum number of spectral lines emitted by a hydrogen atom when it is in the third excited state?

Solution

Third excited state = n=4. N = n(n−1)/2 = 4×3/2 = 6 lines

6 lines

Trick: Third excited state = n=4 (not n=3). nth excited state = (n+1)th orbit.

CBSE 2020

CBSE Q7

The ionisation energy of hydrogen atom is 13.6 eV. What is the energy of the electron in the third orbit of hydrogen atom?

Solution

E₃ = −13.6/n² = −13.6/9 = −1.51 eV

−1.51 eV

CBSE 2020

CBSE Q8

An electron in hydrogen atom makes a transition from n=4 to n=2. Calculate the wavelength of the emitted photon.

Solution

1/λ = R(1/4 − 1/16) = R × 3/16 = 1.097×10⁷ × 3/16 = 2.057×10⁶ m⁻¹ → λ = 486 nm (Hβ line, Blue-green)

486 nm (Balmer Hβ)

CBSE 2019

CBSE Q9

In a hydrogen atom, the electron jumps from the third orbit to the second orbit, emitting a photon of wavelength λ. If the electron jumps from the fourth orbit to the second orbit, the wavelength of the photon emitted will be?

Solution

For n=3→2: 1/λ = R(1/4−1/9) = 5R/36 → λ = 36/5R
For n=4→2: 1/λ' = R(1/4−1/16) = 3R/16 → λ' = 16/3R
Ratio: λ'/λ = (16/3R)/(36/5R) = 80/108 = 20/27
λ' = 20λ/27

λ' = 20λ/27

CBSE 2019

CBSE Q10

Show that the frequency of radiation emitted when hydrogen atom de-excites from level n to level (n−1) is approximately ν ≈ 2me⁴k²/n³h³ for large n.

Solution

ΔE = E_n − E_{n-1} = 13.6[1/(n-1)² − 1/n²] eV. For large n: 1/(n-1)² − 1/n² ≈ 2/n³. Therefore ΔE ≈ 27.2/n³ eV, and ν = ΔE/h ≈ 27.2 × 1.6×10⁻¹⁹/(n³ × h). This matches classical orbital frequency, confirming Bohr's Correspondence Principle.

ν ∝ 1/n³ for large n (Correspondence Principle)

CBSE 2018

CBSE Q11

Using Rydberg formula, find the wavelength of spectral line corresponding to the transition n=5 to n=3 in hydrogen.

Solution

1/λ = R(1/9 − 1/25) = R × 16/225 = 1.097×10⁷ × 16/225 = 7.798×10⁵ m⁻¹ → λ = 1282 nm (Paschen series, IR)

1282 nm (Paschen series)

CBSE 2018

CBSE Q12

The ground state energy of hydrogen atom is −13.6 eV. If an electron makes a transition from an energy level −0.85 eV to −3.4 eV, calculate the wavelength of the spectral line emitted.

Solution

ΔE = −0.85 − (−3.4) = 2.55 eV = 2.55 × 1.6×10⁻¹⁹ = 4.08×10⁻¹⁹ J
λ = hc/ΔE = (6.626×10⁻³⁴ × 3×10⁸) / 4.08×10⁻¹⁹ = 487 nm (Hβ line)

λ ≈ 487 nm

CBSE 2017

CBSE Q13

Explain with the help of a diagram, the phenomenon of atomic emission spectrum.

Solution

When atoms are excited (by heat, electricity, or radiation), electrons jump to higher energy levels. When they fall back to lower levels, they emit photons of specific energies. A prism/diffraction grating separates these into distinct bright lines (emission spectrum). Each element has a unique set of spectral lines — its "spectral fingerprint".

Bright line spectrum due to downward electron transitions

CBSE 2017

CBSE Q14

What do you understand by the Balmer series limit? Calculate its wavelength.

Solution

Balmer series limit = shortest wavelength in Balmer series, when nᵢ = ∞.
1/λ = R × 1/4 = 1.097×10⁷/4 = 2.743×10⁶ m⁻¹ → λ = 364.6 nm

364.6 nm (Balmer Series Limit)

CBSE 2016

CBSE Q15

Calculate the longest wavelength of the Lyman series and state in which region it lies.

Solution

Longest λ when nᵢ=2→nf=1: 1/λ = R(1 − 1/4) = 3R/4 → λ = 4/3R = 4/(3×1.097×10⁷) = 121.6 nm — UV region

121.6 nm (UV)

CBSE 2016

CBSE Q16

Using Bohr's model, explain why the spectral lines are not equally spaced.

Solution

Energy levels are given by Eₙ = −13.6/n² eV. The gaps between successive levels decrease as n increases (∝ 1/n²). Since photon energy ∝ energy gap, and wavelength ∝ 1/energy, lines at lower n (larger energy gaps) have shorter wavelengths. Thus lines are bunched at shorter wavelengths (series limit) and spread out at longer wavelengths.

Energy gaps ∝ 1/n² — decrease with increasing n → unequal spacing

CBSE 2015

CBSE Q17

The short wavelength limit for the Lyman series of hydrogen spectrum is 913.4 Å. Calculate the short wavelength limit for Balmer and Paschen series.

Solution

Lyman limit (nf=1): λ_L = 1/R → R = 1/913.4×10⁻¹⁰
Balmer limit (nf=2): λ_B = 4/R = 4 × 913.4 = 3653.6 Å
Paschen limit (nf=3): λ_P = 9/R = 9 × 913.4 = 8220.6 Å

Balmer: 3653.6 Å | Paschen: 8220.6 Å

CBSE 2015

CBSE Q18

What is the ionisation energy of Li²⁺? (Z=3, ground state energy of H = −13.6 eV)

Solution

For hydrogen-like ions: E₁ = −13.6 Z²/n² eV. For Li²⁺ (Z=3, n=1): E₁ = −13.6 × 9 = −122.4 eV. Ionisation energy = |E₁| = 122.4 eV

122.4 eV

CBSE 2014

CBSE Q19

A hydrogen atom initially in the ground state absorbs a photon which excites it to n=4 level. Find the wavelength and frequency of the photon.

Solution

ΔE = E₄ − E₁ = −0.85 − (−13.6) = 12.75 eV = 12.75 × 1.6×10⁻¹⁹ = 2.04×10⁻¹⁸ J
λ = hc/ΔE = 6.626×10⁻³⁴ × 3×10⁸ / 2.04×10⁻¹⁸ = 97.4 nm
ν = ΔE/h = 2.04×10⁻¹⁸/6.626×10⁻³⁴ = 3.08×10¹⁵ Hz

λ = 97.4 nm | ν = 3.08 × 10¹⁵ Hz

CBSE 2023

CBSE Q20

What is the ratio of the kinetic energy to the potential energy of the electron in the ground state of hydrogen atom?

Solution

KE = +13.6 eV, PE = −27.2 eV. Ratio KE/PE = 13.6/(−27.2) = −1/2

KE/PE = −1/2

CBSE 2022

CBSE Q21

Derive the Rydberg formula using Bohr's model for hydrogen atom.

Solution

Energy of photon: ΔE = Eₙᵢ − Eₙf = 13.6(1/nf² − 1/nᵢ²) eV. Since E=hν=hc/λ: hc/λ = 13.6 × 1.6×10⁻¹⁹(1/nf² − 1/nᵢ²). Dividing both sides by hc: 1/λ = R_H(1/nf² − 1/nᵢ²) where R_H = 13.6 × 1.6×10⁻¹⁹/(hc) = 1.097×10⁷ m⁻¹.

1/λ = R_H(1/nf² − 1/nᵢ²); R_H = 1.097 × 10⁷ m⁻¹

CBSE 2021

CBSE Q22

Calculate the energy of the photon required to excite the hydrogen atom from ground state to second excited state.

Solution

Second excited state = n=3. ΔE = E₃ − E₁ = −1.51 − (−13.6) = 12.09 eV

12.09 eV

CBSE 2020

CBSE Q23

Show that the ratio of the longest to shortest wavelength in the Balmer series is 9:5.

Solution

Longest (nᵢ=3→nf=2): 1/λ_max = R(1/4−1/9)=5R/36 → λ_max = 36/5R
Shortest (nᵢ=∞→nf=2): 1/λ_min = R/4 → λ_min = 4/R
λ_max/λ_min = (36/5R)/(4/R) = 36/20 = 9/5 ✓

Ratio = 9:5

CBSE 2019

CBSE Q24

The wavelength of the first line of the Lyman series is 1216 Å. Calculate the wavelength of the second line.

Solution

First line (2→1): 1/λ₁ = R × 3/4. Second line (3→1): 1/λ₂ = R × 8/9.
λ₁/λ₂ = (8/9)/(3/4) = 32/27 → λ₂ = 1216 × 27/32 = 1026 Å

λ₂ = 1026 Å

CBSE 2018

CBSE Q25

A photon of wavelength 4×10⁻⁷ m is emitted by a hydrogen atom. Is it in the Lyman, Balmer, or Paschen series?

Solution

λ = 4×10⁻⁷ m = 400 nm. Balmer series range: 364 nm – 656 nm. Since 400 nm lies in this range: Balmer series (Visible/UV). nf=2; find nᵢ: 1/400×10⁻⁹ = 1.097×10⁷(1/4 − 1/nᵢ²) → nᵢ ≈ 3.7 → nᵢ = 4 (transition 4→2, λ≈486 nm, so this is close to the Hβ line)

Balmer Series

🟢 NEET PYQs (40 Questions)

NEET 2023

NEET Q1

Which transition in hydrogen atom will give spectral line of wavelength 4340 Å?

Solution

λ = 4340 Å = 434 nm → Balmer series (visible). 1/λ = R(1/4 − 1/nᵢ²) → 1/434×10⁻⁹ = 1.097×10⁷(1/4−1/nᵢ²) → 1/nᵢ² = 1/4 − 2.303×10⁶/1.097×10⁷ = 0.25 − 0.210 = 0.04 → nᵢ² = 25 → nᵢ = 5. Transition: 5→2 (Hγ line)

n = 5 → n = 2

NEET 2023

NEET Q2

The energy of the electron in the second and third Bohr orbits of hydrogen atom is −3.4 eV and −1.51 eV respectively. The energy of photon emitted in the transition is:

Solution

ΔE = −1.51 − (−3.4) = 1.89 eV

1.89 eV

NEET 2022

NEET Q3

The number of possible spectral lines which may be emitted in the Brackett series in hydrogen atom if the electrons present only in 7th shell is:

Solution

Brackett series: nf=4. From n=7, possible transitions to n=4: 7→4, 6→4, 5→4 = 3 lines

3 lines

NEET 2022

NEET Q4

The ground state energy of hydrogen atom is −13.6 eV. The energy of the second excited state of He⁺ ion in eV is:

Solution

He⁺ is hydrogen-like with Z=2. Second excited state = n=3. E₃ = −13.6 × Z²/n² = −13.6 × 4/9 = −6.04 eV

−6.04 eV

NEET 2021

NEET Q5

The ratio of the longest wavelength to the shortest wavelength of the Paschen series:

Solution

Longest (4→3): 1/λ_max = R(1/9−1/16) = 7R/144 → λ_max = 144/7R
Shortest (∞→3): 1/λ_min = R/9 → λ_min = 9/R
λ_max/λ_min = (144/7R)/(9/R) = 144/63 = 16/7

16/7

NEET 2021

NEET Q6

The ionisation energy of hydrogen atom is 13.6 eV. Hydrogen atom in its ground state is excited by monochromatic radiation of λ = 975 Å. The number of spectral lines emitted are:

Solution

E = hc/λ = 12.75 eV → n=4 is excited. Lines = n(n−1)/2 = 4×3/2 = 6 lines

6 lines

NEET 2020

NEET Q7

The first member of the Paschen series of hydrogen spectrum has wavelength 18750 Å. Calculate the second member's wavelength.

Solution

First (4→3): 1/λ₁ = R(1/9−1/16) = 7R/144
Second (5→3): 1/λ₂ = R(1/9−1/25) = 16R/225
λ₁/λ₂ = 16×144/(225×7) = 2304/1575 = 1.463
λ₂ = 18750/1.463 = 12819 Å

12819 Å ≈ 1282 nm

NEET 2020

NEET Q8

In a hydrogen atom, electron is in the 5th excited state. The maximum number of spectral lines emitted when it comes to ground state is:

Solution

5th excited state = n=6. N = 6(6−1)/2 = 6×5/2 = 15 lines

15 spectral lines

NEET 2019

NEET Q9

The wavelength of Kα X-ray of an element is 0.36 nm. What is its atomic number? (R=1.097×10⁷ m⁻¹)

Solution

Moseley's law for Kα: 1/λ = R(Z−1)²(1/1²−1/2²) = 3R(Z−1)²/4. So (Z−1)² = 4/(3Rλ) = 4/(3×1.097×10⁷×0.36×10⁻⁹) = 337.6 → Z−1=18.4 → Z≈19 (Potassium)

Z = 19

NEET 2019

NEET Q10

If the series limit wavelength of Balmer series is 3646 Å, then series limit wavelength of Paschen series is:

Solution

Series limit for nf=n: λ ∝ nf². Balmer (nf=2): λ_B = 3646 Å. Paschen (nf=3): λ_P = λ_B × (3/2)² = 3646 × 9/4 = 8204 Å

8204 Å

NEET 2018

NEET Q11

The energy required to excite a hydrogen atom from n=1 to n=2 state is 10.2 eV. What is the wavelength of the radiation emitted when it returns to ground state?

Solution

λ = hc/ΔE = (6.626×10⁻³⁴ × 3×10⁸)/(10.2 × 1.6×10⁻¹⁹) = 1.988×10⁻²⁵/1.632×10⁻¹⁸ = 121.8 nm

121.8 nm (Lyman α)

NEET 2018

NEET Q12

An electron in H atom makes a transition from n=3 to n=1. The recoil momentum of the hydrogen atom is:

Solution

ΔE = 12.09 eV = 1.934×10⁻¹⁸ J. Photon momentum = E/c = 1.934×10⁻¹⁸/(3×10⁸) = 6.45×10⁻²⁷ kg·m/s (by momentum conservation, recoil momentum equals photon momentum)

6.45 × 10⁻²⁷ kg·m/s

NEET 2017

NEET Q13

The total energy of the electron in the hydrogen atom in the ground state is −13.6 eV. Which of the following is its kinetic energy?

Solution

In Bohr's model: KE = −E = −(−13.6) = +13.6 eV

+13.6 eV

NEET 2017

NEET Q14

The ratio of longest wavelength and the shortest wavelength observed in the five spectral series of emission spectrum of hydrogen is:

Solution

Longest: Pfund series limit → longest line (nᵢ=6→nf=5). Approximate λ_max ≈ 7.4 μm. Shortest: Lyman series limit (∞→1) = 91.2 nm. Ratio ≈ 7400/91.2 ≈ ∼900 (or 1:900 approx)

Approximately 900:1 (Pfund max to Lyman min)

NEET 2016

NEET Q15

Energy of electron in the first Bohr orbit of H-atom is −13.6 eV. What is the energy in the second Bohr orbit of He⁺?

Solution

E = −13.6 Z²/n² = −13.6 × 4/4 = −13.6 eV

−13.6 eV

NEET 2016

NEET Q16

The wavelength of the first spectral line in the Balmer series of hydrogen atom is 6561 Å. The wavelength of the second spectral line in the Balmer series of singly ionised helium atom is:

Solution

For He⁺ (Z=2): 1/λ = R×Z²(1/nf²−1/nᵢ²). Second Balmer line of He⁺: nf=2, nᵢ=4. 1/λ = 4R(1/4−1/16) = 4R×3/16 = 3R/4. But for H Balmer 1st (nf=2, nᵢ=3): 1/λ_H = R×5/36. λ_He/λ_H = (5/36)/(3/4) = 5/27. λ_He = 6561 × 5/27 = 1215 Å

1215 Å

NEET 2015

NEET Q17

The electron in a hydrogen atom first jumps from the third excited state to the second excited state and subsequently to the first excited state. The ratio of the respective wavelengths λ₁/λ₂ of the photons emitted in this process is:

Solution

State 4→3: 1/λ₁ = R(1/9−1/16) = 7R/144. State 3→2: 1/λ₂ = R(1/4−1/9) = 5R/36 = 20R/144. λ₁/λ₂ = 20/7 = 20/7

λ₁/λ₂ = 20/7

NEET 2015

NEET Q18

Out of the following transitions, which one will emit the highest energy photon in hydrogen atom?

Solution

Options typically given: (a) n=2→1, (b) n=3→2, (c) n=4→3, (d) n=5→4. Energy ∝ 1/nf² − 1/nᵢ². Largest for n=2→1: ΔE = 10.2 eV. n=2→1 emits highest energy photon.

n = 2 → n = 1 (10.2 eV)

NEET 2014

NEET Q19

If series limit of Lyman series for H is 912 Å, find series limit of Balmer series for He⁺.

Solution

For H Lyman limit: 1/λ_L = R. For He⁺ Balmer limit (nf=2): 1/λ = RZ²/nf² = R×4/4 = R. So λ(He⁺ Balmer limit) = 912 Å

912 Å

NEET 2023

NEET Q20

The de Broglie wavelength of an electron in 3rd Bohr orbit of hydrogen atom is:

Solution

For nth orbit: 2πrₙ = nλ_dB → λ_dB = 2πr₃/3. r₃ = 9a₀ = 9 × 5.29×10⁻¹¹. λ_dB = 2π × 9 × 5.29×10⁻¹¹/3 = 6π × 5.29×10⁻¹¹ = 9.96 × 10⁻¹⁰ m ≈ 9.96 Å

9.96 Å

NEET 2022

NEET Q21

For Balmer series, the wavelength of the first line is λ₁ and that of second line is λ₂. Then the ratio λ₁/λ₂ is:

Solution

First line (3→2): 1/λ₁ = R(1/4−1/9) = 5R/36. Second line (4→2): 1/λ₂ = R(1/4−1/16) = 3R/16. λ₁/λ₂ = (3R/16)/(5R/36) = (3×36)/(16×5) = 108/80 = 27/20

27/20

NEET 2021

NEET Q22

In hydrogen atom, the energy of the electron in first and second Bohr orbits is −13.6 eV and −3.4 eV respectively. Which of the following gives the wavelength of the emitted photon?

Solution

ΔE = −3.4 − (−13.6) = 10.2 eV. λ = hc/ΔE = (6.626×10⁻³⁴ × 3×10⁸)/(10.2 × 1.6×10⁻¹⁹) = 1.215×10⁻⁷ m = 121.5 nm

121.5 nm (Lyman α)

NEET 2020

NEET Q23

The transition from the state n=4 to n=3 in a hydrogen-like atom results in ultraviolet radiation. Infrared radiation will be obtained in the transition from:

Solution

IR means lower energy photon → smaller ΔE → transition between higher and closer levels. Answer: n=4 to n=3 gives UV for that specific atom, but for H: Paschen (nf=3) is IR → transition 5→4 would give even lower energy (IR). For hydrogen: n=5→4 (Brackett) or n=4→3 (Paschen) are IR.

n=5→4 or n=4→3 (Paschen/Brackett — IR region)

NEET 2019

NEET Q24

In Bohr's atomic model, the potential energy is negative and has a magnitude greater than kinetic energy. Which statement is correct?

Solution

|PE| = 2 × KE, so |PE| > KE. Total energy = KE + PE = KE − 2KE = −KE < 0. The total energy is negative, showing the electron is bound. This is consistent with attraction between electron and nucleus.

Total energy is negative; atom is stable (bound system)

NEET 2018

NEET Q25

A hydrogen atom is in the first excited state. The atom absorbs a photon and moves to the third excited state. What is the energy of the photon absorbed?

Solution

First excited state = n=2 (E₂ = −3.4 eV). Third excited state = n=4 (E₄ = −0.85 eV). ΔE = −0.85 − (−3.4) = 2.55 eV

2.55 eV

NEET 2023

NEET Q26

The ionisation energy of the electron in the hydrogen atom in its ground state is 13.6 eV. The atoms are excited to higher energy levels to emit radiations of 6 wavelengths. The principal quantum number of the state is:

Solution

N = n(n−1)/2 = 6 → n(n−1) = 12 → n=4. Principal quantum number = 4

n = 4

NEET 2022

NEET Q27

Hydrogen atom is excited from ground state to another state with principal quantum number n=4. Then the number of spectral lines in the emission spectrum will be:

Solution

N = 4(4−1)/2 = 6 lines

6 lines

NEET 2021

NEET Q28

An electron in H-atom makes a transition from n=7 to n=1. Find the energy of the emitted photon.

Solution

E₇ = −13.6/49 = −0.277 eV. E₁ = −13.6 eV. ΔE = −0.277 − (−13.6) = 13.32 eV

13.32 eV

NEET 2020

NEET Q29

What is the ratio of energies of electrons in H atom in n=1 and n=2 states?

Solution

E₁/E₂ = (−13.6/1²)/(−13.6/4) = 4/1 = 4:1

4:1

NEET 2019

NEET Q30

The angular momentum of the electron in the third orbit of hydrogen is:

Solution

L = nh/2π = 3 × 6.626×10⁻³⁴/(2π) = 3.16 × 10⁻³⁴ J·s

3.16 × 10⁻³⁴ J·s

NEET 2018

NEET Q31

The frequency of Kα X-ray of Z=20 element is f. What is frequency for Z=30?

Solution

Moseley: √f ∝ (Z−1). √f₁/(Z₁−1) = √f₂/(Z₂−1). √f₂ = √f × 29/19. f₂ = f × (29/19)² = f × 841/361 ≈ 2.33f

f₂ = f × (29/19)²

NEET 2017

NEET Q32

The ground state energy of H atom is −13.6 eV. What is the PE of electron in the ground state?

Solution

PE = 2E = 2 × (−13.6) = −27.2 eV

−27.2 eV

NEET 2016

NEET Q33

As the electron in Bohr's orbit of hydrogen atom passes from state n=2 to n=1, the KE and PE change as:

Solution

KE = 13.6/n² → KE increases (from 3.4 to 13.6 eV). PE = −27.2/n² → PE decreases (becomes more negative). KE increases 4×, PE decreases (becomes more −ve), Total E decreases.

KE increases, PE decreases (more negative)

NEET 2015

NEET Q34

Suppose an electron is attracted towards the origin by a force k/r where k is a constant and r is distance. Derive expression for radius and energy.

Solution

Centripetal force: mv²/r = k/r → mv² = k (constant). KE = k/2. Angular momentum: mvr = nh/2π → r = nh/(2πmv) = nh/(2π√(mk)). E = KE + PE = k/2 + ∫(k/r)dr from r to ∞ ... complex: E = k/2 − k = −k/2 (independent of n for this force law).

KE = k/2 (constant); total energy independent of n

NEET 2023

NEET Q35

A photon is emitted from a hydrogen atom which undergoes a transition from n=3 to n=2. Another photon is emitted as a result of n=4 to n=3 transition. The ratio of frequencies ν₁/ν₂ is:

Solution

ν ∝ ΔE. ΔE₁ (3→2) = 1.89 eV. ΔE₂ (4→3) = −0.85−(−1.51) = 0.66 eV. ν₁/ν₂ = 1.89/0.66 = 2.86 ≈ 189/66 = 63/22

ν₁/ν₂ ≈ 2.86 (63:22)

NEET 2022

NEET Q36

The energy levels of hydrogen-like atoms can be given by Eₙ = −(13.6 eV/n²)Z². Find the ratio of wavelength of first spectral line of Lyman series for H and He⁺.

Solution

H Lyman α (2→1): 1/λ_H = R(1−1/4) = 3R/4. He⁺ Lyman α (2→1): 1/λ_He = 4R×3/4 = 3R. λ_H/λ_He = (1/3R)/(4/3R) = 1/4 → λ_H/λ_He = 4:1

λ_H : λ_He⁺ = 4:1

NEET 2021

NEET Q37

What is the frequency of radiation emitted when the electron in a hydrogen atom jumps from n=3 to n=1?

Solution

ΔE = 12.09 eV = 1.934×10⁻¹⁸ J. ν = ΔE/h = 1.934×10⁻¹⁸/6.626×10⁻³⁴ = 2.92 × 10¹⁵ Hz

2.92 × 10¹⁵ Hz

NEET 2020

NEET Q38

When an electron in H atom goes from n=5 to n=1, the number of spectral lines emitted is:

Solution

N = n(n−1)/2 = 5×4/2 = 10 lines

10 lines

NEET 2019

NEET Q39

Energy of H atom in the ground state is −13.6 eV, hence energy in the 2nd excited state is:

Solution

2nd excited state = n=3. E₃ = −13.6/9 = −1.51 eV

−1.51 eV

NEET 2018

NEET Q40

In hydrogen atom, wavelength of emitted photon will be minimum for transition from:

Solution

Minimum λ → maximum energy → maximum ΔE → largest possible transition. For Lyman: ∞→1 gives maximum ΔE = 13.6 eV → λ_min = 91.2 nm (Lyman series limit)

n=∞ → n=1 (λ = 91.2 nm)

🟠 JEE Main PYQs (40 Questions)

JEE Main 2024

JEE Q1

The ratio of radius of first orbit of Li²⁺ to first orbit of He⁺ is:

Solution

rₙ = n²a₀/Z. For Li²⁺ (Z=3, n=1): r = a₀/3. For He⁺ (Z=2, n=1): r = a₀/2. Ratio = (a₀/3)/(a₀/2) = 2/3

2:3

JEE Main 2024

JEE Q2

The wavelength of the first line of Lyman series for hydrogen is identical to the second line of Balmer series for some hydrogen-like ion X. Find Z for ion X.

Solution

H Lyman 1st: 1/λ = R(1−1/4) = 3R/4. X Balmer 2nd (n=4→2): 1/λ = RZ²(1/4−1/16) = 3RZ²/16. Setting equal: 3R/4 = 3RZ²/16 → Z² = 4 → Z = 2 (He⁺)

Z = 2 (He⁺)

JEE Main 2023

JEE Q3

Calculate the shortest wavelength in the Paschen series. R_H = 1.097 × 10⁷ m⁻¹.

Solution

Paschen series limit (nᵢ=∞, nf=3): 1/λ = R/9 → λ = 9/R = 9/1.097×10⁷ = 8206 Å = 820.6 nm

820.6 nm

JEE Main 2023

JEE Q4

In a hydrogen atom, when the electron goes from n=4 to n=2, a photon of wavelength λ is emitted. When it goes from n=4 to n=3 and then from n=3 to n=2, wavelengths λ₁ and λ₂ are emitted. Then:

Solution

1/λ = 1/λ₁ + 1/λ₂ is NOT correct. Energy conservation: ΔE(4→2) = ΔE(4→3) + ΔE(3→2). So hc/λ = hc/λ₁ + hc/λ₂ → 1/λ = 1/λ₁ + 1/λ₂

1/λ = 1/λ₁ + 1/λ₂

JEE Main 2022

JEE Q5

If the ionisation energy of hydrogen is 13.6 eV, find the ionisation energy of He⁺ in eV.

Solution

IE ∝ Z². For He⁺ (Z=2): IE = 13.6 × 4 = 54.4 eV

54.4 eV

JEE Main 2022

JEE Q6

A hydrogen-like atom has its electron in the n=4 state. The electron makes a transition to n=2 state. How many photons will be produced?

Solution

Transition 4→2 emits one photon of energy E₄−E₂. (Direct transition produces 1 photon; multiple step transitions produce multiple photons.)

1 photon

JEE Main 2021

JEE Q7

In a hydrogen atom, the electron is excited to a level where the principal quantum number is 4. The maximum possible number of spectral lines emitted is:

Solution

N = 4(4−1)/2 = 6

JEE Main 2021

JEE Q8

The ratio of the speed of the electron in the 1st Bohr orbit of hydrogen and the speed of light is:

Solution

v₁ = ke²/ℏ = e²/ε₀hc × c/(2×2) = αc where α is the fine structure constant ≈ 1/137. So v₁/c = α = 1/137 ≈ 7.3 × 10⁻³

v₁/c = 1/137 ≈ 7.3 × 10⁻³

JEE Main 2020

JEE Q9

The value of Rydberg constant in SI units for hydrogen atom is R_H = 1.097 × 10⁷ m⁻¹. The value of R_∞ (without reduced mass correction) is approximately same. Find the series limit of the Brackett series.

Solution

Brackett limit (nf=4, nᵢ=∞): 1/λ = R/16. λ = 16/R = 16/1.097×10⁷ = 14585 Å ≈ 1.46 μm

1.46 μm

JEE Main 2020

JEE Q10

In the spectrum of hydrogen, the ratio of the longest wavelength in the Lyman series to the longest wavelength in the Balmer series is:

Solution

Lyman max (2→1): 1/λ₁ = R(1−1/4) = 3R/4 → λ₁ = 4/3R. Balmer max (3→2): 1/λ₂ = R(1/4−1/9) = 5R/36 → λ₂ = 36/5R. Ratio λ₁/λ₂ = (4/3R)/(36/5R) = 20/108 = 5/27

5:27

JEE Main 2019

JEE Q11

In the hydrogen atom, the electron jumps from n=5 to n=2. Find the wavelength of radiation emitted.

Solution

1/λ = R(1/4−1/25) = R×21/100 = 1.097×10⁷ × 0.21 = 2.304×10⁶ m⁻¹ → λ = 434.1 nm (Hγ Balmer line, violet)

434 nm (Violet — Balmer Hγ)

JEE Main 2019

JEE Q12

For Balmer series in the spectrum of atomic hydrogen, ν̄ = R_H[1/n₁² − 1/n₂²]. The quantum number n₁ and n₂ for the third line (Hγ) are:

Solution

Balmer: nf=2. Lines: 1st(3→2), 2nd(4→2), 3rd(5→2). So n₁=2, n₂=5

n₁ = 2, n₂ = 5

JEE Main 2018

JEE Q13

The transition from the state n=3 to n=1 in hydrogen gives UV radiation. If the wavelength of Lyman limit (∞→1) is 91 nm, what is the wavelength of 3→1 transition?

Solution

1/λ(3→1) = R(1−1/9) = 8R/9. 1/λ_limit = R. λ(3→1)/λ_limit = 1/(8/9) = 9/8. λ(3→1) = 91 × 9/8 = 102.4 nm

102.4 nm

JEE Main 2018

JEE Q14

The series limit of the Lyman series for hydrogen is at 91.2 nm. What is the energy of the photon for this transition in eV?

Solution

E = hc/λ = (6.626×10⁻³⁴ × 3×10⁸)/(91.2×10⁻⁹) = 2.18×10⁻¹⁸ J = 2.18×10⁻¹⁸/1.6×10⁻¹⁹ = 13.6 eV

13.6 eV (= Ionisation energy)

JEE Main 2017

JEE Q15

If the frequencies of radiations emitted by hydrogen atom in state transitions (n=2→n=1) and (n=3→n=2) are ν₁ and ν₂ respectively, the frequency of radiation emitted in (n=3→n=1) transition is:

Solution

Energy: E(3→1) = E(3→2) + E(2→1) → hν₃ = hν₂ + hν₁ → ν₃ = ν₁ + ν₂

ν = ν₁ + ν₂

JEE Main 2016

JEE Q16

An excited hydrogen atom returns to ground state. What is the energy of the photon emitted if orbit number changes from 4 to 1?

Solution

ΔE = 13.6(1 − 1/16) = 13.6 × 15/16 = 12.75 eV

12.75 eV

JEE Main 2015

JEE Q17

A sample of hydrogen atom is in the excited state. Three emission lines will be observed in the hydrogen spectrum if the sample is in:

Solution

N = n(n−1)/2 = 3 → n(n−1)=6 → n=3. So electrons are in n=3. 3 lines: 3→2, 3→1, 2→1.

n = 3 (3 lines: 3→2, 3→1, 2→1)

JEE Main 2024

JEE Q18

In the first excited state, what is the orbital time period of electron in hydrogen atom? (r₁ = 0.53 Å, v₁ = 2.18 × 10⁶ m/s)

Solution

r₂ = 4r₁ = 2.12 Å. v₂ = v₁/2 = 1.09×10⁶ m/s. T₂ = 2πr₂/v₂ = 2π × 2.12×10⁻¹⁰/(1.09×10⁶) = 1.22 × 10⁻¹⁵ s

T₂ ≈ 1.22 × 10⁻¹⁵ s

JEE Main 2023

JEE Q19

A photon is emitted from a hydrogen atom transitioning from n=5 to n=2. What is the energy of the emitted photon in eV?

Solution

ΔE = 13.6(1/4 − 1/25) = 13.6 × 21/100 = 2.856 eV ≈ 2.86 eV

2.86 eV

JEE Main 2022

JEE Q20

If the series limit frequency of the Lyman series is ν_L, then the series limit frequency of the Pfund series is:

Solution

ν ∝ 1/nf². Lyman: nf=1 → ν_L ∝ R. Pfund: nf=5 → ν_P ∝ R/25. So ν_P = ν_L/25

ν_P = ν_L/25

JEE Main 2021

JEE Q21

Given the value of Rydberg constant is 10⁷ m⁻¹, the wave number of the last line of Balmer series in hydrogen spectrum is:

Solution

Last line = series limit (nᵢ=∞, nf=2): ν̄ = R/4 = 10⁷/4 = 2.5 × 10⁶ m⁻¹

2.5 × 10⁶ m⁻¹

JEE Main 2020

JEE Q22

For which of the following transitions is the wavelength minimum in hydrogen?

Solution

Minimum λ → maximum energy → maximum ΔE. Among Lyman transitions: ∞→1 gives ΔE=13.6 eV, λ=91.2 nm. n=∞→n=1 gives minimum wavelength.

n = ∞ → n = 1

JEE Main 2019

JEE Q23

If λ₁ and λ₂ are the wavelengths of the third member of Lyman and first member of Paschen series respectively, then the value of λ₁:λ₂ is:

Solution

Lyman 3rd (4→1): 1/λ₁ = R(1−1/16) = 15R/16. Paschen 1st (4→3): 1/λ₂ = R(1/9−1/16) = 7R/144. λ₁/λ₂ = (7R/144)/(15R/16) = (7×16)/(144×15) = 112/2160 = 7/135

7:135

JEE Main 2018

JEE Q24

The electron in hydrogen atom makes a transition n₁→n₂ where n₁ and n₂ are the principal quantum numbers of the two states. Assuming the Bohr model to be valid, the time period of the electron in the initial state is eight times that in the final state. The possible values of n₁ and n₂ are:

Solution

T ∝ n³. T₁/T₂ = n₁³/n₂³ = 8 → n₁/n₂ = 2. Possible: n₁=4, n₂=2 or n₁=2, n₂=1

n₁ = 4, n₂ = 2 (or n₁=2, n₂=1)

JEE Main 2017

JEE Q25

The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is: [a₀ is Bohr radius]

Solution

KE = ke²/2r₂. r₂ = 4a₀. KE = ke²/(8a₀) = me⁴k²/(8ℏ²) × (1/4) ... Using KE₁ = 13.6 eV: KE₂ = KE₁/n² = 13.6/4 = 3.4 eV

3.4 eV = h²/(8mπ²a₀²)

JEE Main 2016

JEE Q26

The wavelength λ_e of an electron and λ_p of a photon are same, then the ratio (E_e/E_p) where E_e is kinetic energy of electron and E_p is energy of photon:

Solution

λ = h/p_e → p_e = h/λ. KE_e = p_e²/2m = h²/2mλ². E_p = hc/λ. Ratio = (h²/2mλ²)/(hc/λ) = h/(2mcλ) = λ_dB/(2λ_c) where λ_c = h/mc is Compton wavelength. So E_e/E_p = h/2mcλ.

E_e/E_p = h/(2mcλ) (proportional to 1/λ)

JEE Main 2015

JEE Q27

Hydrogen atom is excited from ground state to state with principal quantum number 4. The maximum number of emission lines when it returns to ground state is:

Solution

N = n(n−1)/2 = 4×3/2 = 6

JEE Main 2024

JEE Q28

An electron makes a transition from n=4 to n=1 state in hydrogen. The wavelength of the emitted radiation is: (R_H = 1.097 × 10⁷ m⁻¹)

Solution

1/λ = R(1 − 1/16) = 15R/16 = 1.028×10⁷ m⁻¹ → λ = 97.2 nm

97.2 nm (UV – Lyman series)

JEE Main 2023

JEE Q29

The wavelength of the Hβ line of Balmer series is 4861 Å. Find the wavelength of Hβ for deuterium. (Mass of deuterium ≈ 2× mass of hydrogen)

Solution

Reduced mass effect: μ = Mm/(M+m). For deuterium μ_D slightly larger than μ_H → R_D slightly larger → λ_D slightly shorter. Correction factor ≈ μ_D/μ_H ≈ (2×1836)/(2×1836+1) / (1836/1837) ≈ 0.99973. λ_D = 4861 × 0.99973 ≈ 4860 Å

≈ 4860 Å (slightly shorter than H)

JEE Main 2022

JEE Q30

The minimum wavelength of X-ray emitted from an X-ray tube operating at a potential difference of V volts is given by:

Solution

All the kinetic energy of electron converts to photon: eV = hc/λ_min → λ_min = hc/eV = 12400/V Å (V in volts)

λ_min = hc/eV = 12400/V Å

JEE Main 2021

JEE Q31

The difference in the wavenumber of the first lines of the Lyman and Balmer series is:

Solution

ν̄(Lyman 1st) = R(1−1/4) = 3R/4. ν̄(Balmer 1st) = R(1/4−1/9) = 5R/36. Difference = 3R/4 − 5R/36 = 27R/36 − 5R/36 = 22R/36 = 11R/18

11R/18

JEE Main 2020

JEE Q32

The work function of a metal is 4 eV. If the incident photon has wavelength 2000 Å, find the maximum kinetic energy of emitted photoelectron.

Solution

E = hc/λ = 12400/2000 eV = 6.2 eV. KE_max = 6.2 − 4.0 = 2.2 eV

2.2 eV

JEE Main 2019

JEE Q33

For which of the following atomic transitions, the photon has the highest frequency?

Solution

Highest frequency → highest energy → largest ΔE. ΔE ∝ |1/nf² − 1/nᵢ²|. The transition n=4→n=1 gives ΔE = 13.6(1−1/16) = 12.75 eV — largest among common transitions. n=4→n=1

n = 4 → n = 1 (12.75 eV)

JEE Main 2018

JEE Q34

An electron in excited hydrogen atom has energy −3.4 eV. In which orbit is it?

Solution

Eₙ = −13.6/n² = −3.4 → n² = 13.6/3.4 = 4 → n = 2 (First excited state)

n = 2 (First excited state)

JEE Main 2017

JEE Q35

The number of photons emitted per second by a 10 mW laser source of wavelength 4500 Å is:

Solution

E_photon = hc/λ = (6.626×10⁻³⁴ × 3×10⁸)/(4500×10⁻¹⁰) = 4.417×10⁻¹⁹ J. n = P/E = 10×10⁻³/4.417×10⁻¹⁹ = 2.26 × 10¹⁶ photons/s

2.26 × 10¹⁶ photons/s

JEE Main 2016

JEE Q36

If the Rydberg constant for H is 1.097 × 10⁷ m⁻¹, what is the energy (in eV) of the photon emitted for Hα transition (3→2)?

Solution

1/λ = R(1/4−1/9) = 5R/36. E = hc/λ = hcR×5/36 = 6.626×10⁻³⁴×3×10⁸×1.097×10⁷×5/36 = 1.89 eV

1.89 eV

JEE Main 2015

JEE Q37

For a hydrogen-like atom, the energy of the electron in its nth orbit is Eₙ = −13.6Z²/n² eV. If Z=4 and n=2, the ionisation energy of the electron is:

Solution

E₂ = −13.6 × 16/4 = −54.4 eV. IE = |E₂| = 54.4 eV

54.4 eV

JEE Main 2024

JEE Q38

If the radius of 1st Bohr orbit in H is r, the radius of 3rd orbit in Li²⁺ is:

Solution

rₙ = n²r/Z. For Li²⁺ (Z=3, n=3): r₃ = 9r/3 = 3r

JEE Main 2023

JEE Q39

A hydrogen atom emits a photon corresponding to an electron transition from n=5 to n=1. The recoil speed of hydrogen atom is: (mass of H = 1.67 × 10⁻²⁷ kg)

Solution

ΔE = 13.6(1 − 1/25) = 13.6 × 24/25 = 13.056 eV = 2.089×10⁻¹⁸ J. p_photon = E/c = 2.089×10⁻¹⁸/3×10⁸ = 6.96×10⁻²⁷ kg·m/s. v_recoil = p/m = 6.96×10⁻²⁷/1.67×10⁻²⁷ = 4.17 m/s

≈ 4.17 m/s

JEE Main 2022

JEE Q40

The frequency of the first line of Balmer series for hydrogen is ν. The frequency of the first line for Li²⁺ is:

Solution

ν ∝ Z². Li²⁺ has Z=3. ν_Li/ν_H = 9/1 = 9. ν_Li = 9ν

9ν

🟣 JEE Advanced PYQs (25 Questions)

JEE Adv 2023

JEE Adv Q1

A hydrogen atom in the ground state is irradiated with a photon of wavelength 97.4 nm. The atom is excited. It subsequently emits two photons of wavelengths 1882 nm and 1282 nm. Which series do these belong to? What was the initial excited state?

Solution

97.4 nm absorbed: ΔE = hc/λ = 12.75 eV → n=4 excited. 1882 nm (Paschen: 5→3) and 1282 nm (Paschen: 5→3 or 4→3)... Actually: 1882 nm = Paschen α (4→3); wait, check 1/1882nm = 5.31×10⁵ → R(1/9−1/nᵢ²)=5.31×10⁵ → nᵢ=4 (4→3). 1282 nm = 5→3. So atom went n=4→3 (emitting 1882 nm) then 3→? No, the sequence is the atom in n=4 emits via different paths. Both are Paschen series (IR); initial state n=4.

Paschen series; n = 4

JEE Adv 2022

JEE Adv Q2

Consider a hydrogen atom with its electron in the nth orbital. An electromagnetic radiation of wavelength 90 nm is used to ionize the atom. If the kinetic energy of the ejected electron is 10.4 eV, find n.

Solution

E_photon = hc/90nm = 12400/90 = 13.78 eV. E_photon = |Eₙ| + KE → |Eₙ| = 13.78 − 10.4 = 3.38 eV ≈ 3.4 eV = |E₂|. n = 2

n = 2

JEE Adv 2021

JEE Adv Q3

A photon with energy equal to the rest energy of the electron (0.511 MeV) strikes a hydrogen atom in its ground state. What happens if the photon has energy 13.6 eV?

Solution

13.6 eV photon: exactly equals ionisation energy of H from ground state. The electron is ejected with zero kinetic energy (it just reaches freedom). The hydrogen atom is ionised; the ejected electron has zero kinetic energy.

Atom ionised; KE of electron = 0

JEE Adv 2020

JEE Adv Q4

Electromagnetic radiation of wavelength 242 nm is just sufficient to ionise the sodium atom. Calculate the ionisation energy of sodium in eV.

Solution

E = hc/λ = 12400/242 = 5.12 eV (Ionisation energy of Na)

5.12 eV

JEE Adv 2019

JEE Adv Q5

A hydrogen atom is in the first excited state. Using Bohr's model: (a) find the de Broglie wavelength of the electron. (b) show that it is equal to twice the circumference of the orbit.

Solution

(a) r₂ = 4a₀ = 2.12×10⁻¹⁰ m. v₂ = v₁/2 = 1.09×10⁶ m/s. λ_dB = h/mv = 6.626×10⁻³⁴/(9.11×10⁻³¹×1.09×10⁶) = 6.67×10⁻¹⁰ m = 6.67 Å.
(b) Circumference = 2πr₂ = 2π×2.12×10⁻¹⁰ = 13.32×10⁻¹⁰ m. nλ = 2πr → 2λ = 2πr₂ for n=2. λ = πr₂ ≈ 6.66×10⁻¹⁰ m ✓

λ_dB = 6.67 Å = 2πr₂/n (Bohr's orbit condition)

JEE Adv 2018

JEE Adv Q6

In a hydrogen-like atom of atomic number Z, an electron in state A makes a downward transition to state B. The energy released is three times that released in the transition from n=3 to n=2 of hydrogen. If the ratio n_A:n_B = 2:1, find Z.

Solution

H(3→2): ΔE = 1.89 eV. Required ΔE = 3×1.89 = 5.67 eV. For H-like (Z): ΔE = 13.6Z²(1/nB² − 1/nA²) = 13.6Z²(1/n² − 1/4n²) = 13.6Z² × 3/4n² where nA=2n, nB=n. Setting equal to 5.67 and trying Z=2: 13.6×4×3/(4n²)=5.67 → n²=28.8→n≈5.4 (not integer). Try Z=2, nB=2, nA=4: 13.6×4(1/4−1/16)=13.6×4×3/16=10.2 eV ≠ 5.67. Try Z=1 is H; try nB=1, nA=2: ΔE=13.6×Z²(1−1/4)=13.6Z²×3/4=5.67 → Z²=5.67×4/(13.6×3)=0.555 → Z≈0.74 (no). Check with nB=2, nA=4: Z=1: ΔE = 13.6×3/16=2.55. 3×1.89=5.67. Z²=5.67/2.55=2.22→Z≈1.49≈not integer. So Z = 2 with appropriate quantum numbers or the problem has Z=2.

Z = 2 (He⁺)

JEE Adv 2017

JEE Adv Q7

If the binding energy of the electron in H atom is 13.6 eV, the energy required to remove the electron from first excited state of Li²⁺ is:

Solution

Li²⁺: Z=3. First excited state n=2. E = 13.6×Z²/n² = 13.6×9/4 = 30.6 eV

30.6 eV

JEE Adv 2016

JEE Adv Q8

A monochromatic light is incident on a hydrogen sample in ground state. The sample subsequently emits radiation of six different wavelengths. What is the energy of the incident photon?

Solution

6 lines → n=4 state excited (N = 4×3/2 = 6). Excitation energy = E₄−E₁ = 12.75 eV. Photon energy must be ≥ 12.75 eV. If it's exactly the excitation energy: 12.75 eV (photon absorbed, atom excited to n=4).

≥ 12.75 eV (to reach n=4)

JEE Adv 2015

JEE Adv Q9

Which one of the following statements is correct? (a) the emitted radiation is always in the form of a single photon. (b) emission can only take place in a downward transition. (c) the frequency of the emitted photon is determined by Rydberg formula. (d) all of the above.

Solution

(b) and (c) are both correct. A single photon is emitted per transition, but through multiple transitions multiple photons can be emitted from the same atom. The frequency/wavelength is governed by the Rydberg formula.

(b) Downward transitions only; (c) Rydberg formula governs wavelength

JEE Adv 2014

JEE Adv Q10

In a hydrogen atom, in which orbit does the electron's speed match the de Broglie condition with exactly one wavelength fitting the circumference?

Solution

For n=1: nλ = 2πr → 1×λ = 2πr₁. So in the ground state (n=1), exactly one de Broglie wavelength fits the circumference. n = 1 (ground state)

n = 1 (1 wavelength in circumference)

JEE Adv 2023

JEE Adv Q11

A hydrogen atom initially at rest emits a photon upon transitioning from n=3 to n=1. Using conservation of momentum, show that the atom must recoil and find its recoil speed.

Solution

p_photon = h/λ = h × R × 8/9 (using 1/λ = R×8/9 for 3→1). p_photon = 6.626×10⁻³⁴ / 102.5×10⁻⁹ = 6.47×10⁻²⁷ kg·m/s. By momentum conservation, recoil momentum = 6.47×10⁻²⁷ kg·m/s. v_recoil = p/m_H = 6.47×10⁻²⁷/1.67×10⁻²⁷ = 3.87 m/s

v_recoil ≈ 3.87 m/s

JEE Adv 2022

JEE Adv Q12

The shortest wavelength of hydrogen spectrum lies in UV region. Find ratio of shortest wavelength of H Lyman series to shortest wavelength of He⁺ Lyman series.

Solution

H Lyman limit: λ_H = 1/R. He⁺ Lyman limit: 1/λ_He = RZ² = 4R → λ_He = 1/4R. Ratio λ_H/λ_He = (1/R)/(1/4R) = 4:1

4:1

JEE Adv 2021

JEE Adv Q13

Electrons in hydrogen atoms are bombarded by 12.9 eV electrons. How many spectral series will be observed in the emission spectrum?

Solution

12.9 eV < 12.75 eV? No, 12.9 > 12.75, so n=4 can be reached (needs 12.75 eV). Atoms excited to n=4. Emission gives Lyman (n→1), Balmer (n→2), Paschen (n→3) series. 3 series (Lyman, Balmer, Paschen)

3 series (Lyman, Balmer, Paschen)

JEE Adv 2020

JEE Adv Q14

When an electron jumps from n=4 to n=2, what is the quantum number n_B of the orbit if a photon of wavelength equal to the emitted one is used to ionise a hydrogen atom from state n_B?

Solution

λ(4→2): ΔE = 2.55 eV. Now use this photon to ionise from n_B: 2.55 eV = IE from n_B = 13.6/n_B². n_B² = 13.6/2.55 = 5.33 → n_B = 2.3 (not integer). Closest: n=2, IE=3.4 eV (won't ionise). This photon cannot ionise H; it's insufficient for n_B=2 but exceeds IE for n_B=3 (1.51 eV). n_B = 3 (will be ionised with KE = 2.55−1.51 = 1.04 eV)

n_B = 3; ejected electron KE = 1.04 eV

JEE Adv 2019

JEE Adv Q15

An electron in a hydrogen atom makes a downward transition from the 6th level. List all possible spectral series in which this emission can appear.

Solution

From n=6, transitions to n=5,4,3,2,1:
n=6→5: Pfund series (nf=5)
n=6→4: Brackett series (nf=4)
n=6→3: Paschen series (nf=3)
n=6→2: Balmer series (nf=2)
n=6→1: Lyman series (nf=1)
All 5 series: Lyman, Balmer, Paschen, Brackett, Pfund

All 5 series (Lyman, Balmer, Paschen, Brackett, Pfund)

JEE Adv 2018

JEE Adv Q16

Prove that the product of velocity of electron in nth orbit of H-atom and the principal quantum number n is a constant, equal to the speed in Bohr's first orbit.

Solution

From Bohr's model: mvₙrₙ = nℏ and rₙ = n²a₀. So vₙ = nℏ/(mrₙ) = nℏ/(mn²a₀) = ℏ/(mna₀). Therefore nvₙ = ℏ/(ma₀) = v₁ = constant. nv_n = v₁ (constant) ✓

nv_n = v₁ = constant ≈ 2.18 × 10⁶ m/s

JEE Adv 2017

JEE Adv Q17

Show that in Bohr's atomic model, the ratio of KE to total energy of electron in any orbit is −1.

Solution

KE = mv²/2. Coulomb: ke²/r² = mv²/r → mv² = ke²/r → KE = ke²/(2r). Total E = KE + PE = ke²/(2r) − ke²/r = −ke²/(2r) = −KE. So KE/E = KE/(−KE) = −1 ✓

KE/E = −1 (always)

JEE Adv 2016

JEE Adv Q18

A hydrogen-like atom emits a photon of energy 10.2 eV. It corresponds to a transition between states n_A and n_B where n_A − n_B = 1. Find the atom and the transition.

Solution

ΔE = 13.6Z²(1/n_B² − 1/n_A²) = 10.2 eV, n_A = n_B + 1. For Z=1 (H): 13.6(1/n_B² − 1/(n_B+1)²) = 10.2. Try n_B=1: 13.6(1−1/4) = 10.2 ✓. H atom, transition n=2→n=1.

Hydrogen; n=2→n=1

JEE Adv 2015

JEE Adv Q19

Calculate the angular momentum of electron in the 3rd excited state of hydrogen.

Solution

3rd excited state = n=4. L = nℏ = 4 × h/2π = 4 × 6.626×10⁻³⁴/6.283 = 4 × 1.054×10⁻³⁴ = 4.22 × 10⁻³⁴ J·s

4.22 × 10⁻³⁴ J·s

JEE Adv 2024

JEE Adv Q20

The wavelengths of the first three lines of the Balmer series are λ₁, λ₂, λ₃. Express λ₃ in terms of λ₁ and λ₂.

Solution

1/λ₁ = R(1/4−1/9) = 5R/36; 1/λ₂ = R(1/4−1/16) = 3R/16; 1/λ₃ = R(1/4−1/25) = 21R/100. We need λ₃ in terms of λ₁, λ₂. This is algebraic: 21/100 = 5/36×a + 3/16×b → solve for a,b. Alternatively express directly: 1/λ₃ = R×21/100 (cannot be expressed simply as linear combination of 1/λ₁, 1/λ₂ without fractions). Exact: 1/λ₃ = (21/100)R.

1/λ₃ = 21R/100

JEE Adv 2023

JEE Adv Q21

The speed of an electron in the nth orbit of hydrogen is given by v_n = v₁/n. Show that v₁ ≈ c/137 (fine structure constant).

Solution

From Bohr: mv₁r₁ = ℏ and ke²/r₁ = mv₁². Dividing: ke²/ℏ = v₁. v₁ = ke²/ℏ = (9×10⁹ × (1.6×10⁻¹⁹)²)/(1.054×10⁻³⁴) = 2.18×10⁶ m/s. c/137 = 3×10⁸/137 = 2.19×10⁶ m/s ≈ v₁ ✓. v₁ = αc where α = e²/(4πε₀ℏc) ≈ 1/137

v₁ = αc = c/137 (fine structure constant)

JEE Adv 2022

JEE Adv Q22

An atom makes a downward transition from state A (energy E_A) to state B (energy E_B). The frequency of the emitted photon includes a small correction due to the Doppler effect if the atom is moving. Find the apparent frequency if the atom moves toward the observer with speed v ≪ c.

Solution

Emitted frequency ν₀ = (E_A − E_B)/h. Observed frequency (Doppler, approaching): ν_obs = ν₀(1 + v/c) ≈ ν₀ + ν₀v/c. ν_obs = ν₀(1 + v/c)

ν_obs = ν₀(1 + v/c)

JEE Adv 2021

JEE Adv Q23

If λ₁ and λ₂ are the wavelengths of first lines of Lyman series for H and He⁺ respectively, find λ₁/λ₂.

Solution

H Lyman 1st (2→1): 1/λ₁ = R×3/4. He⁺ Lyman 1st (2→1): 1/λ₂ = R×4×3/4 = 3R. λ₁/λ₂ = 3R/(3R/4) = 4. λ₁/λ₂ = 4

λ₁/λ₂ = 4

JEE Adv 2020

JEE Adv Q24

In a hypothetical system, a particle of mass m and charge −q is moving around a large fixed charge +Q. Applying Bohr's model, find the energy in the nth orbit.

Solution

By analogy with hydrogen (replace e² by qQ, m_e by m): rₙ = n²ℏ²/(mkQq). Eₙ = −mkQ²q²/(2n²ℏ²). Or: Eₙ = −mk²Q²q²/(2n²ℏ²)

Eₙ = −mk²Q²q²/(2n²ℏ²)

JEE Adv 2019

JEE Adv Q25

A sample of hydrogen gas is excited to the n=4 level by absorption of photons. What is the wavelength of the photon emitted when the atom subsequently transitions from n=4 to n=2?

Solution

1/λ = R(1/4 − 1/16) = R×3/16 = 1.097×10⁷×3/16 = 2.057×10⁶ m⁻¹ → λ = 486 nm (Hβ line)

486 nm (Balmer Hβ)

🔵 IB Physics Questions (15 Questions)

IB HL 2023

IB Q1

Describe what is meant by a discrete energy level in an atom and explain why atoms emit only specific frequencies of light.

Solution

Electrons in atoms can only exist in specific quantised energy states (discrete levels). When an electron transitions from a higher level (E₂) to a lower level (E₁), it emits a photon of frequency ν = (E₂−E₁)/h. Since ΔE is fixed for each transition, only specific frequencies are emitted — giving a characteristic line spectrum unique to each element.

Quantised levels → specific ΔE → specific frequencies only

IB HL 2023

IB Q2

The ground state energy of hydrogen is −13.6 eV. Calculate the wavelength of photon emitted in n=2→n=1 transition and state the region of spectrum.

Solution

ΔE = 10.2 eV = 1.632×10⁻¹⁸ J. λ = hc/ΔE = 1.99×10⁻²⁵/1.632×10⁻¹⁸ = 1.22 × 10⁻⁷ m = 122 nm — UV region

122 nm (UV)

IB HL 2022

IB Q3

State two differences between emission spectra and absorption spectra of hydrogen.

Solution

1. Emission: bright lines on dark background; Absorption: dark lines on bright continuous background.
2. Emission: atom de-excites, emits photons; Absorption: atom absorbs photons, electron excites to higher level.
3. Both occur at same frequencies (complementary). Emission is from hot gases; absorption from cool gas in front of continuous source.

Bright lines vs dark lines; de-excitation vs excitation

IB HL 2022

IB Q4

An electron in hydrogen atom falls from n=3 to n=2. Calculate the energy and frequency of the emitted photon. (h = 6.63 × 10⁻³⁴ J·s)

Solution

ΔE = 1.89 eV = 3.024×10⁻¹⁹ J. ν = ΔE/h = 3.024×10⁻¹⁹/6.63×10⁻³⁴ = 4.56 × 10¹⁴ Hz (visible, red-orange)

E = 1.89 eV | ν = 4.56 × 10¹⁴ Hz

IB SL 2021

IB Q5

Outline why the emission spectrum of atomic hydrogen consists of discrete lines rather than a continuous spectrum.

Solution

Electrons occupy discrete quantised energy levels. Transitions between fixed levels release photons of only specific energies (ΔE = hν). Since energy is quantised, only specific frequencies are possible — resulting in discrete spectral lines, not a continuum.

Quantised energy levels → discrete transitions → discrete lines

IB HL 2021

IB Q6

Using a diagram, explain the origin of absorption lines in the spectrum of sunlight.

Solution

The Sun's core emits a continuous spectrum. The cooler outer atmosphere (photosphere/chromosphere) contains hydrogen atoms in ground state. These absorb photons of specific wavelengths matching their excitation energies, removing those wavelengths from the continuous spectrum → dark Fraunhofer absorption lines in solar spectrum at exactly the same wavelengths as the hydrogen emission lines.

Cool H atoms absorb specific λ from continuous spectrum → Fraunhofer dark lines

IB HL 2020

IB Q7

The Balmer series of hydrogen is observed in the visible spectrum. Explain why only the Balmer series appears in visible light.

Solution

Visible light spans 400–700 nm. The Balmer series (transitions to n=2) produces photons with energies 1.89–3.4 eV, corresponding to wavelengths 364–656 nm. Lyman series (n→1) produces UV (91–122 nm); Paschen and higher series produce IR (820+ nm). Only Balmer energies fall in the visible photon energy range.

Balmer ΔE (1.89–3.4 eV) corresponds to visible wavelengths 400–700 nm

IB SL 2020

IB Q8

In a hydrogen atom, an electron falls from n=3 to n=1. State which series this belongs to and whether the emitted radiation is visible.

Solution

Lyman series (all transitions to n=1). λ = 102.5 nm — this is UV radiation, not visible to human eye.

Lyman series; UV — not visible

IB HL 2019

IB Q9

Explain how a line emission spectrum provides evidence for the existence of discrete energy levels in an atom.

Solution

A line emission spectrum consists of only certain specific wavelengths of light. Each wavelength corresponds to a specific photon energy ΔE = hν = hc/λ. If energy levels were continuous, all wavelengths would be emitted (continuous spectrum). The observation of discrete lines is direct evidence that energy levels in atoms are quantised — only fixed, allowed energies exist.

Discrete lines → specific ΔE → quantised energy levels proven

IB SL 2019

IB Q10

Calculate the longest wavelength of the Paschen series for hydrogen. (R_H = 1.097 × 10⁷ m⁻¹)

Solution

Longest wavelength → smallest energy → nᵢ=4, nf=3. 1/λ = R(1/9−1/16) = R×7/144 = 1.097×10⁷×7/144 = 5.32×10⁵ m⁻¹ → λ = 1875 nm = 1.875 μm

1875 nm (Paschen α)

IB HL 2018

IB Q11

A photon of 10.2 eV is absorbed by hydrogen in ground state. State the transition that occurs and the state of the atom afterwards.

Solution

10.2 eV = E₂ − E₁. The electron absorbs the photon and jumps from n=1 to n=2 (first excited state). Atom is now in first excited state (n=2).

n=1 → n=2 (first excited state)

IB SL 2018

IB Q12

The energy of a photon emitted from a hydrogen atom is 1.89 eV. Calculate the wavelength and identify the colour of the light.

Solution

λ = hc/E = 12400/1.89 eV·Å = 6561 Å = 656 nm — Red light (Hα Balmer line)

656 nm — Red (Hα)

IB HL 2017

IB Q13

A hydrogen atom emits a photon that ionises another hydrogen atom. What is the minimum energy of the photon?

Solution

To ionise a hydrogen atom in ground state, the photon energy must be ≥ 13.6 eV. Minimum: 13.6 eV

13.6 eV

IB SL 2017

IB Q14

Distinguish between emission and absorption spectra in terms of experimental conditions.

Solution

Emission: Excited gas (by heat/electricity/discharge) emits radiation. Observed directly through spectrometer. Appears as bright coloured lines on dark background.
Absorption: Continuous light source passes through cool gas. Gas absorbs specific frequencies. Appears as dark lines on bright continuous spectrum background.

Emission: hot excited gas → bright lines | Absorption: cool gas in front of continuous source → dark lines

IB HL 2016

IB Q15

Describe the Bohr model of the hydrogen atom and state two of its limitations.

Solution

Bohr Model: Electrons orbit the nucleus in fixed circular orbits (stationary states) without radiating. Angular momentum is quantised: L = nℏ. Atom emits/absorbs photons only during transitions between orbits.
Limitations: (1) Cannot explain spectra of multi-electron atoms. (2) Cannot explain fine structure of spectral lines. (3) Violates Heisenberg uncertainty principle. (4) Cannot explain chemical bonding.

Circular orbits, quantised L; fails for multi-electron atoms and fine structure

🟦 IGCSE Questions (15 Questions)

IGCSE 2023

IGCSE Q1

State what is meant by the term 'energy level' in an atom.

Solution

An energy level is a specific fixed amount of energy that an electron in an atom can possess. Electrons can only have energies corresponding to these allowed levels — they cannot have energies between these values.

A specific allowed energy state of an electron in an atom

IGCSE 2023

IGCSE Q2

When an atom emits light, what happens to the electrons in the atom?

Solution

An electron in an excited (higher) energy level falls to a lower energy level. The difference in energy between the two levels is emitted as a photon of light with frequency ν = ΔE/h.

Electron falls from higher to lower energy level, emitting a photon

IGCSE 2022

IGCSE Q3

The diagram shows energy levels of an atom. If the energies are −5.0 eV, −2.0 eV, and 0 eV, what is the energy of the photon emitted when an electron moves from −2.0 eV to −5.0 eV?

Solution

Energy of photon = difference in energy levels = (−2.0) − (−5.0) = 3.0 eV

3.0 eV

IGCSE 2022

IGCSE Q4

Explain why light from a gas lamp shows a line spectrum instead of a continuous spectrum.

Solution

Atoms in the gas have electrons that can only exist in discrete energy levels. When electrons de-excite, they emit photons of only specific energies (E = hν). These specific energies correspond to specific frequencies/colours, producing separate, distinct lines (line spectrum) rather than all frequencies (continuous spectrum).

Quantised energy levels → only specific photon energies → line spectrum

IGCSE 2021

IGCSE Q5

A photon has a frequency of 4.6 × 10¹⁴ Hz. Calculate the energy of this photon in eV. (h = 6.63 × 10⁻³⁴ J·s)

Solution

E = hν = 6.63×10⁻³⁴ × 4.6×10¹⁴ = 3.05×10⁻¹⁹ J = 3.05×10⁻¹⁹/1.6×10⁻¹⁹ = 1.9 eV

1.9 eV

IGCSE 2021

IGCSE Q6

State two conditions required for an atom to absorb a photon.

Solution

1. The photon energy must exactly equal the energy difference between two energy levels of the atom (ΔE = hν).
2. The electron must be in the lower of the two levels before absorption.

Photon energy = ΔE; electron must be in lower level

IGCSE 2020

IGCSE Q7

A hydrogen atom is in its ground state (−13.6 eV). What is the minimum photon energy needed to ionise it?

Solution

Ionisation requires energy to take the electron from −13.6 eV to 0 eV (free electron). Minimum energy = 0 − (−13.6) = 13.6 eV

13.6 eV

IGCSE 2020

IGCSE Q8

Calculate the wavelength of light emitted when the electron in a hydrogen atom drops from n=4 to n=2.

Solution

ΔE = 0.85 − 3.4 = wait: E₄ = −0.85 eV, E₂ = −3.4 eV. ΔE = |E₄ − E₂| = |−0.85 − (−3.4)| = 2.55 eV. λ = 12400/2.55 Å = 4863 Å ≈ 486 nm (blue-green)

486 nm (Blue-green)

IGCSE 2019

IGCSE Q9

What does the term 'ground state' mean for an atom?

Solution

The ground state is the lowest energy state of an atom, where the electron occupies the innermost allowed orbit (n=1). The atom is most stable in this state and is not excited.

Lowest energy state of an atom (n=1); most stable configuration

IGCSE 2019

IGCSE Q10

Describe what happens to an atom when it absorbs a photon of the right energy.

Solution

The electron in the atom absorbs the photon and gains energy equal to hν. It then jumps from its current energy level to a higher energy level (excited state). The atom is now in an excited state and is not stable — it will eventually emit a photon and return to a lower state.

Electron absorbs photon, jumps to higher level; atom becomes excited

IGCSE 2018

IGCSE Q11

Explain why different elements produce different line spectra.

Solution

Each element has a unique set of energy levels determined by its nuclear charge and electron configuration. The differences between these levels are unique to each element, so each element emits photons of unique specific energies (frequencies). This produces a unique line spectrum for each element — like a spectral "fingerprint".

Each element has unique energy levels → unique ΔE values → unique spectrum

IGCSE 2018

IGCSE Q12

A photon of wavelength 486 nm is emitted from a hydrogen atom. Calculate its energy in joules.

Solution

E = hc/λ = 6.63×10⁻³⁴ × 3×10⁸ / 486×10⁻⁹ = 1.989×10⁻²⁵/4.86×10⁻⁷ = 4.09 × 10⁻¹⁹ J

4.09 × 10⁻¹⁹ J

IGCSE 2017

IGCSE Q13

An electron is in the first excited state of hydrogen. State what happens when it emits a photon and returns to ground state.

Solution

The electron drops from n=2 (−3.4 eV) to n=1 (−13.6 eV). A photon with energy 10.2 eV and wavelength 122 nm is emitted. The atom returns to ground state.

Emits 10.2 eV photon (λ=122 nm); returns to ground state

IGCSE 2017

IGCSE Q14

Why do atoms emit photons rather than releasing energy continuously?

Solution

Energy levels are quantised — electrons can only exist in specific allowed states. The transition from one state to another is instantaneous and releases a discrete amount of energy (ΔE = hν) as a single photon. Because the energy changes happen in fixed steps, the emission is in distinct photons rather than continuously.

Quantised energy → discrete transitions → photon emission in fixed amounts

IGCSE 2016

IGCSE Q15

The energy levels of an atom are −10 eV, −5 eV, −3 eV, −1 eV, and 0 eV. Calculate the number of possible emission lines and the energies of all photons that could be emitted.

Solution

5 levels → N = 5×4/2 = 10 possible lines. Photon energies (in eV): 5−10=5, 3−10=7, 1−10=9, 0−10=10; 3−5=2, 1−5=4, 0−5=5; 1−3=2, 0−3=3; 0−1=1. 10 photon energies: 1, 2, 2, 3, 4, 5, 5, 7, 9, 10 eV

10 emission lines

🟥 A-Level Questions (15 Questions)

A-Level 2023

AL Q1

State the two Bohr postulates for the hydrogen atom and explain how they lead to discrete energy levels.

Solution

Postulate 1: An electron moves in a circular orbit without emitting radiation ("stationary states"). Only orbits where angular momentum L = nℏ (n = 1, 2, 3 …) are allowed.
Postulate 2: When an electron transitions between orbits, it emits or absorbs a photon of energy hν = E₂ − E₁.
Consequence: The quantisation condition L = nℏ limits orbits to specific radii, each with a specific (quantised) energy.

L = nℏ (orbit quantisation) + ΔE = hν (transition) → discrete levels

A-Level 2023

AL Q2

Derive, from Bohr's postulates, that the radius of the nth orbit in hydrogen is rₙ = n²a₀, where a₀ = 0.529 Å.

Solution

Bohr postulate: mvr = nℏ → v = nℏ/mr. Coulomb force = centripetal: ke²/r² = mv²/r → v² = ke²/mr. Substituting: (nℏ/mr)² = ke²/mr → r = n²ℏ²/mke² = n²a₀ where a₀ = ℏ²/mke² = 5.29×10⁻¹¹ m. rₙ = n²a₀ ✓

rₙ = n²ℏ²/mke² = n²a₀

A-Level 2022

AL Q3

A hydrogen atom is in the state n=3. List all possible photon energies that can be emitted as it relaxes to ground state.

Solution

From n=3: Possible transitions: 3→2 (1.89 eV), 3→1 (12.09 eV), 2→1 (10.2 eV). Three photon energies: 1.89 eV, 10.2 eV, 12.09 eV

1.89 eV (3→2), 10.2 eV (2→1), 12.09 eV (3→1)

A-Level 2022

AL Q4

Calculate the first ionisation energy of He⁺ (singly ionised helium) and state in which region of the spectrum the series limit of its Lyman series lies.

Solution

IE = 13.6 × Z² = 13.6 × 4 = 54.4 eV. Series limit photon: E = 54.4 eV → λ = hc/E = 12400/54.4 ≈ 228 Å = 22.8 nm → EUV/far UV region

54.4 eV; far UV region (22.8 nm)

A-Level 2021

AL Q5

Explain the Correspondence Principle as applied to atomic spectra.

Solution

Bohr's Correspondence Principle states that for very large quantum numbers (n → ∞), the quantum mechanical predictions must agree with classical physics. For large n, the frequency of emitted radiation (during n → n−1 transition) equals the classical orbital frequency of the electron. This validates Bohr's model in the classical limit.

For large n: quantum predictions → classical results; ν_emission → ν_orbital

A-Level 2021

AL Q6

The Balmer formula for hydrogen is 1/λ = R(1/4 − 1/n²) for n = 3, 4, 5 … Derive a formula for the Brackett series.

Solution

Brackett series: transitions to nf = 4. Using Rydberg: 1/λ = R(1/16 − 1/n²) for n = 5, 6, 7 … The first line (5→4): 1/λ = R(1/16 − 1/25) = 9R/400.

1/λ = R(1/16 − 1/n²) for n ≥ 5

A-Level 2020

AL Q7

Photons of wavelength 100 nm are incident on hydrogen atoms in the ground state. Find the kinetic energy of the emitted electrons.

Solution

E_photon = hc/λ = 12400/100 eV = 12.4 eV. IE (ground state) = 13.6 eV. Since 12.4 < 13.6 eV, the photon cannot ionise hydrogen in ground state. It can excite to n=? Check: 12.4 eV < 12.75 eV (n=4), so can excite to n=3 only (12.09 eV). Atom is excited to n=3 with no free electron.

Cannot ionise from ground state; atom excited to n=3 (absorbs 12.09 eV)

A-Level 2020

AL Q8

Calculate the shortest and longest wavelength of the Lyman series for hydrogen (R = 1.097 × 10⁷ m⁻¹).

Solution

Longest (nᵢ=2→1): 1/λ_max = R×3/4 → λ_max = 4/(3R) = 121.6 nm
Shortest (nᵢ=∞→1): 1/λ_min = R → λ_min = 1/R = 91.2 nm

Longest: 121.6 nm | Shortest: 91.2 nm

A-Level 2019

AL Q9

A hydrogen atom emits a photon of wavelength 434 nm (Hγ line). To which transition does this correspond?

Solution

1/λ = 1/434×10⁻⁹ = 2.304×10⁶ m⁻¹ = R(1/4−1/nᵢ²). 1/nᵢ² = 1/4 − 2.304×10⁶/1.097×10⁷ = 0.25 − 0.21 = 0.04 → nᵢ = 5. Transition: n=5 → n=2 (Balmer Hγ)

n = 5 → n = 2 (Balmer Hγ)

A-Level 2019

AL Q10

What is meant by the 'series limit' in atomic spectra? Why does a series converge?

Solution

The series limit is the wavelength of the photon emitted when the electron falls from n=∞ (barely ionised) to the final state nf. As n → ∞, the energy levels converge (get closer together), so the photon energies approach a maximum value and the wavelengths approach a minimum. The series limit represents complete ionisation from state nf.

Limit = min λ, from n=∞→nf; levels converge as n→∞

A-Level 2018

AL Q11

Use energy conservation to show that if an atom emitting two photons successively (n₃→n₂ and n₂→n₁), the frequency of photon from n₃→n₁ equals ν₁ + ν₂.

Solution

E₃−E₁ = (E₃−E₂) + (E₂−E₁). So h×ν(3→1) = h×ν(3→2) + h×ν(2→1). Dividing by h: ν(3→1) = ν(3→2) + ν(2→1) = ν₁ + ν₂ ✓ (Ritz combination principle)

ν₃₁ = ν₃₂ + ν₂₁ (Ritz Combination Principle)

A-Level 2018

AL Q12

Why is the potential energy of an electron in a hydrogen atom taken as negative?

Solution

The convention is that PE = 0 when the electron is at infinity (completely separated from the nucleus). Since the electron is attracted to the nucleus, work is done by the electric force as the electron approaches — meaning the system releases energy. Therefore PE at any finite distance is less than zero — it is negative. This reflects that the electron-proton system is bound.

PE=0 at infinity; electron is attracted → PE decreases (negative) as it approaches nucleus

A-Level 2017

AL Q13

Explain why hydrogen atoms in the chromosphere of the Sun produce dark lines in the solar spectrum.

Solution

The Sun's core emits a continuous spectrum. The cooler chromosphere contains hydrogen atoms that absorb photons of specific wavelengths (matching allowed transitions from ground/lower states). These absorbed wavelengths are missing from the transmitted spectrum, appearing as dark (Fraunhofer) lines when observed from Earth.

H atoms absorb specific wavelengths from continuous spectrum → Fraunhofer absorption lines

A-Level 2017

AL Q14

Calculate the de Broglie wavelength of an electron in the ground state of hydrogen and compare it to the circumference of the first Bohr orbit.

Solution

v₁ = 2.18×10⁶ m/s. λ_dB = h/mv = 6.626×10⁻³⁴/(9.11×10⁻³¹×2.18×10⁶) = 3.33×10⁻¹⁰ m. Circumference = 2πr₁ = 2π×5.29×10⁻¹¹ = 3.32×10⁻¹⁰ m. λ_dB = circumference ✓ — exactly one wavelength fits (n=1).

λ_dB = 2πr₁ ≈ 3.33 Å (one wavelength per orbit)

A-Level 2016

AL Q15

A hydrogen atom in state n=3 undergoes stimulated emission to n=2. Describe the process and calculate the wavelength of the photon involved.

Solution

Stimulated emission: an incident photon of exactly the right frequency (1.89 eV) triggers the atom to emit a second, identical photon. The incident and emitted photons are in phase (coherent). λ = 12400/1.89 = 6561 Å = 656 nm (Hα, red). This is the basis of LASER operation.

λ = 656 nm (Hα); stimulated emission = basis of LASER

🇬🇧 British Curriculum Questions (15 Questions)

British 2023

BC Q1

A student states: "All atoms of the same element produce the same spectral lines." Explain why this is true, using the concept of energy levels.

Solution

All atoms of the same element have identical nuclear structure (same proton number Z), so their electrons occupy identical sets of discrete energy levels. The energy differences between levels are identical for all atoms of the same element. Since photon frequency ν = ΔE/h, the emitted frequencies are the same for all atoms of that element.

Same Z → same energy levels → same ΔE → same frequencies

British 2023

BC Q2

Calculate the photon energy required to excite a hydrogen atom from n=1 to n=3. Give your answer in both eV and joules.

Solution

ΔE = E₃ − E₁ = −1.51 − (−13.6) = 12.09 eV = 12.09 × 1.6×10⁻¹⁹ = 12.09 eV = 1.93 × 10⁻¹⁸ J

12.09 eV = 1.93 × 10⁻¹⁸ J

British 2022

BC Q3

Explain the difference between ionisation energy and excitation energy for a hydrogen atom.

Solution

Excitation energy: Energy required to move an electron from the ground state to a higher energy level (electron remains bound to atom).
Ionisation energy: Energy required to completely remove an electron from the atom (electron escapes to infinity, n→∞, E=0). For hydrogen in ground state: excitation to n=2 needs 10.2 eV; ionisation needs 13.6 eV.

Excitation: electron remains bound (higher level); Ionisation: electron completely removed

British 2022

BC Q4

Describe an experiment that demonstrates the existence of discrete energy levels in atoms.

Solution

Franck-Hertz Experiment: Electrons are accelerated through mercury vapour by a variable voltage. Current is measured after electrons pass through the vapour. Current drops sharply at specific voltages (4.9 V multiples), showing that electrons lose specific amounts of energy (exactly 4.9 eV = first excitation energy of Hg) in collisions with atoms. This proves discrete energy levels exist.

Franck-Hertz experiment: sharp current drops at specific voltages → discrete energy levels

British 2021

BC Q5

The lines in the Balmer series become closer together towards shorter wavelengths. Explain why, using energy level diagrams.

Solution

Energy levels get closer together as n increases (Eₙ ∝ 1/n², so the gaps decrease). Balmer series involves transitions to nf=2. As nᵢ increases (from 3 to ∞), the energy gaps become progressively closer (smaller ΔE differences between successive transitions). Smaller ΔE means longer wavelength; as ΔE differences decrease, the wavelength changes become smaller — lines converge towards the series limit at 364 nm.

Energy gaps ∝ 1/n³ → decrease with n → lines converge at series limit

British 2021

BC Q6

State Bohr's quantum condition for angular momentum and explain why this leads to stable orbits.

Solution

Bohr's condition: L = mvr = nℏ (n = 1, 2, 3 …). This means only orbits where an integer number of de Broglie wavelengths fit the circumference (2πr = nλ_dB) are allowed. In these orbits, the electron wave interferes constructively with itself, producing a stable standing wave. Other orbits would cause destructive interference — the electron would not persist in them.

mvr = nℏ; stable when nλ = 2πr (constructive interference of de Broglie wave)

British 2020

BC Q7

A hydrogen atom emits a photon of wavelength 121.6 nm. To which transition does this correspond and in which region of the spectrum does it lie?

Solution

E = hc/λ = 12400/121.6 ≈ 10.2 eV. This equals E₂−E₁ = 10.2 eV → transition n=2→n=1 (Lyman α line). Region: Ultraviolet (UV)

n=2→n=1 (Lyman α); UV region

British 2020

BC Q8

Explain what is meant by the photoelectric effect and how it provides evidence for the photon model of light.

Solution

The photoelectric effect: light incident on a metal surface ejects electrons. Key observations: (1) electrons emitted only above a threshold frequency regardless of intensity; (2) KE of electrons depends on frequency, not intensity; (3) emission is instantaneous. Wave theory cannot explain (1) or (2). Einstein's photon model: light comes in packets (photons) of energy E=hν. Each photon interacts with one electron; if hν ≥ work function, electron is ejected with KE = hν − φ. This perfectly explains all observations.

Light comes in photons (E=hν); KE = hν−φ; threshold frequency → wave model fails

British 2019

BC Q9

Show that the Rydberg constant R_H = me⁴/(8ε₀²h³c) and calculate its value.

Solution

Energy levels: Eₙ = −me⁴/(8ε₀²h²n²). For transition nᵢ→nf: hν = Eₙᵢ − Eₙf = me⁴/(8ε₀²h²)(1/nf²−1/nᵢ²). Since ν = c/λ: 1/λ = me⁴/(8ε₀²h³c)(1/nf²−1/nᵢ²) = R_H(1/nf²−1/nᵢ²). So R_H = me⁴/(8ε₀²h³c) = 1.097×10⁷ m⁻¹ ✓

R_H = me⁴/(8ε₀²h³c) = 1.097 × 10⁷ m⁻¹

British 2019

BC Q10

An atom has energy levels at −12 eV, −5 eV, −2 eV, and 0 eV. Draw the energy level diagram and identify which transition emits the highest frequency photon.

Solution

Energy level diagram: four levels with above values. Transitions and energies: −2→−12 = 10 eV; −5→−12 = 7 eV; 0→−12 = 12 eV; −2→−5 = 3 eV; 0→−5 = 5 eV; 0→−2 = 2 eV. Highest energy = 12 eV (0→−12) → highest frequency. ν = 12×1.6×10⁻¹⁹/6.63×10⁻³⁴ = 2.9×10¹⁵ Hz.

0 → −12 eV (12 eV photon = highest frequency)

British 2018

BC Q11

Explain what is meant by stimulated emission in atomic physics and state one application.

Solution

Stimulated emission: An incoming photon of exactly the right energy causes an excited atom to emit a second photon that is identical (same frequency, phase, direction, and polarisation) to the incoming photon. This doubles the photon count coherently.
Application: LASER (Light Amplification by Stimulated Emission of Radiation). Lasers use population inversion + stimulated emission to produce coherent, monochromatic, collimated light.

Stimulated emission: incident photon triggers identical emitted photon; Application: LASER

British 2018

BC Q12

A student claims hydrogen atoms can emit photons of any energy. Critique this statement.

Solution

This is incorrect. Hydrogen atoms can only emit photons of specific energies corresponding to differences between allowed (quantised) energy levels: ΔE = Eₙᵢ − Eₙf where nᵢ > nf. These energies form discrete series (Lyman, Balmer, Paschen, etc.). The energy values form a specific set — not continuous. A hydrogen atom cannot emit a photon of, say, 5 eV because no transition corresponds to exactly 5 eV.

Incorrect — only specific ΔE values (between quantised levels) are allowed

British 2017

BC Q13

State Heisenberg's uncertainty principle and explain how it relates to electron orbits in Bohr's model.

Solution

Heisenberg uncertainty principle: Δx·Δp ≥ ℏ/2. Precise simultaneous knowledge of position and momentum is impossible. Bohr's model assigns electrons to precise circular orbits with definite position and momentum — violating the uncertainty principle. This is a key limitation of the Bohr model. Quantum mechanics (wavefunctions, probability clouds) correctly represents the electron's position as a probability distribution.

ΔxΔp ≥ ℏ/2; Bohr's precise orbits violate this — key limitation of Bohr model

British 2017

BC Q14

Calculate the frequency of radiation emitted when an electron in hydrogen makes a transition from n=6 to n=3 (Paschen series). (R_H = 1.097 × 10⁷ m⁻¹)

Solution

1/λ = R(1/9 − 1/36) = R × 3/36 = R/12 = 1.097×10⁷/12 = 9.14×10⁵ m⁻¹ → λ = 1094 nm. ν = c/λ = 3×10⁸/1094×10⁻⁹ = 2.74 × 10¹⁴ Hz (IR)

ν = 2.74 × 10¹⁴ Hz (IR, Paschen series)

British 2016

BC Q15

A gas discharge tube contains hydrogen at low pressure. Explain the origin of the light and why it appears as coloured lines.

Solution

Electric discharge provides energy to the hydrogen gas. Electrons collide with hydrogen atoms, transferring energy and exciting electrons to higher energy levels. The excited electrons then fall back to lower levels, emitting photons. Only certain transitions are allowed (between quantised energy levels), so only specific photon energies/frequencies are emitted. These appear as distinct coloured lines when passed through a diffraction grating — the Balmer series lines (Hα: red 656 nm, Hβ: blue-green 486 nm, etc.) are visible.

Electron collisions → excitation → de-excitation → specific photons → coloured lines

13 Assertion Reason Questions

    Instructions:

    (A) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.

    (B) Both Assertion and Reason are true but Reason is NOT the correct explanation of Assertion.

    (C) Assertion is true but Reason is false.

    (D) Both Assertion and Reason are false.

Assertion (1): The spectral lines of hydrogen are not equally spaced in wavelength.

Reason: The energy levels of hydrogen are not equally spaced (Eₙ ∝ 1/n²).

Options: (A) (B) (C) (D)

Answer: (A)

Energy gaps decrease as n increases (∝ 1/n³), so photon energies (and thus wavelengths) are not equally spaced. Reason correctly explains Assertion.

Assertion (2): The Lyman series of hydrogen lies in the UV region.

Reason: The Lyman series involves transitions to n=1, which is the ground state, resulting in high-energy (short wavelength) photons.

Options: (A) (B) (C) (D)

Answer: (A)

n=1 has the largest energy gap from higher levels. ΔE for n=2→1 is 10.2 eV → λ = 122 nm (UV). Reason correctly explains Assertion.

Assertion (3): Balmer series is the only visible series in hydrogen spectrum.

Reason: Balmer series photons have energies in the range 1.89 eV to 3.4 eV, corresponding to visible wavelengths.

Options: (A) (B) (C) (D)

Answer: (A)

Visible light: 400–700 nm (1.77–3.1 eV). Balmer series (nf=2) energies fall in this range. All other series fall outside the visible range.

Assertion (4): An electron in an atom cannot stay in an excited state indefinitely.

Reason: The excited states are unstable; the electron spontaneously returns to a lower state by emitting a photon (lifetime ~10⁻⁸ s).

Options: (A) (B) (C) (D)

Answer: (A)

Excited states are metastable at best. The average lifetime is ~10⁻⁸ s before spontaneous emission returns the electron to a lower level.

Assertion (5): The ionisation energy of H from n=2 is 3.4 eV.

Reason: Ionisation energy from state n = 13.6/n² eV.

Options: (A) (B) (C) (D)

Answer: (A)

E₂ = −3.4 eV; IE = 0 − (−3.4) = 3.4 eV. Formula: IE = 13.6/n² = 13.6/4 = 3.4 eV ✓.

Assertion (6): When an electron jumps from n=2 to n=1, it absorbs a photon of 10.2 eV.

Reason: A downward transition releases energy; energy is emitted, not absorbed.

Options: (A) (B) (C) (D)

Answer: (D)

The assertion is WRONG — n=2→n=1 is a downward (emission) transition, not absorption. The reason correctly states this. So both Assertion (as stated) is false (wrong premise) and Reason is true. Actually Assertion is false, Reason is true → (C). Correction: Assertion is FALSE (should be "emits"), Reason is TRUE → Answer: (C).

Assertion (7): The number of spectral lines emitted by hydrogen in the n=4 state is 6.

Reason: The number of spectral lines = n(n-1)/2.

Options: (A) (B) (C) (D)

Answer: (A)

N = 4(4−1)/2 = 6. Formula is correct and explains the assertion.

Assertion (8): KE of an electron in an atom increases as it moves to higher orbits.

Reason: KE = 13.6/n² eV, which decreases as n increases.

Options: (A) (B) (C) (D)

Answer: (D) → Actually (C)

Assertion is FALSE — KE decreases as n increases (KE ∝ 1/n²). Reason is TRUE. Answer: (C) — Assertion false, Reason true.

Assertion (9): The Paschen series of hydrogen lies in the infrared region.

Reason: Paschen series involves transitions to n=3, giving photons of energies 0.66 eV to 1.51 eV (wavelengths 820–1875 nm).

Options: (A) (B) (C) (D)

Answer: (A)

All Paschen lines (820 nm to ∞) are in the infrared region (IR: 700 nm to 1 mm). Reason correctly explains why.

Assertion (10): Photon absorption can only occur if the photon energy exactly matches the energy gap.

Reason: Energy levels are quantised; partial absorption of a photon is not possible.

Options: (A) (B) (C) (D)

Answer: (A)

Atoms can only absorb photons of exact energies matching the available transitions. A photon with slightly different energy cannot be partially absorbed — it passes through without interaction.

Assertion (11): The total energy of the electron in hydrogen atom is always negative.

Reason: The potential energy is more negative than the kinetic energy for a bound electron.

Options: (A) (B) (C) (D)

Answer: (A)

E = KE + PE = +ke²/2r − ke²/r = −ke²/2r < 0. Since |PE| = 2×KE, total E = −KE < 0. Reason correctly explains.

Assertion (12): Hydrogen atoms that absorb photons will subsequently emit photons of the same wavelength.

Reason: The atom returns to its ground state through the same transition that caused absorption.

Options: (A) (B) (C) (D)

Answer: (B)

The assertion is true — the same wavelength CAN be emitted. But the reason is not always correct: the atom may return via multiple steps (emitting photons of different wavelengths). Both A and R are true, but R is not the complete/correct explanation.

Assertion (13): The frequency of radiation emitted in n=3→n=1 transition equals the sum of frequencies of n=3→n=2 and n=2→n=1 transitions.

Reason: Energy is conserved: ΔE(3→1) = ΔE(3→2) + ΔE(2→1).

Options: (A) (B) (C) (D)

Answer: (A)

E(3→1) = E(3→2) + E(2→1). Since E = hν, ν(3→1) = ν(3→2) + ν(2→1). This is the Ritz Combination Principle.

Assertion (14): The radius of the electron orbit in hydrogen is proportional to n².

Reason: Bohr's model gives rₙ = n²a₀, where a₀ is the Bohr radius.

Options: (A) (B) (C) (D)

Answer: (A)

From Bohr's model: rₙ = n²ℏ²/mke² = n²a₀. Reason is correct and explains the assertion.

Assertion (15): Bohr's model successfully explains the spectrum of helium.

Reason: Bohr's model is valid only for single-electron systems.

Options: (A) (B) (C) (D)

Answer: (D)

Assertion is FALSE — Bohr's model fails for helium (2 electrons). Reason is TRUE. Answer: (C). Bohr's model works only for hydrogen and hydrogen-like ions (He⁺, Li²⁺, etc.).

Assertion (16): The energy of the electron in n=2 state of He⁺ equals the energy in n=1 state of H.

Reason: For hydrogen-like ions: E = −13.6Z²/n². For He⁺ (Z=2, n=2): E = −13.6×4/4 = −13.6 eV = E₁ of H.

Options: (A) (B) (C) (D)

Answer: (A)

Both are −13.6 eV. Reason correctly derives and explains the assertion.

Assertion (17): In Bohr's model, the angular momentum of the electron in any orbit is an integer multiple of ℏ.

Reason: This is Bohr's quantisation condition: L = nℏ = nh/2π.

Options: (A) (B) (C) (D)

Answer: (A)

Bohr's postulate directly states L = nℏ. Reason is the quantisation condition that explains the assertion.

Assertion (18): When a hydrogen atom is excited from n=1 to n=2, it emits a photon of 10.2 eV.

Reason: Excitation requires absorption of a photon, not emission.

Options: (A) (B) (C) (D)

Answer: (D) — actually (C)

Assertion is FALSE (excitation involves ABSORPTION, not emission). Reason is TRUE. Answer: (C).

Assertion (19): The shortest wavelength of the Lyman series is 91.2 nm.

Reason: It corresponds to the transition from n=∞ to n=1, giving maximum photon energy of 13.6 eV.

Options: (A) (B) (C) (D)

Answer: (A)

λ = hc/E = 12400 eV·Å / 13.6 eV = 912 Å = 91.2 nm. Reason correctly identifies the transition.

Assertion (20): The speed of the electron in hydrogen atom decreases as the quantum number n increases.

Reason: From Bohr's model: vₙ = v₁/n, so v ∝ 1/n.

Options: (A) (B) (C) (D)

Answer: (A)

v₁ = 2.18×10⁶ m/s. v₂ = 1.09×10⁶ m/s. Speed decreases as n increases. Reason gives the correct formula.

Assertion (21): Bohr's model is consistent with de Broglie's hypothesis.

Reason: Bohr's condition L = nℏ is equivalent to fitting n de Broglie wavelengths in the circumference of the orbit.

Options: (A) (B) (C) (D)

Answer: (A)

nλ_dB = 2πr → n(h/mv) = 2πr → mvr = nh/2π = nℏ, which is exactly Bohr's quantisation condition. Perfect consistency.

Assertion (22): An atom can exist in excited states for an unlimited time if not disturbed.

Reason: Excited states decay spontaneously with a mean lifetime of ~10⁻⁸ s.

Options: (A) (B) (C) (D)

Answer: (C)

Assertion is FALSE — excited states are metastable and decay spontaneously. The reason correctly states the mean lifetime. Answer: (C) — Assertion false, Reason true.

Assertion (23): The Brackett and Pfund series of hydrogen are not observable by the naked eye.

Reason: These series lie in the far infrared region (wavelengths > 1.4 μm), outside the visible range (400–700 nm).

Options: (A) (B) (C) (D)

Answer: (A)

Brackett: 1.46–4.05 μm; Pfund: 2.28–7.46 μm — both in far IR, well beyond the visible spectrum. Reason correctly explains.

Assertion (24): Increasing the temperature of a gas causes it to emit more spectral lines.

Reason: Higher temperature provides more energy, exciting electrons to higher quantum numbers, enabling more downward transitions.

Options: (A) (B) (C) (D)

Answer: (A)

At room temperature, most hydrogen atoms are in ground state (only Lyman series visible). Higher temperature populates higher levels, enabling Balmer, Paschen, etc. to appear. Reason correctly explains.

Assertion (25): Hydrogen atom can absorb photon of energy 10.2 eV but not 10.0 eV.

Reason: Only photons whose energies exactly equal the difference between energy levels can be absorbed; 10.0 eV does not match any energy difference in hydrogen.

Options: (A) (B) (C) (D)

Answer: (A)

10.2 eV exactly matches E₂−E₁. 10.0 eV does not match any transition in hydrogen, so cannot be absorbed.

14 Case Study Questions

Case Study 1: Hydrogen Atom Energy Levels

Niels Bohr proposed in 1913 that the electron in a hydrogen atom can only occupy certain allowed circular orbits without radiating energy. The energy of the electron in the nth orbit is given by Eₙ = −13.6/n² eV. When an electron transitions from a higher orbit (nᵢ) to a lower orbit (nf), it emits a photon of energy hν = Eₙᵢ − Eₙf. The ground state (n=1) has the lowest energy of −13.6 eV, and the ionisation energy (complete removal of electron) equals 13.6 eV.

Case 1, Q1

What is the energy of the electron in the 3rd orbit of hydrogen?

Solution

E₃ = −13.6/9 = −1.51 eV

−1.51 eV

Case 1, Q2

Calculate the energy required to excite a hydrogen atom from ground state to 2nd excited state.

Solution

2nd excited state = n=3. ΔE = E₃ − E₁ = −1.51 − (−13.6) = 12.09 eV

12.09 eV

Case 1, Q3

If the ionisation energy of H is 13.6 eV, find the ionisation energy for Li²⁺ (Z=3) from ground state.

Solution

IE = 13.6 × Z² = 13.6 × 9 = 122.4 eV

122.4 eV

Case 1, Q4

In which orbit is an electron if its energy in hydrogen atom is −0.85 eV?

Solution

−13.6/n² = −0.85 → n² = 16 → n = 4

n = 4

Case Study 2: Spectral Series of Hydrogen

The spectral lines of hydrogen are grouped into series based on the final orbit of the electron transition. The Rydberg formula 1/λ = R_H(1/nf² − 1/nᵢ²) gives the wavelength of each spectral line. The Lyman series (nf=1) lies in UV, Balmer series (nf=2) in visible, and Paschen series (nf=3) in infrared. The first line of each series has the longest wavelength; the series limit (nᵢ→∞) gives the shortest wavelength. R_H = 1.097 × 10⁷ m⁻¹.

Case 2, Q1

Calculate the wavelength of the first line of the Lyman series.

Solution

nf=1, nᵢ=2: 1/λ = R×(1−1/4) = 3R/4 → λ = 4/3R = 4/(3×1.097×10⁷) = 121.6 nm

121.6 nm

Case 2, Q2

In which region of the electromagnetic spectrum does the Balmer series lie and why?

Solution

Balmer series (nf=2) photon energies range from 1.89 eV (3→2) to 3.4 eV (∞→2), corresponding to wavelengths 364–656 nm. This range overlaps with visible light (400–700 nm), so the Balmer series lies primarily in the visible region.

Visible region (wavelengths 364–656 nm)

Case 2, Q3

Find the ratio of the wavelengths of the series limit of Lyman series to Balmer series.

Solution

Lyman limit (nf=1): λ_L = 1/R. Balmer limit (nf=2): λ_B = 4/R. Ratio λ_L/λ_B = 1/4. λ_L : λ_B = 1 : 4

1:4

Case 2, Q4

Explain why the Paschen series is not visible to the human eye.

Solution

Paschen series (nf=3) photon energies: 0.66 eV (4→3) to 1.51 eV (∞→3), wavelengths 820–1875 nm. Human eye can detect only 400–700 nm. Since all Paschen wavelengths exceed 820 nm, they fall in the infrared (IR) — beyond the sensitivity of human eyes.

Paschen wavelengths (820–1875 nm) exceed visible range (400–700 nm)

Case Study 3: Photon Emission in Hydrogen

A hydrogen atom in its 4th excited state (n=5) can de-excite to the ground state through various paths, emitting photons at each transition. The total energy released is always the same regardless of the path taken, as energy is conserved. Each individual photon carries energy equal to the energy difference between the two levels involved in that specific transition.

Case 3, Q1

How many different spectral lines can be emitted when a hydrogen atom in n=5 state returns to ground state?

Solution

N = n(n−1)/2 = 5×4/2 = 10 lines

10 spectral lines

Case 3, Q2

What is the total energy released when the electron goes from n=5 to n=1?

Solution

ΔE = E₅ − E₁ = −0.54 − (−13.6) = 13.06 eV

13.06 eV

Case 3, Q3

If the atom de-excites 5→3→1, calculate the energy of each photon emitted.

Solution

5→3: ΔE = −0.54−(−1.51) = 0.97 eV. 3→1: ΔE = −1.51−(−13.6) = 12.09 eV. Total = 0.97+12.09 = 13.06 eV ✓

0.97 eV + 12.09 eV = 13.06 eV total

Case 3, Q4

Which photon (emitted in the above 2-step process) falls in the Paschen series?

Solution

Paschen series: transitions to nf=3. The transition 5→3 goes to nf=3 → Paschen series. Energy = 0.97 eV, λ = 12400/0.97 ≈ 1278 nm (IR).

5→3 transition (Paschen series, 0.97 eV, 1278 nm)

Case Study 4: Absorption Spectrum

When white light (continuous spectrum) passes through hydrogen gas at room temperature, some wavelengths are absorbed. At room temperature, most hydrogen atoms are in the ground state (n=1). Therefore, the absorption spectrum of hydrogen at room temperature shows primarily Lyman series absorption lines (dark bands in the UV region corresponding to transitions from n=1 to higher levels). When the gas is heated, atoms get excited and absorption from higher levels is also seen.

Case 4, Q1

Why does the absorption spectrum of hydrogen at room temperature show only Lyman series lines?

Solution

At room temperature, most H atoms are in the ground state (n=1). Absorption can only occur when photon energy matches a transition from the current state of the electron. Since electrons are in n=1, only upward transitions from n=1 (i.e., Lyman series transitions) can be absorbed.

Electrons in n=1 (ground state) at room temp → only Lyman (n=1→higher) absorption

Case 4, Q2

Calculate the wavelength of the Lyman α absorption line.

Solution

n=1→2: ΔE = 10.2 eV. λ = hc/ΔE = 12400/10.2 Å = 1216 Å = 121.6 nm

121.6 nm (UV)

Case 4, Q3

If the hydrogen gas is heated to a high temperature, which additional spectral series might appear in absorption?

Solution

At high temperature, atoms are thermally excited to n=2, 3, etc. Absorption from n=2 gives Balmer series absorption lines (visible). From n=3 gives Paschen series (IR). The higher the temperature, the more series appear in absorption.

Balmer series (from n=2) and Paschen series (from n=3) appear additionally

Case 4, Q4

The solar absorption spectrum shows dark lines corresponding to hydrogen. What does this tell us about the solar atmosphere?

Solution

Dark Fraunhofer lines indicate that hydrogen is present in the cooler outer layers of the Sun (photosphere). These hydrogen atoms absorb specific wavelengths from the Sun's continuous spectrum. The presence of both Lyman and Balmer absorption lines tells us there is hydrogen at both low and moderately excited states in the solar atmosphere.

Hydrogen present in solar atmosphere; absorbs specific wavelengths from core's continuous emission

Case Study 5: Ionisation and Excitation

In a gas discharge tube, hydrogen atoms are bombarded by electrons accelerated through a potential difference V. When the electrons have enough kinetic energy (eV), they can either excite hydrogen atoms to higher energy levels (excitation) or completely remove the electron from the atom (ionisation). The minimum energy for excitation to the nth level from ground state is E₁ₙ = Eₙ − E₁. Ionisation requires a minimum of 13.6 eV.

Case 5, Q1

What is the minimum potential difference needed to excite hydrogen from n=1 to n=3?

Solution

ΔE = 12.09 eV. The bombarding electron must have KE ≥ 12.09 eV → V ≥ 12.09 V. Minimum V = 12.09 V

V = 12.09 V

Case 5, Q2

An electron beam of 12.5 eV bombards hydrogen in ground state. What is the highest level that can be excited? What spectral lines will be emitted?

Solution

12.5 eV < 12.75 eV (needed for n=4). Can reach n=3 (needs 12.09 eV). Highest level = n=3. Spectral lines: 3→2 (Balmer, 656.3 nm), 3→1 (Lyman, 102.5 nm), 2→1 (Lyman, 121.6 nm).

n=3 excited; 3 spectral lines (Lyman + Balmer)

Case 5, Q3

If electrons are accelerated through 14 V, will hydrogen be ionised? What will be the KE of the ejected electron?

Solution

eV = 14 eV > 13.6 eV (ionisation energy). Yes, hydrogen will be ionised. KE of ejected electron = 14 − 13.6 = 0.4 eV

Yes, ionised; KE = 0.4 eV

Case 5, Q4

A hydrogen atom in n=3 state is irradiated by UV light of wavelength 60 nm. Will it be ionised? Find the kinetic energy of the ejected electron.

Solution

E_photon = 12400/60 = 206.7 eV. IE from n=3 = 1.51 eV. Since 206.7 >> 1.51, yes, ionised. KE = 206.7 − 1.51 = 205.2 eV

Yes, ionised; KE = 205.2 eV

Case Study 6: Rydberg Formula Applications

The Rydberg formula (1/λ = R_H(1/nf² − 1/nᵢ²)) was empirically discovered before Bohr's model and was later derived from first principles. It applies to all spectral series of hydrogen. For hydrogen-like ions, the formula is modified: 1/λ = R_H × Z² × (1/nf² − 1/nᵢ²), where Z is the atomic number.

Case 6, Q1

Using the Rydberg formula, find the wave number (1/λ) of the 3rd line of the Balmer series.

Solution

3rd Balmer line: nf=2, nᵢ=5. 1/λ = R(1/4−1/25) = R×21/100 = 1.097×10⁷×0.21 = 2.304×10⁶ m⁻¹

2.304 × 10⁶ m⁻¹

Case 6, Q2

Calculate the wavelength of the first Balmer line for He⁺ (Z=2).

Solution

1/λ = R×4×(1/4−1/9) = 4R×5/36 = 5R/9 = 5×1.097×10⁷/9 = 6.09×10⁶ m⁻¹ → λ = 164 nm

164 nm

Case 6, Q3

Show that the series limit of any series of hydrogen is given by λ_limit = nf²/R_H.

Solution

At series limit nᵢ→∞: 1/λ_limit = R(1/nf² − 0) = R/nf² → λ_limit = nf²/R. For Lyman (nf=1): λ=1/R=91.2 nm. For Balmer (nf=2): λ=4/R=364.6 nm ✓

λ_limit = nf²/R_H (proven)

Case 6, Q4

For which hydrogen-like ion does the first Lyman line have the same wavelength as the first Balmer line of H?

Solution

H Balmer 1st (3→2): 1/λ_B = R×5/36. For ion X Lyman 1st (2→1): 1/λ_L = R×Z²×3/4. Setting equal: Z²×3/4 = 5/36 → Z² = 5/(36×3/4) = 5/27 (not integer). No hydrogen-like ion has this property.

No hydrogen-like ion satisfies this condition

Case Studies 7–15: Quick Reference

Case 7: de Broglie & Bohr

Bohr's condition L=nℏ is equivalent to nλ_dB=2πr. For n=2 in H: 2λ=2πr₂. This shows quantum mechanics underpins Bohr's model.

Case 8: Correspondence Principle

For large n (n→∞), quantum frequency of emission equals classical orbital frequency. This validates Bohr's model in the classical limit.

Case 9: Laser Basics

Stimulated emission: incoming photon triggers identical photon. Population inversion: more atoms in excited state than ground state. Basis of LASER operation.

Case 10: Franck-Hertz

Electrons passing through Hg vapour lose exactly 4.9 eV at threshold voltages. Proves discrete energy levels exist experimentally.

Case 11: Solar Spectrum

Fraunhofer dark lines in solar spectrum arise from absorption by cooler solar atmosphere. H, He, Ca, Na lines identified — reveals stellar composition.

Case 12: Hydrogen-like Ions

He⁺, Li²⁺, Be³⁺ etc. have E = −13.6Z²/n² eV. Their spectra are shifted to shorter wavelengths (higher Z → larger energy gaps).

Case 13: Multiple Emissions

From n=4: possible 6 transitions. Maximum spectral lines = n(n−1)/2. All transitions obey Rydberg formula.

Case 14: Photoelectric Effect

KE_max = hν − φ. Threshold frequency: ν₀ = φ/h. Proves photons are quantised — wave theory fails to explain threshold frequency and instantaneous emission.

Case 15: X-Ray Spectra

Moseley's law: √ν = a(Z−b). Characteristic X-rays arise from inner shell transitions. Minimum λ of X-rays: λ_min = hc/eV (bremsstrahlung cutoff).

Still Confused? 🎓

If you are searching for a Physics Tutor and still do not understand Energy Levels, Electron Transitions, Spectral Series, Excitation Energy, Ionisation Energy, Numericals or PYQs — contact Kumar Sir for one-to-one online Physics classes.

⭐ 30+ Years Teaching Experience FIITJEE Aakash IIT-JEE NEET AP Physics IB Physics IGCSE A-Level British Curriculum CBSE Class 12

📞 +91-9958461445 📧 kumarsirphysics@gmail.com 🌐 kumarphysicsclasses.com