Hydrogen Spectrum | Complete Notes, Numericals & PYQs | Kumar Physics Classes
Atoms · Chapter Page 03

Hydrogen
Spectrum

Study line spectrum, emission spectrum, absorption spectrum, spectral series, Rydberg formula and important numericals.

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01

Introduction to Hydrogen Spectrum

Continuous, emission and absorption spectra — foundation of atomic spectroscopy.

Continuous Spectrum

A continuous spectrum contains all wavelengths without gaps. It is produced by hot solids, liquids and dense gases at high temperature — e.g. incandescent bulb filament or sunlight through a prism (before selective absorption).

Emission Spectrum

When atoms in an excited gas de-excite, they emit photons of specific energies. The result is a bright line spectrum — coloured lines on a dark background.

Absorption Spectrum

When white light passes through cool gas, atoms absorb photons matching their transition energies. The continuous spectrum shows dark lines at those wavelengths.

Why Hydrogen Shows Line Spectrum

Hydrogen has discrete quantised energy levels (Bohr model). Electrons can only absorb or emit photons with energy exactly equal to the difference between two allowed levels — hence discrete spectral lines.

Importance in Atomic Physics: Hydrogen spectrum provided the first precise evidence for quantised atomic energy levels. Balmer's empirical formula (1885) and Rydberg's generalisation led directly to Bohr's model and modern quantum mechanics. Astronomers use hydrogen lines to identify stellar composition and radial velocity.
+ZeNucleuse⁻, mrv
Fig. 12.9 Bohr's model of hydrogen atom.
02

Line Spectrum of Hydrogen

Discrete lines from quantised electron transitions.

Definition

A line spectrum consists of sharp, discrete wavelengths (or frequencies) rather than a continuous spread. For hydrogen, each line corresponds to a unique electron transition between two energy levels.

Origin of Spectral Lines

When an electron moves from a higher orbit n₂ to a lower orbit n₁, energy is released as a photon: hν = En₂ − En₁. Because only certain orbits are allowed, only certain photon energies — and hence wavelengths — are observed.

Electron Transitions

  • Emission: electron falls from higher to lower level → bright line
  • Absorption: electron rises from lower to higher level → dark line
  • Each transition is labelled by quantum numbers (n₁, n₂)
Energyn=1n=2n=3−13.6 eV−3.4 eV−1.5 eVEmission
Electron transition from n=3 to n=2 with photon emission (ΔE = hν).
03

Emission Spectrum

Bright line spectrum from de-excitation of hydrogen atoms.

Excited Electron

Electrical discharge, heat or photon absorption raises electrons to higher energy states (n = 2, 3, 4 …).

Return to Lower Orbit

The excited state is unstable. The electron spontaneously falls to a lower permitted orbit.

Photon Emission

The energy difference appears as a photon: ΔE = hν = hc/λ.

Bright Line Spectrum

Each transition produces one spectral line. The collection of all possible transitions gives the complete emission spectrum of hydrogen.

NucleusElectronPhoton12
Fig. 12.10 Emission of a spectral line by a hydrogen atom.
04

Absorption Spectrum

Dark lines in a continuous background.

When white light (continuous spectrum) passes through hydrogen gas at low pressure:

  1. Photons with energies matching allowed transitions are absorbed
  2. Electrons are excited from lower to higher levels
  3. Those wavelengths are missing from the transmitted light
  4. Dark lines appear at exactly the same positions as emission lines
Kirchhoff's Law: An atom that emits a wavelength at a given temperature will also absorb that same wavelength at a lower temperature.
White light sourceH₂ gas tubeAbsorbedwavelengthsContinuous spectrumwith dark absorption linesAbsorptionn=1 → n=2Excited atom absorbs photon of matching energy
Absorption spectrum: dark lines appear where hydrogen gas absorbs specific wavelengths from white light.
05

Rydberg Formula

Unified expression for all hydrogen spectral lines.

1/λ = R [ 1/n₁² − 1/n₂² ]
R = Rydberg constant = 1.097 × 10⁷ m⁻¹  |  n₁ = lower (final) level  |  n₂ = higher (initial) level, n₂ > n₁

Derivation Outline

From Bohr's energy formula Eₙ = −13.6/n² eV, the photon energy for a transition is:

ΔE = 13.6 (1/n₁² − 1/n₂²) eV = hc/λ

Converting to wave number (1/λ) gives the Rydberg formula with R = me⁴/(8ε₀²h³c) ≈ 1.097×10⁷ m⁻¹.

Special Cases — Spectral Series

Lyman

n₁ = 1

Ultraviolet region. Shortest λ ≈ 91.2 nm (series limit).

Balmer

n₁ = 2

Visible region. Hα = 656 nm, Hβ = 486 nm, Hγ = 434 nm.

Paschen

n₁ = 3

Infrared region. λ > 820 nm.

Brackett

n₁ = 4

Infrared. Discovered 1922.

Pfund

n₁ = 5

Far infrared. Discovered 1924.

06

Lyman Series

Transitions to ground state — ultraviolet.

1/λ = R (1 − 1/n₂²)    where n₂ = 2, 3, 4, …

Wavelength Range

91.2 nm (series limit) to 121.6 nm (Lyman-α, n=2→1)

Region

Ultraviolet (UV). Not visible to the naked eye.

Applications

Stellar UV spectroscopy, interstellar hydrogen detection, laboratory discharge tube studies.

07

Balmer Series

Transitions to n = 2 — visible hydrogen lines.

1/λ = R (1/4 − 1/n₂²)    where n₂ = 3, 4, 5, …

n=3→2
656.3 nm
Red line

n=4→2
486.1 nm
Blue-green

n=5→2
434.0 nm
Violet

Applications

Astronomical red-shift measurement, hydrogen identification in stars, spectroscopy labs.

08

Paschen Series

Transitions to n = 3 — infrared.

1/λ = R (1/9 − 1/n₂²)    where n₂ = 4, 5, 6, …

Wavelength Range

820 nm (series limit) to 1875 nm (first line, n=4→3). All lines lie in the infrared region.

Applications

Infrared astronomy, molecular physics laboratories, semiconductor industry spectroscopy.

09

Spectral Series of Hydrogen

NCERT diagrams — circular orbits and energy level diagram.

n=1 n=2 n=3 n=4 n=5 n=6 Lyman series(ultraviolet)Balmer seriesPaschen series(infrared)Brackett seriesPfund series
Fig. 12.11 Spectral series of hydrogen atom.
−13.6−3.4−1.5−0.85−0.54−0.38−0.280n=1n=2n=3n=4n=5n=6n=7n=∞Energy (eV) →Lyman seriesBalmer seriesPaschen seriesBrackett seriesPfund seriesUltravioletVisibleInfraredInfrared
Fig. 12.12 Energy level diagram of hydrogen atom and its various spectral series.
10

Formula Sheet

All important hydrogen spectrum formulae.

Rydberg Formula

1/λ = R(1/n₁² − 1/n₂²)

Frequency Formula

ν = Rc(1/n₁² − 1/n₂²)

Photon Energy

E = hν = hc/λ

Wavelength Formula

λ = hc/ΔE = 1240/ΔE (nm)

Lyman Series

1/λ = R(1 − 1/n²)

Balmer Series

1/λ = R(1/4 − 1/n²)

Paschen Series

1/λ = R(1/9 − 1/n²)

Transition Energy

ΔE = 13.6(1/n₁² − 1/n₂²) eV

Hydrogen Energy

Eₙ = −13.6/n² eV
11

Solved Numericals

Basic, medium and advanced — wavelength, frequency, photon energy, spectral series.

Wavelength Numerical 1: Calculate wavelength of Lyman line for transition n=2 → n=1.

Given

n₂=2, n₁=1, R=1.097×10⁷ m⁻¹

Formula

1/λ = R(1/n₁² − 1/n₂²)

Solution

1/λ = 1.097×10⁷(1 − 1/4) = 8.228e+06 m⁻¹

Final Answer

λ = 121.5 nm

Exam Tip

Lyman series lies in ultraviolet region.

Wavelength Numerical 2: Calculate wavelength of Lyman line for transition n=3 → n=1.

Given

n₂=3, n₁=1, R=1.097×10⁷ m⁻¹

Formula

1/λ = R(1/n₁² − 1/n₂²)

Solution

1/λ = 1.097×10⁷(1 − 1/9) = 9.751e+06 m⁻¹

Final Answer

λ = 102.6 nm

Exam Tip

Lyman series lies in ultraviolet region.

Wavelength Numerical 3: Calculate wavelength of Lyman line for transition n=4 → n=1.

Given

n₂=4, n₁=1, R=1.097×10⁷ m⁻¹

Formula

1/λ = R(1/n₁² − 1/n₂²)

Solution

1/λ = 1.097×10⁷(1 − 1/16) = 1.028e+07 m⁻¹

Final Answer

λ = 97.2 nm

Exam Tip

Lyman series lies in ultraviolet region.

Wavelength Numerical 4: Calculate wavelength of Lyman line for transition n=5 → n=1.

Given

n₂=5, n₁=1, R=1.097×10⁷ m⁻¹

Formula

1/λ = R(1/n₁² − 1/n₂²)

Solution

1/λ = 1.097×10⁷(1 − 1/25) = 1.053e+07 m⁻¹

Final Answer

λ = 95.0 nm

Exam Tip

Lyman series lies in ultraviolet region.

Wavelength Numerical 5: Calculate wavelength of Lyman line for transition n=6 → n=1.

Given

n₂=6, n₁=1, R=1.097×10⁷ m⁻¹

Formula

1/λ = R(1/n₁² − 1/n₂²)

Solution

1/λ = 1.097×10⁷(1 − 1/36) = 1.067e+07 m⁻¹

Final Answer

λ = 93.8 nm

Exam Tip

Lyman series lies in ultraviolet region.

Wavelength Numerical 6: Calculate wavelength of Lyman line for transition n=7 → n=1.

Given

n₂=7, n₁=1, R=1.097×10⁷ m⁻¹

Formula

1/λ = R(1/n₁² − 1/n₂²)

Solution

1/λ = 1.097×10⁷(1 − 1/49) = 1.075e+07 m⁻¹

Final Answer

λ = 93.1 nm

Exam Tip

Lyman series lies in ultraviolet region.

Wavelength Numerical 7: Find wavelength for hydrogen transition n=3 → n=2.

Given

n₂=3, n₁=2

Formula

ΔE=13.6(1/n₁²−1/n₂²) eV; λ(nm)=1240/ΔE

Solution

ΔE=1.889 eV

Final Answer

λ ≈ 656.3 nm (Balmer series)

Exam Tip

This is a Balmer series line.

Wavelength Numerical 8: Find wavelength for hydrogen transition n=4 → n=2.

Given

n₂=4, n₁=2

Formula

ΔE=13.6(1/n₁²−1/n₂²) eV; λ(nm)=1240/ΔE

Solution

ΔE=2.550 eV

Final Answer

λ ≈ 486.1 nm (Balmer series)

Exam Tip

This is a Balmer series line.

Wavelength Numerical 9: Find wavelength for hydrogen transition n=5 → n=2.

Given

n₂=5, n₁=2

Formula

ΔE=13.6(1/n₁²−1/n₂²) eV; λ(nm)=1240/ΔE

Solution

ΔE=2.856 eV

Final Answer

λ ≈ 434.0 nm (Balmer series)

Exam Tip

This is a Balmer series line.

Wavelength Numerical 10: Find wavelength for hydrogen transition n=6 → n=2.

Given

n₂=6, n₁=2

Formula

ΔE=13.6(1/n₁²−1/n₂²) eV; λ(nm)=1240/ΔE

Solution

ΔE=3.022 eV

Final Answer

λ ≈ 410.2 nm (Balmer series)

Exam Tip

This is a Balmer series line.

Wavelength Numerical 11: Find wavelength for hydrogen transition n=7 → n=2.

Given

n₂=7, n₁=2

Formula

ΔE=13.6(1/n₁²−1/n₂²) eV; λ(nm)=1240/ΔE

Solution

ΔE=3.122 eV

Final Answer

λ ≈ 397.0 nm (Balmer series)

Exam Tip

This is a Balmer series line.

Wavelength Numerical 12: Find wavelength for hydrogen transition n=4 → n=3.

Given

n₂=4, n₁=3

Formula

ΔE=13.6(1/n₁²−1/n₂²) eV; λ(nm)=1240/ΔE

Solution

ΔE=0.661 eV

Final Answer

λ ≈ 1875.1 nm (Paschen series)

Exam Tip

This is a Paschen series line.

Wavelength Numerical 13: Find wavelength for hydrogen transition n=5 → n=3.

Given

n₂=5, n₁=3

Formula

ΔE=13.6(1/n₁²−1/n₂²) eV; λ(nm)=1240/ΔE

Solution

ΔE=0.967 eV

Final Answer

λ ≈ 1281.8 nm (Paschen series)

Exam Tip

This is a Paschen series line.

Wavelength Numerical 14: Find wavelength for hydrogen transition n=6 → n=3.

Given

n₂=6, n₁=3

Formula

ΔE=13.6(1/n₁²−1/n₂²) eV; λ(nm)=1240/ΔE

Solution

ΔE=1.133 eV

Final Answer

λ ≈ 1093.8 nm (Paschen series)

Exam Tip

This is a Paschen series line.

Wavelength Numerical 15: Find wavelength for hydrogen transition n=5 → n=4.

Given

n₂=5, n₁=4

Formula

ΔE=13.6(1/n₁²−1/n₂²) eV; λ(nm)=1240/ΔE

Solution

ΔE=0.306 eV

Final Answer

λ ≈ 4051.2 nm (Brackett series)

Exam Tip

This is a Brackett series line.

Wavelength Numerical 16: Find wavelength for hydrogen transition n=6 → n=4.

Given

n₂=6, n₁=4

Formula

ΔE=13.6(1/n₁²−1/n₂²) eV; λ(nm)=1240/ΔE

Solution

ΔE=0.472 eV

Final Answer

λ ≈ 2625.1 nm (Brackett series)

Exam Tip

This is a Brackett series line.

Wavelength Numerical 17: Find wavelength for hydrogen transition n=6 → n=5.

Given

n₂=6, n₁=5

Formula

ΔE=13.6(1/n₁²−1/n₂²) eV; λ(nm)=1240/ΔE

Solution

ΔE=0.166 eV

Final Answer

λ ≈ 7457.8 nm (Pfund series)

Exam Tip

This is a Pfund series line.

Photon Energy Numerical 18: Photon energy for Balmer transition n=2 → n=2.

Given

n₂=2, n₁=2

Formula

ΔE = 13.6(1/4 − 1/n²) eV

Solution

ΔE = 13.6(1/4 − 1/4)

Final Answer

0.000 eV

Exam Tip

Balmer series is in visible region for lower members.

Photon Energy Numerical 19: Photon energy for Balmer transition n=3 → n=2.

Given

n₂=3, n₁=2

Formula

ΔE = 13.6(1/4 − 1/n²) eV

Solution

ΔE = 13.6(1/4 − 1/9)

Final Answer

1.889 eV

Exam Tip

Balmer series is in visible region for lower members.

Photon Energy Numerical 20: Photon energy for Balmer transition n=4 → n=2.

Given

n₂=4, n₁=2

Formula

ΔE = 13.6(1/4 − 1/n²) eV

Solution

ΔE = 13.6(1/4 − 1/16)

Final Answer

2.550 eV

Exam Tip

Balmer series is in visible region for lower members.

Photon Energy Numerical 21: Photon energy for Balmer transition n=5 → n=2.

Given

n₂=5, n₁=2

Formula

ΔE = 13.6(1/4 − 1/n²) eV

Solution

ΔE = 13.6(1/4 − 1/25)

Final Answer

2.856 eV

Exam Tip

Balmer series is in visible region for lower members.

Photon Energy Numerical 22: Photon energy for Balmer transition n=6 → n=2.

Given

n₂=6, n₁=2

Formula

ΔE = 13.6(1/4 − 1/n²) eV

Solution

ΔE = 13.6(1/4 − 1/36)

Final Answer

3.022 eV

Exam Tip

Balmer series is in visible region for lower members.

Photon Energy Numerical 23: Photon energy for Balmer transition n=7 → n=2.

Given

n₂=7, n₁=2

Formula

ΔE = 13.6(1/4 − 1/n²) eV

Solution

ΔE = 13.6(1/4 − 1/49)

Final Answer

3.122 eV

Exam Tip

Balmer series is in visible region for lower members.

Frequency Numerical 24: Frequency of Lyman line n=2 → n=1.

Given

n₂=2, n₁=1

Formula

ν = Rc(1/n₁² − 1/n₂²)

Solution

ν = 1.097×10⁷ × 3×10⁸ × (1 − 1/4)

Final Answer

ν = 2.468e+15 Hz

Exam Tip

Use c = 3×10⁸ m/s.

Frequency Numerical 25: Frequency of Lyman line n=3 → n=1.

Given

n₂=3, n₁=1

Formula

ν = Rc(1/n₁² − 1/n₂²)

Solution

ν = 1.097×10⁷ × 3×10⁸ × (1 − 1/9)

Final Answer

ν = 2.925e+15 Hz

Exam Tip

Use c = 3×10⁸ m/s.

Frequency Numerical 26: Frequency of Lyman line n=4 → n=1.

Given

n₂=4, n₁=1

Formula

ν = Rc(1/n₁² − 1/n₂²)

Solution

ν = 1.097×10⁷ × 3×10⁸ × (1 − 1/16)

Final Answer

ν = 3.085e+15 Hz

Exam Tip

Use c = 3×10⁸ m/s.

Frequency Numerical 27: Frequency of Lyman line n=5 → n=1.

Given

n₂=5, n₁=1

Formula

ν = Rc(1/n₁² − 1/n₂²)

Solution

ν = 1.097×10⁷ × 3×10⁸ × (1 − 1/25)

Final Answer

ν = 3.159e+15 Hz

Exam Tip

Use c = 3×10⁸ m/s.

Frequency Numerical 28: Frequency of Lyman line n=6 → n=1.

Given

n₂=6, n₁=1

Formula

ν = Rc(1/n₁² − 1/n₂²)

Solution

ν = 1.097×10⁷ × 3×10⁸ × (1 − 1/36)

Final Answer

ν = 3.200e+15 Hz

Exam Tip

Use c = 3×10⁸ m/s.

Frequency Numerical 29: Frequency of Lyman line n=7 → n=1.

Given

n₂=7, n₁=1

Formula

ν = Rc(1/n₁² − 1/n₂²)

Solution

ν = 1.097×10⁷ × 3×10⁸ × (1 − 1/49)

Final Answer

ν = 3.224e+15 Hz

Exam Tip

Use c = 3×10⁸ m/s.

Spectral Series Numerical 30: Series limit wavelength for Lyman series.

Given

n₁=1, n₂=∞

Formula

λ_limit = 1240 / (13.6/n₁²) nm

Solution

λ = 1240 / 13.600

Final Answer

≈ 91.2 nm

Exam Tip

Lyman series terminates at n₁=1.

Spectral Series Numerical 31: Series limit wavelength for Balmer series.

Given

n₁=2, n₂=∞

Formula

λ_limit = 1240 / (13.6/n₁²) nm

Solution

λ = 1240 / 3.400

Final Answer

≈ 364.7 nm

Exam Tip

Balmer series terminates at n₁=2.

Spectral Series Numerical 32: Series limit wavelength for Paschen series.

Given

n₁=3, n₂=∞

Formula

λ_limit = 1240 / (13.6/n₁²) nm

Solution

λ = 1240 / 1.511

Final Answer

≈ 820.6 nm

Exam Tip

Paschen series terminates at n₁=3.

Spectral Series Numerical 33: Series limit wavelength for Brackett series.

Given

n₁=4, n₂=∞

Formula

λ_limit = 1240 / (13.6/n₁²) nm

Solution

λ = 1240 / 0.850

Final Answer

≈ 1458.8 nm

Exam Tip

Brackett series terminates at n₁=4.

Spectral Series Numerical 34: Series limit wavelength for Pfund series.

Given

n₁=5, n₂=∞

Formula

λ_limit = 1240 / (13.6/n₁²) nm

Solution

λ = 1240 / 0.544

Final Answer

≈ 2279.4 nm

Exam Tip

Pfund series terminates at n₁=5.

Hydrogen Transition Numerical 35: How many spectral lines when electron de-excites from n=7?

Given

n = 7

Formula

N = n(n−1)/2

Solution

N = 7×6/2

Final Answer

21 lines

Exam Tip

Count all possible downward transitions.

Hydrogen Transition Numerical 36: How many spectral lines when electron de-excites from n=3?

Given

n = 3

Formula

N = n(n−1)/2

Solution

N = 3×2/2

Final Answer

3 lines

Exam Tip

Count all possible downward transitions.

Hydrogen Transition Numerical 37: How many spectral lines when electron de-excites from n=4?

Given

n = 4

Formula

N = n(n−1)/2

Solution

N = 4×3/2

Final Answer

6 lines

Exam Tip

Count all possible downward transitions.

Hydrogen Transition Numerical 38: How many spectral lines when electron de-excites from n=5?

Given

n = 5

Formula

N = n(n−1)/2

Solution

N = 5×4/2

Final Answer

10 lines

Exam Tip

Count all possible downward transitions.

Hydrogen Transition Numerical 39: How many spectral lines when electron de-excites from n=6?

Given

n = 6

Formula

N = n(n−1)/2

Solution

N = 6×5/2

Final Answer

15 lines

Exam Tip

Count all possible downward transitions.

Hydrogen Transition Numerical 40: How many spectral lines when electron de-excites from n=7?

Given

n = 7

Formula

N = n(n−1)/2

Solution

N = 7×6/2

Final Answer

21 lines

Exam Tip

Count all possible downward transitions.

Hydrogen Transition Numerical 41: How many spectral lines when electron de-excites from n=7?

Given

n = 7

Formula

N = n(n−1)/2

Solution

N = 7×6/2

Final Answer

21 lines

Exam Tip

Count all possible downward transitions.

Hydrogen Transition Numerical 42: How many spectral lines when electron de-excites from n=3?

Given

n = 3

Formula

N = n(n−1)/2

Solution

N = 3×2/2

Final Answer

3 lines

Exam Tip

Count all possible downward transitions.

Hydrogen Transition Numerical 43: How many spectral lines when electron de-excites from n=4?

Given

n = 4

Formula

N = n(n−1)/2

Solution

N = 4×3/2

Final Answer

6 lines

Exam Tip

Count all possible downward transitions.

Hydrogen Transition Numerical 44: How many spectral lines when electron de-excites from n=5?

Given

n = 5

Formula

N = n(n−1)/2

Solution

N = 5×4/2

Final Answer

10 lines

Exam Tip

Count all possible downward transitions.

Hydrogen Transition Numerical 45: How many spectral lines when electron de-excites from n=6?

Given

n = 6

Formula

N = n(n−1)/2

Solution

N = 6×5/2

Final Answer

15 lines

Exam Tip

Count all possible downward transitions.

Hydrogen Transition Numerical 46: How many spectral lines when electron de-excites from n=7?

Given

n = 7

Formula

N = n(n−1)/2

Solution

N = 7×6/2

Final Answer

21 lines

Exam Tip

Count all possible downward transitions.

Hydrogen Transition Numerical 47: How many spectral lines when electron de-excites from n=7?

Given

n = 7

Formula

N = n(n−1)/2

Solution

N = 7×6/2

Final Answer

21 lines

Exam Tip

Count all possible downward transitions.

Hydrogen Transition Numerical 48: How many spectral lines when electron de-excites from n=3?

Given

n = 3

Formula

N = n(n−1)/2

Solution

N = 3×2/2

Final Answer

3 lines

Exam Tip

Count all possible downward transitions.

Hydrogen Transition Numerical 49: How many spectral lines when electron de-excites from n=4?

Given

n = 4

Formula

N = n(n−1)/2

Solution

N = 4×3/2

Final Answer

6 lines

Exam Tip

Count all possible downward transitions.

Hydrogen Transition Numerical 50: How many spectral lines when electron de-excites from n=5?

Given

n = 5

Formula

N = n(n−1)/2

Solution

N = 5×4/2

Final Answer

10 lines

Exam Tip

Count all possible downward transitions.

Hydrogen Transition Numerical 51: How many spectral lines when electron de-excites from n=6?

Given

n = 6

Formula

N = n(n−1)/2

Solution

N = 6×5/2

Final Answer

15 lines

Exam Tip

Count all possible downward transitions.

Hydrogen Transition Numerical 52: How many spectral lines when electron de-excites from n=7?

Given

n = 7

Formula

N = n(n−1)/2

Solution

N = 7×6/2

Final Answer

21 lines

Exam Tip

Count all possible downward transitions.

12

CBSE PYQs

1 mark, 2 mark, 3 mark, 5 mark and case study questions.

1. [1 mark] Define emission spectrum.

Answer / Solution

Spectrum produced when atoms emit radiation at specific wavelengths.

Exam Tip

Write series name and wavelength range in CBSE answers.

2. [1 mark] What is a line spectrum?

Answer / Solution

Spectrum consisting of discrete bright or dark lines.

Exam Tip

Write series name and wavelength range in CBSE answers.

3. [1 mark] State Rydberg constant value.

Answer / Solution

R = 1.097 × 10⁷ m⁻¹.

Exam Tip

Write series name and wavelength range in CBSE answers.

4. [1 mark] Which series is in visible region?

Answer / Solution

Balmer series.

Exam Tip

Write series name and wavelength range in CBSE answers.

5. [2 mark] Distinguish continuous and line spectrum.

Answer / Solution

Continuous: all wavelengths; Line: discrete wavelengths only.

Exam Tip

Write series name and wavelength range in CBSE answers.

6. [2 mark] Why does hydrogen show line spectrum?

Answer / Solution

Due to quantised energy levels and discrete transitions.

Exam Tip

Write series name and wavelength range in CBSE answers.

7. [3 mark] Explain emission spectrum of hydrogen.

Answer / Solution

Excited electrons fall to lower levels emitting photons of definite energy.

Exam Tip

Write series name and wavelength range in CBSE answers.

8. [3 mark] Write Rydberg formula and explain terms.

Answer / Solution

1/λ=R(1/n₁²−1/n₂²); n₁ lower, n₂ higher level.

Exam Tip

Write series name and wavelength range in CBSE answers.

9. [5 mark] Derive expression for energy of nth orbit.

Answer / Solution

From Bohr model: Eₙ=−13.6/n² eV for hydrogen.

Exam Tip

Write series name and wavelength range in CBSE answers.

10. [Case Study] A student observes Balmer lines. Identify transitions.

Answer / Solution

All transitions ending at n=2.

Exam Tip

Write series name and wavelength range in CBSE answers.

11. [1 mark] Define emission spectrum.

Answer / Solution

Spectrum produced when atoms emit radiation at specific wavelengths.

Exam Tip

Write series name and wavelength range in CBSE answers.

12. [1 mark] What is a line spectrum?

Answer / Solution

Spectrum consisting of discrete bright or dark lines.

Exam Tip

Write series name and wavelength range in CBSE answers.

13. [1 mark] State Rydberg constant value.

Answer / Solution

R = 1.097 × 10⁷ m⁻¹.

Exam Tip

Write series name and wavelength range in CBSE answers.

14. [1 mark] Which series is in visible region?

Answer / Solution

Balmer series.

Exam Tip

Write series name and wavelength range in CBSE answers.

15. [2 mark] Distinguish continuous and line spectrum.

Answer / Solution

Continuous: all wavelengths; Line: discrete wavelengths only.

Exam Tip

Write series name and wavelength range in CBSE answers.

16. [2 mark] Why does hydrogen show line spectrum?

Answer / Solution

Due to quantised energy levels and discrete transitions.

Exam Tip

Write series name and wavelength range in CBSE answers.

17. [3 mark] Explain emission spectrum of hydrogen.

Answer / Solution

Excited electrons fall to lower levels emitting photons of definite energy.

Exam Tip

Write series name and wavelength range in CBSE answers.

18. [3 mark] Write Rydberg formula and explain terms.

Answer / Solution

1/λ=R(1/n₁²−1/n₂²); n₁ lower, n₂ higher level.

Exam Tip

Write series name and wavelength range in CBSE answers.

19. [5 mark] Derive expression for energy of nth orbit.

Answer / Solution

From Bohr model: Eₙ=−13.6/n² eV for hydrogen.

Exam Tip

Write series name and wavelength range in CBSE answers.

20. [Case Study] A student observes Balmer lines. Identify transitions.

Answer / Solution

All transitions ending at n=2.

Exam Tip

Write series name and wavelength range in CBSE answers.

21. [1 mark] Define emission spectrum.

Answer / Solution

Spectrum produced when atoms emit radiation at specific wavelengths.

Exam Tip

Write series name and wavelength range in CBSE answers.

22. [1 mark] What is a line spectrum?

Answer / Solution

Spectrum consisting of discrete bright or dark lines.

Exam Tip

Write series name and wavelength range in CBSE answers.

23. [1 mark] State Rydberg constant value.

Answer / Solution

R = 1.097 × 10⁷ m⁻¹.

Exam Tip

Write series name and wavelength range in CBSE answers.

24. [1 mark] Which series is in visible region?

Answer / Solution

Balmer series.

Exam Tip

Write series name and wavelength range in CBSE answers.

25. [2 mark] Distinguish continuous and line spectrum.

Answer / Solution

Continuous: all wavelengths; Line: discrete wavelengths only.

Exam Tip

Write series name and wavelength range in CBSE answers.

26. [2 mark] Why does hydrogen show line spectrum?

Answer / Solution

Due to quantised energy levels and discrete transitions.

Exam Tip

Write series name and wavelength range in CBSE answers.

27. [3 mark] Explain emission spectrum of hydrogen.

Answer / Solution

Excited electrons fall to lower levels emitting photons of definite energy.

Exam Tip

Write series name and wavelength range in CBSE answers.

28. [3 mark] Write Rydberg formula and explain terms.

Answer / Solution

1/λ=R(1/n₁²−1/n₂²); n₁ lower, n₂ higher level.

Exam Tip

Write series name and wavelength range in CBSE answers.

29. [5 mark] Derive expression for energy of nth orbit.

Answer / Solution

From Bohr model: Eₙ=−13.6/n² eV for hydrogen.

Exam Tip

Write series name and wavelength range in CBSE answers.

30. [Case Study] A student observes Balmer lines. Identify transitions.

Answer / Solution

All transitions ending at n=2.

Exam Tip

Write series name and wavelength range in CBSE answers.

31. [1 mark] Define emission spectrum.

Answer / Solution

Spectrum produced when atoms emit radiation at specific wavelengths.

Exam Tip

Write series name and wavelength range in CBSE answers.

32. [1 mark] What is a line spectrum?

Answer / Solution

Spectrum consisting of discrete bright or dark lines.

Exam Tip

Write series name and wavelength range in CBSE answers.

33. [1 mark] State Rydberg constant value.

Answer / Solution

R = 1.097 × 10⁷ m⁻¹.

Exam Tip

Write series name and wavelength range in CBSE answers.

34. [1 mark] Which series is in visible region?

Answer / Solution

Balmer series.

Exam Tip

Write series name and wavelength range in CBSE answers.

35. [2 mark] Distinguish continuous and line spectrum.

Answer / Solution

Continuous: all wavelengths; Line: discrete wavelengths only.

Exam Tip

Write series name and wavelength range in CBSE answers.

36. [2 mark] Why does hydrogen show line spectrum?

Answer / Solution

Due to quantised energy levels and discrete transitions.

Exam Tip

Write series name and wavelength range in CBSE answers.

37. [3 mark] Explain emission spectrum of hydrogen.

Answer / Solution

Excited electrons fall to lower levels emitting photons of definite energy.

Exam Tip

Write series name and wavelength range in CBSE answers.

38. [3 mark] Write Rydberg formula and explain terms.

Answer / Solution

1/λ=R(1/n₁²−1/n₂²); n₁ lower, n₂ higher level.

Exam Tip

Write series name and wavelength range in CBSE answers.

39. [5 mark] Derive expression for energy of nth orbit.

Answer / Solution

From Bohr model: Eₙ=−13.6/n² eV for hydrogen.

Exam Tip

Write series name and wavelength range in CBSE answers.

40. [Case Study] A student observes Balmer lines. Identify transitions.

Answer / Solution

All transitions ending at n=2.

Exam Tip

Write series name and wavelength range in CBSE answers.

41. [1 mark] Define emission spectrum.

Answer / Solution

Spectrum produced when atoms emit radiation at specific wavelengths.

Exam Tip

Write series name and wavelength range in CBSE answers.

42. [1 mark] What is a line spectrum?

Answer / Solution

Spectrum consisting of discrete bright or dark lines.

Exam Tip

Write series name and wavelength range in CBSE answers.

43. [1 mark] State Rydberg constant value.

Answer / Solution

R = 1.097 × 10⁷ m⁻¹.

Exam Tip

Write series name and wavelength range in CBSE answers.

44. [1 mark] Which series is in visible region?

Answer / Solution

Balmer series.

Exam Tip

Write series name and wavelength range in CBSE answers.

45. [2 mark] Distinguish continuous and line spectrum.

Answer / Solution

Continuous: all wavelengths; Line: discrete wavelengths only.

Exam Tip

Write series name and wavelength range in CBSE answers.

46. [2 mark] Why does hydrogen show line spectrum?

Answer / Solution

Due to quantised energy levels and discrete transitions.

Exam Tip

Write series name and wavelength range in CBSE answers.

47. [3 mark] Explain emission spectrum of hydrogen.

Answer / Solution

Excited electrons fall to lower levels emitting photons of definite energy.

Exam Tip

Write series name and wavelength range in CBSE answers.

48. [3 mark] Write Rydberg formula and explain terms.

Answer / Solution

1/λ=R(1/n₁²−1/n₂²); n₁ lower, n₂ higher level.

Exam Tip

Write series name and wavelength range in CBSE answers.

49. [5 mark] Derive expression for energy of nth orbit.

Answer / Solution

From Bohr model: Eₙ=−13.6/n² eV for hydrogen.

Exam Tip

Write series name and wavelength range in CBSE answers.

50. [Case Study] A student observes Balmer lines. Identify transitions.

Answer / Solution

All transitions ending at n=2.

Exam Tip

Write series name and wavelength range in CBSE answers.

13

NEET PYQs

Last 20 years important questions — 100 with solutions.

1. Wavelength of Lyman-alpha is approximately?

Answer / Solution

121.6 nm

Exam Tip

NEET often asks series identification and wavelength.

2. Balmer series lies in?

Answer / Solution

Visible region

Exam Tip

NEET often asks series identification and wavelength.

3. Paschen series region?

Answer / Solution

Infrared

Exam Tip

NEET often asks series identification and wavelength.

4. Rydberg constant R equals?

Answer / Solution

1.097×10⁷ m⁻¹

Exam Tip

NEET often asks series identification and wavelength.

5. Energy of photon for 656 nm line?

Answer / Solution

≈1.89 eV

Exam Tip

NEET often asks series identification and wavelength.

6. Transition for Hα line?

Answer / Solution

n=3 to n=2

Exam Tip

NEET often asks series identification and wavelength.

7. Series ending at n=1?

Answer / Solution

Lyman

Exam Tip

NEET often asks series identification and wavelength.

8. Absorption spectrum shows?

Answer / Solution

Dark lines at same wavelengths as emission

Exam Tip

NEET often asks series identification and wavelength.

9. Shortest Balmer wavelength transition?

Answer / Solution

n=∞ to n=2

Exam Tip

NEET often asks series identification and wavelength.

10. Number of lines from n=4?

Answer / Solution

6

Exam Tip

NEET often asks series identification and wavelength.

11. Wavelength of Lyman-alpha is approximately?

Answer / Solution

121.6 nm

Exam Tip

NEET often asks series identification and wavelength.

12. Balmer series lies in?

Answer / Solution

Visible region

Exam Tip

NEET often asks series identification and wavelength.

13. Paschen series region?

Answer / Solution

Infrared

Exam Tip

NEET often asks series identification and wavelength.

14. Rydberg constant R equals?

Answer / Solution

1.097×10⁷ m⁻¹

Exam Tip

NEET often asks series identification and wavelength.

15. Energy of photon for 656 nm line?

Answer / Solution

≈1.89 eV

Exam Tip

NEET often asks series identification and wavelength.

16. Transition for Hα line?

Answer / Solution

n=3 to n=2

Exam Tip

NEET often asks series identification and wavelength.

17. Series ending at n=1?

Answer / Solution

Lyman

Exam Tip

NEET often asks series identification and wavelength.

18. Absorption spectrum shows?

Answer / Solution

Dark lines at same wavelengths as emission

Exam Tip

NEET often asks series identification and wavelength.

19. Shortest Balmer wavelength transition?

Answer / Solution

n=∞ to n=2

Exam Tip

NEET often asks series identification and wavelength.

20. Number of lines from n=4?

Answer / Solution

6

Exam Tip

NEET often asks series identification and wavelength.

21. Wavelength of Lyman-alpha is approximately?

Answer / Solution

121.6 nm

Exam Tip

NEET often asks series identification and wavelength.

22. Balmer series lies in?

Answer / Solution

Visible region

Exam Tip

NEET often asks series identification and wavelength.

23. Paschen series region?

Answer / Solution

Infrared

Exam Tip

NEET often asks series identification and wavelength.

24. Rydberg constant R equals?

Answer / Solution

1.097×10⁷ m⁻¹

Exam Tip

NEET often asks series identification and wavelength.

25. Energy of photon for 656 nm line?

Answer / Solution

≈1.89 eV

Exam Tip

NEET often asks series identification and wavelength.

26. Transition for Hα line?

Answer / Solution

n=3 to n=2

Exam Tip

NEET often asks series identification and wavelength.

27. Series ending at n=1?

Answer / Solution

Lyman

Exam Tip

NEET often asks series identification and wavelength.

28. Absorption spectrum shows?

Answer / Solution

Dark lines at same wavelengths as emission

Exam Tip

NEET often asks series identification and wavelength.

29. Shortest Balmer wavelength transition?

Answer / Solution

n=∞ to n=2

Exam Tip

NEET often asks series identification and wavelength.

30. Number of lines from n=4?

Answer / Solution

6

Exam Tip

NEET often asks series identification and wavelength.

31. Wavelength of Lyman-alpha is approximately?

Answer / Solution

121.6 nm

Exam Tip

NEET often asks series identification and wavelength.

32. Balmer series lies in?

Answer / Solution

Visible region

Exam Tip

NEET often asks series identification and wavelength.

33. Paschen series region?

Answer / Solution

Infrared

Exam Tip

NEET often asks series identification and wavelength.

34. Rydberg constant R equals?

Answer / Solution

1.097×10⁷ m⁻¹

Exam Tip

NEET often asks series identification and wavelength.

35. Energy of photon for 656 nm line?

Answer / Solution

≈1.89 eV

Exam Tip

NEET often asks series identification and wavelength.

36. Transition for Hα line?

Answer / Solution

n=3 to n=2

Exam Tip

NEET often asks series identification and wavelength.

37. Series ending at n=1?

Answer / Solution

Lyman

Exam Tip

NEET often asks series identification and wavelength.

38. Absorption spectrum shows?

Answer / Solution

Dark lines at same wavelengths as emission

Exam Tip

NEET often asks series identification and wavelength.

39. Shortest Balmer wavelength transition?

Answer / Solution

n=∞ to n=2

Exam Tip

NEET often asks series identification and wavelength.

40. Number of lines from n=4?

Answer / Solution

6

Exam Tip

NEET often asks series identification and wavelength.

41. Wavelength of Lyman-alpha is approximately?

Answer / Solution

121.6 nm

Exam Tip

NEET often asks series identification and wavelength.

42. Balmer series lies in?

Answer / Solution

Visible region

Exam Tip

NEET often asks series identification and wavelength.

43. Paschen series region?

Answer / Solution

Infrared

Exam Tip

NEET often asks series identification and wavelength.

44. Rydberg constant R equals?

Answer / Solution

1.097×10⁷ m⁻¹

Exam Tip

NEET often asks series identification and wavelength.

45. Energy of photon for 656 nm line?

Answer / Solution

≈1.89 eV

Exam Tip

NEET often asks series identification and wavelength.

46. Transition for Hα line?

Answer / Solution

n=3 to n=2

Exam Tip

NEET often asks series identification and wavelength.

47. Series ending at n=1?

Answer / Solution

Lyman

Exam Tip

NEET often asks series identification and wavelength.

48. Absorption spectrum shows?

Answer / Solution

Dark lines at same wavelengths as emission

Exam Tip

NEET often asks series identification and wavelength.

49. Shortest Balmer wavelength transition?

Answer / Solution

n=∞ to n=2

Exam Tip

NEET often asks series identification and wavelength.

50. Number of lines from n=4?

Answer / Solution

6

Exam Tip

NEET often asks series identification and wavelength.

51. Wavelength of Lyman-alpha is approximately?

Answer / Solution

121.6 nm

Exam Tip

NEET often asks series identification and wavelength.

52. Balmer series lies in?

Answer / Solution

Visible region

Exam Tip

NEET often asks series identification and wavelength.

53. Paschen series region?

Answer / Solution

Infrared

Exam Tip

NEET often asks series identification and wavelength.

54. Rydberg constant R equals?

Answer / Solution

1.097×10⁷ m⁻¹

Exam Tip

NEET often asks series identification and wavelength.

55. Energy of photon for 656 nm line?

Answer / Solution

≈1.89 eV

Exam Tip

NEET often asks series identification and wavelength.

56. Transition for Hα line?

Answer / Solution

n=3 to n=2

Exam Tip

NEET often asks series identification and wavelength.

57. Series ending at n=1?

Answer / Solution

Lyman

Exam Tip

NEET often asks series identification and wavelength.

58. Absorption spectrum shows?

Answer / Solution

Dark lines at same wavelengths as emission

Exam Tip

NEET often asks series identification and wavelength.

59. Shortest Balmer wavelength transition?

Answer / Solution

n=∞ to n=2

Exam Tip

NEET often asks series identification and wavelength.

60. Number of lines from n=4?

Answer / Solution

6

Exam Tip

NEET often asks series identification and wavelength.

61. Wavelength of Lyman-alpha is approximately?

Answer / Solution

121.6 nm

Exam Tip

NEET often asks series identification and wavelength.

62. Balmer series lies in?

Answer / Solution

Visible region

Exam Tip

NEET often asks series identification and wavelength.

63. Paschen series region?

Answer / Solution

Infrared

Exam Tip

NEET often asks series identification and wavelength.

64. Rydberg constant R equals?

Answer / Solution

1.097×10⁷ m⁻¹

Exam Tip

NEET often asks series identification and wavelength.

65. Energy of photon for 656 nm line?

Answer / Solution

≈1.89 eV

Exam Tip

NEET often asks series identification and wavelength.

66. Transition for Hα line?

Answer / Solution

n=3 to n=2

Exam Tip

NEET often asks series identification and wavelength.

67. Series ending at n=1?

Answer / Solution

Lyman

Exam Tip

NEET often asks series identification and wavelength.

68. Absorption spectrum shows?

Answer / Solution

Dark lines at same wavelengths as emission

Exam Tip

NEET often asks series identification and wavelength.

69. Shortest Balmer wavelength transition?

Answer / Solution

n=∞ to n=2

Exam Tip

NEET often asks series identification and wavelength.

70. Number of lines from n=4?

Answer / Solution

6

Exam Tip

NEET often asks series identification and wavelength.

71. Wavelength of Lyman-alpha is approximately?

Answer / Solution

121.6 nm

Exam Tip

NEET often asks series identification and wavelength.

72. Balmer series lies in?

Answer / Solution

Visible region

Exam Tip

NEET often asks series identification and wavelength.

73. Paschen series region?

Answer / Solution

Infrared

Exam Tip

NEET often asks series identification and wavelength.

74. Rydberg constant R equals?

Answer / Solution

1.097×10⁷ m⁻¹

Exam Tip

NEET often asks series identification and wavelength.

75. Energy of photon for 656 nm line?

Answer / Solution

≈1.89 eV

Exam Tip

NEET often asks series identification and wavelength.

76. Transition for Hα line?

Answer / Solution

n=3 to n=2

Exam Tip

NEET often asks series identification and wavelength.

77. Series ending at n=1?

Answer / Solution

Lyman

Exam Tip

NEET often asks series identification and wavelength.

78. Absorption spectrum shows?

Answer / Solution

Dark lines at same wavelengths as emission

Exam Tip

NEET often asks series identification and wavelength.

79. Shortest Balmer wavelength transition?

Answer / Solution

n=∞ to n=2

Exam Tip

NEET often asks series identification and wavelength.

80. Number of lines from n=4?

Answer / Solution

6

Exam Tip

NEET often asks series identification and wavelength.

81. Wavelength of Lyman-alpha is approximately?

Answer / Solution

121.6 nm

Exam Tip

NEET often asks series identification and wavelength.

82. Balmer series lies in?

Answer / Solution

Visible region

Exam Tip

NEET often asks series identification and wavelength.

83. Paschen series region?

Answer / Solution

Infrared

Exam Tip

NEET often asks series identification and wavelength.

84. Rydberg constant R equals?

Answer / Solution

1.097×10⁷ m⁻¹

Exam Tip

NEET often asks series identification and wavelength.

85. Energy of photon for 656 nm line?

Answer / Solution

≈1.89 eV

Exam Tip

NEET often asks series identification and wavelength.

86. Transition for Hα line?

Answer / Solution

n=3 to n=2

Exam Tip

NEET often asks series identification and wavelength.

87. Series ending at n=1?

Answer / Solution

Lyman

Exam Tip

NEET often asks series identification and wavelength.

88. Absorption spectrum shows?

Answer / Solution

Dark lines at same wavelengths as emission

Exam Tip

NEET often asks series identification and wavelength.

89. Shortest Balmer wavelength transition?

Answer / Solution

n=∞ to n=2

Exam Tip

NEET often asks series identification and wavelength.

90. Number of lines from n=4?

Answer / Solution

6

Exam Tip

NEET often asks series identification and wavelength.

91. Wavelength of Lyman-alpha is approximately?

Answer / Solution

121.6 nm

Exam Tip

NEET often asks series identification and wavelength.

92. Balmer series lies in?

Answer / Solution

Visible region

Exam Tip

NEET often asks series identification and wavelength.

93. Paschen series region?

Answer / Solution

Infrared

Exam Tip

NEET often asks series identification and wavelength.

94. Rydberg constant R equals?

Answer / Solution

1.097×10⁷ m⁻¹

Exam Tip

NEET often asks series identification and wavelength.

95. Energy of photon for 656 nm line?

Answer / Solution

≈1.89 eV

Exam Tip

NEET often asks series identification and wavelength.

96. Transition for Hα line?

Answer / Solution

n=3 to n=2

Exam Tip

NEET often asks series identification and wavelength.

97. Series ending at n=1?

Answer / Solution

Lyman

Exam Tip

NEET often asks series identification and wavelength.

98. Absorption spectrum shows?

Answer / Solution

Dark lines at same wavelengths as emission

Exam Tip

NEET often asks series identification and wavelength.

99. Shortest Balmer wavelength transition?

Answer / Solution

n=∞ to n=2

Exam Tip

NEET often asks series identification and wavelength.

100. Number of lines from n=4?

Answer / Solution

6

Exam Tip

NEET often asks series identification and wavelength.

14

JEE Main PYQs

Single correct, numerical value type and integer type.

1. [Single Correct] λ for n=4→2 transition?

Answer / Solution

486.1 nm

2. [Numerical Value] Photon energy (eV) for 121.6 nm?

Answer / Solution

10.2

3. [Integer Type] Lines from n=5 de-excitation?

Answer / Solution

10

4. [Single Correct] Series limit of Lyman in nm?

Answer / Solution

91.2

5. [Numerical Value] Rydberg constant ×10⁷?

Answer / Solution

1.097

6. [Single Correct] λ for n=4→2 transition?

Answer / Solution

486.1 nm

7. [Numerical Value] Photon energy (eV) for 121.6 nm?

Answer / Solution

10.2

8. [Integer Type] Lines from n=5 de-excitation?

Answer / Solution

10

9. [Single Correct] Series limit of Lyman in nm?

Answer / Solution

91.2

10. [Numerical Value] Rydberg constant ×10⁷?

Answer / Solution

1.097

11. [Single Correct] λ for n=4→2 transition?

Answer / Solution

486.1 nm

12. [Numerical Value] Photon energy (eV) for 121.6 nm?

Answer / Solution

10.2

13. [Integer Type] Lines from n=5 de-excitation?

Answer / Solution

10

14. [Single Correct] Series limit of Lyman in nm?

Answer / Solution

91.2

15. [Numerical Value] Rydberg constant ×10⁷?

Answer / Solution

1.097

16. [Single Correct] λ for n=4→2 transition?

Answer / Solution

486.1 nm

17. [Numerical Value] Photon energy (eV) for 121.6 nm?

Answer / Solution

10.2

18. [Integer Type] Lines from n=5 de-excitation?

Answer / Solution

10

19. [Single Correct] Series limit of Lyman in nm?

Answer / Solution

91.2

20. [Numerical Value] Rydberg constant ×10⁷?

Answer / Solution

1.097

21. [Single Correct] λ for n=4→2 transition?

Answer / Solution

486.1 nm

22. [Numerical Value] Photon energy (eV) for 121.6 nm?

Answer / Solution

10.2

23. [Integer Type] Lines from n=5 de-excitation?

Answer / Solution

10

24. [Single Correct] Series limit of Lyman in nm?

Answer / Solution

91.2

25. [Numerical Value] Rydberg constant ×10⁷?

Answer / Solution

1.097

26. [Single Correct] λ for n=4→2 transition?

Answer / Solution

486.1 nm

27. [Numerical Value] Photon energy (eV) for 121.6 nm?

Answer / Solution

10.2

28. [Integer Type] Lines from n=5 de-excitation?

Answer / Solution

10

29. [Single Correct] Series limit of Lyman in nm?

Answer / Solution

91.2

30. [Numerical Value] Rydberg constant ×10⁷?

Answer / Solution

1.097

31. [Single Correct] λ for n=4→2 transition?

Answer / Solution

486.1 nm

32. [Numerical Value] Photon energy (eV) for 121.6 nm?

Answer / Solution

10.2

33. [Integer Type] Lines from n=5 de-excitation?

Answer / Solution

10

34. [Single Correct] Series limit of Lyman in nm?

Answer / Solution

91.2

35. [Numerical Value] Rydberg constant ×10⁷?

Answer / Solution

1.097

36. [Single Correct] λ for n=4→2 transition?

Answer / Solution

486.1 nm

37. [Numerical Value] Photon energy (eV) for 121.6 nm?

Answer / Solution

10.2

38. [Integer Type] Lines from n=5 de-excitation?

Answer / Solution

10

39. [Single Correct] Series limit of Lyman in nm?

Answer / Solution

91.2

40. [Numerical Value] Rydberg constant ×10⁷?

Answer / Solution

1.097

41. [Single Correct] λ for n=4→2 transition?

Answer / Solution

486.1 nm

42. [Numerical Value] Photon energy (eV) for 121.6 nm?

Answer / Solution

10.2

43. [Integer Type] Lines from n=5 de-excitation?

Answer / Solution

10

44. [Single Correct] Series limit of Lyman in nm?

Answer / Solution

91.2

45. [Numerical Value] Rydberg constant ×10⁷?

Answer / Solution

1.097

46. [Single Correct] λ for n=4→2 transition?

Answer / Solution

486.1 nm

47. [Numerical Value] Photon energy (eV) for 121.6 nm?

Answer / Solution

10.2

48. [Integer Type] Lines from n=5 de-excitation?

Answer / Solution

10

49. [Single Correct] Series limit of Lyman in nm?

Answer / Solution

91.2

50. [Numerical Value] Rydberg constant ×10⁷?

Answer / Solution

1.097

51. [Single Correct] λ for n=4→2 transition?

Answer / Solution

486.1 nm

52. [Numerical Value] Photon energy (eV) for 121.6 nm?

Answer / Solution

10.2

53. [Integer Type] Lines from n=5 de-excitation?

Answer / Solution

10

54. [Single Correct] Series limit of Lyman in nm?

Answer / Solution

91.2

55. [Numerical Value] Rydberg constant ×10⁷?

Answer / Solution

1.097

56. [Single Correct] λ for n=4→2 transition?

Answer / Solution

486.1 nm

57. [Numerical Value] Photon energy (eV) for 121.6 nm?

Answer / Solution

10.2

58. [Integer Type] Lines from n=5 de-excitation?

Answer / Solution

10

59. [Single Correct] Series limit of Lyman in nm?

Answer / Solution

91.2

60. [Numerical Value] Rydberg constant ×10⁷?

Answer / Solution

1.097

61. [Single Correct] λ for n=4→2 transition?

Answer / Solution

486.1 nm

62. [Numerical Value] Photon energy (eV) for 121.6 nm?

Answer / Solution

10.2

63. [Integer Type] Lines from n=5 de-excitation?

Answer / Solution

10

64. [Single Correct] Series limit of Lyman in nm?

Answer / Solution

91.2

65. [Numerical Value] Rydberg constant ×10⁷?

Answer / Solution

1.097

66. [Single Correct] λ for n=4→2 transition?

Answer / Solution

486.1 nm

67. [Numerical Value] Photon energy (eV) for 121.6 nm?

Answer / Solution

10.2

68. [Integer Type] Lines from n=5 de-excitation?

Answer / Solution

10

69. [Single Correct] Series limit of Lyman in nm?

Answer / Solution

91.2

70. [Numerical Value] Rydberg constant ×10⁷?

Answer / Solution

1.097

71. [Single Correct] λ for n=4→2 transition?

Answer / Solution

486.1 nm

72. [Numerical Value] Photon energy (eV) for 121.6 nm?

Answer / Solution

10.2

73. [Integer Type] Lines from n=5 de-excitation?

Answer / Solution

10

74. [Single Correct] Series limit of Lyman in nm?

Answer / Solution

91.2

75. [Numerical Value] Rydberg constant ×10⁷?

Answer / Solution

1.097

15

JEE Advanced PYQs

50 difficult questions — single, multiple correct, numerical, matrix match.

1. [Single Correct] Ratio of energies of Lyman limit to Balmer limit?

Answer / Solution

4

2. [Multiple Correct] Which are Balmer lines?

Answer / Solution

656 nm, 486 nm, 434 nm

3. [Numerical] ν for n=3→1 in Hz order?

Answer / Solution

2.47×10¹⁵

4. [Matrix Match] Match series with n₁

Answer / Solution

Lyman→1, Balmer→2, Paschen→3

5. [Single Correct] Ratio of energies of Lyman limit to Balmer limit?

Answer / Solution

4

6. [Multiple Correct] Which are Balmer lines?

Answer / Solution

656 nm, 486 nm, 434 nm

7. [Numerical] ν for n=3→1 in Hz order?

Answer / Solution

2.47×10¹⁵

8. [Matrix Match] Match series with n₁

Answer / Solution

Lyman→1, Balmer→2, Paschen→3

9. [Single Correct] Ratio of energies of Lyman limit to Balmer limit?

Answer / Solution

4

10. [Multiple Correct] Which are Balmer lines?

Answer / Solution

656 nm, 486 nm, 434 nm

11. [Numerical] ν for n=3→1 in Hz order?

Answer / Solution

2.47×10¹⁵

12. [Matrix Match] Match series with n₁

Answer / Solution

Lyman→1, Balmer→2, Paschen→3

13. [Single Correct] Ratio of energies of Lyman limit to Balmer limit?

Answer / Solution

4

14. [Multiple Correct] Which are Balmer lines?

Answer / Solution

656 nm, 486 nm, 434 nm

15. [Numerical] ν for n=3→1 in Hz order?

Answer / Solution

2.47×10¹⁵

16. [Matrix Match] Match series with n₁

Answer / Solution

Lyman→1, Balmer→2, Paschen→3

17. [Single Correct] Ratio of energies of Lyman limit to Balmer limit?

Answer / Solution

4

18. [Multiple Correct] Which are Balmer lines?

Answer / Solution

656 nm, 486 nm, 434 nm

19. [Numerical] ν for n=3→1 in Hz order?

Answer / Solution

2.47×10¹⁵

20. [Matrix Match] Match series with n₁

Answer / Solution

Lyman→1, Balmer→2, Paschen→3

21. [Single Correct] Ratio of energies of Lyman limit to Balmer limit?

Answer / Solution

4

22. [Multiple Correct] Which are Balmer lines?

Answer / Solution

656 nm, 486 nm, 434 nm

23. [Numerical] ν for n=3→1 in Hz order?

Answer / Solution

2.47×10¹⁵

24. [Matrix Match] Match series with n₁

Answer / Solution

Lyman→1, Balmer→2, Paschen→3

25. [Single Correct] Ratio of energies of Lyman limit to Balmer limit?

Answer / Solution

4

26. [Multiple Correct] Which are Balmer lines?

Answer / Solution

656 nm, 486 nm, 434 nm

27. [Numerical] ν for n=3→1 in Hz order?

Answer / Solution

2.47×10¹⁵

28. [Matrix Match] Match series with n₁

Answer / Solution

Lyman→1, Balmer→2, Paschen→3

29. [Single Correct] Ratio of energies of Lyman limit to Balmer limit?

Answer / Solution

4

30. [Multiple Correct] Which are Balmer lines?

Answer / Solution

656 nm, 486 nm, 434 nm

31. [Numerical] ν for n=3→1 in Hz order?

Answer / Solution

2.47×10¹⁵

32. [Matrix Match] Match series with n₁

Answer / Solution

Lyman→1, Balmer→2, Paschen→3

33. [Single Correct] Ratio of energies of Lyman limit to Balmer limit?

Answer / Solution

4

34. [Multiple Correct] Which are Balmer lines?

Answer / Solution

656 nm, 486 nm, 434 nm

35. [Numerical] ν for n=3→1 in Hz order?

Answer / Solution

2.47×10¹⁵

36. [Matrix Match] Match series with n₁

Answer / Solution

Lyman→1, Balmer→2, Paschen→3

37. [Single Correct] Ratio of energies of Lyman limit to Balmer limit?

Answer / Solution

4

38. [Multiple Correct] Which are Balmer lines?

Answer / Solution

656 nm, 486 nm, 434 nm

39. [Numerical] ν for n=3→1 in Hz order?

Answer / Solution

2.47×10¹⁵

40. [Matrix Match] Match series with n₁

Answer / Solution

Lyman→1, Balmer→2, Paschen→3

41. [Single Correct] Ratio of energies of Lyman limit to Balmer limit?

Answer / Solution

4

42. [Multiple Correct] Which are Balmer lines?

Answer / Solution

656 nm, 486 nm, 434 nm

43. [Numerical] ν for n=3→1 in Hz order?

Answer / Solution

2.47×10¹⁵

44. [Matrix Match] Match series with n₁

Answer / Solution

Lyman→1, Balmer→2, Paschen→3

45. [Single Correct] Ratio of energies of Lyman limit to Balmer limit?

Answer / Solution

4

46. [Multiple Correct] Which are Balmer lines?

Answer / Solution

656 nm, 486 nm, 434 nm

47. [Numerical] ν for n=3→1 in Hz order?

Answer / Solution

2.47×10¹⁵

48. [Matrix Match] Match series with n₁

Answer / Solution

Lyman→1, Balmer→2, Paschen→3

49. [Single Correct] Ratio of energies of Lyman limit to Balmer limit?

Answer / Solution

4

50. [Multiple Correct] Which are Balmer lines?

Answer / Solution

656 nm, 486 nm, 434 nm

16

IB Physics Questions

SL, HL, structured and data-based — 30 questions.

1. [SL] Define spectral line.

Answer / Solution

Discrete wavelength from atomic transition.

2. [HL] Explain Rydberg formula.

Answer / Solution

Predicts hydrogen wavelengths from quantum numbers.

3. [Structured] Compare emission and absorption.

Answer / Solution

Emission: bright lines; Absorption: dark lines in continuous spectrum.

4. [Data Based] Given λ=656 nm, find ΔE.

Answer / Solution

≈1.89 eV

5. [SL] Define spectral line.

Answer / Solution

Discrete wavelength from atomic transition.

6. [HL] Explain Rydberg formula.

Answer / Solution

Predicts hydrogen wavelengths from quantum numbers.

7. [Structured] Compare emission and absorption.

Answer / Solution

Emission: bright lines; Absorption: dark lines in continuous spectrum.

8. [Data Based] Given λ=656 nm, find ΔE.

Answer / Solution

≈1.89 eV

9. [SL] Define spectral line.

Answer / Solution

Discrete wavelength from atomic transition.

10. [HL] Explain Rydberg formula.

Answer / Solution

Predicts hydrogen wavelengths from quantum numbers.

11. [Structured] Compare emission and absorption.

Answer / Solution

Emission: bright lines; Absorption: dark lines in continuous spectrum.

12. [Data Based] Given λ=656 nm, find ΔE.

Answer / Solution

≈1.89 eV

13. [SL] Define spectral line.

Answer / Solution

Discrete wavelength from atomic transition.

14. [HL] Explain Rydberg formula.

Answer / Solution

Predicts hydrogen wavelengths from quantum numbers.

15. [Structured] Compare emission and absorption.

Answer / Solution

Emission: bright lines; Absorption: dark lines in continuous spectrum.

16. [Data Based] Given λ=656 nm, find ΔE.

Answer / Solution

≈1.89 eV

17. [SL] Define spectral line.

Answer / Solution

Discrete wavelength from atomic transition.

18. [HL] Explain Rydberg formula.

Answer / Solution

Predicts hydrogen wavelengths from quantum numbers.

19. [Structured] Compare emission and absorption.

Answer / Solution

Emission: bright lines; Absorption: dark lines in continuous spectrum.

20. [Data Based] Given λ=656 nm, find ΔE.

Answer / Solution

≈1.89 eV

21. [SL] Define spectral line.

Answer / Solution

Discrete wavelength from atomic transition.

22. [HL] Explain Rydberg formula.

Answer / Solution

Predicts hydrogen wavelengths from quantum numbers.

23. [Structured] Compare emission and absorption.

Answer / Solution

Emission: bright lines; Absorption: dark lines in continuous spectrum.

24. [Data Based] Given λ=656 nm, find ΔE.

Answer / Solution

≈1.89 eV

25. [SL] Define spectral line.

Answer / Solution

Discrete wavelength from atomic transition.

26. [HL] Explain Rydberg formula.

Answer / Solution

Predicts hydrogen wavelengths from quantum numbers.

27. [Structured] Compare emission and absorption.

Answer / Solution

Emission: bright lines; Absorption: dark lines in continuous spectrum.

28. [Data Based] Given λ=656 nm, find ΔE.

Answer / Solution

≈1.89 eV

29. [SL] Define spectral line.

Answer / Solution

Discrete wavelength from atomic transition.

30. [HL] Explain Rydberg formula.

Answer / Solution

Predicts hydrogen wavelengths from quantum numbers.

17

IGCSE / ICSE Questions

MCQ, structured and long answer — 30 questions.

1. [MCQ] Hydrogen spectrum is?

Answer / Solution

Line spectrum

2. [Structured] Why discrete lines?

Answer / Solution

Quantised energy levels

3. [Long Answer] Explain Balmer series.

Answer / Solution

Transitions to n=2, visible lines

4. [MCQ] Hydrogen spectrum is?

Answer / Solution

Line spectrum

5. [Structured] Why discrete lines?

Answer / Solution

Quantised energy levels

6. [Long Answer] Explain Balmer series.

Answer / Solution

Transitions to n=2, visible lines

7. [MCQ] Hydrogen spectrum is?

Answer / Solution

Line spectrum

8. [Structured] Why discrete lines?

Answer / Solution

Quantised energy levels

9. [Long Answer] Explain Balmer series.

Answer / Solution

Transitions to n=2, visible lines

10. [MCQ] Hydrogen spectrum is?

Answer / Solution

Line spectrum

11. [Structured] Why discrete lines?

Answer / Solution

Quantised energy levels

12. [Long Answer] Explain Balmer series.

Answer / Solution

Transitions to n=2, visible lines

13. [MCQ] Hydrogen spectrum is?

Answer / Solution

Line spectrum

14. [Structured] Why discrete lines?

Answer / Solution

Quantised energy levels

15. [Long Answer] Explain Balmer series.

Answer / Solution

Transitions to n=2, visible lines

16. [MCQ] Hydrogen spectrum is?

Answer / Solution

Line spectrum

17. [Structured] Why discrete lines?

Answer / Solution

Quantised energy levels

18. [Long Answer] Explain Balmer series.

Answer / Solution

Transitions to n=2, visible lines

19. [MCQ] Hydrogen spectrum is?

Answer / Solution

Line spectrum

20. [Structured] Why discrete lines?

Answer / Solution

Quantised energy levels

21. [Long Answer] Explain Balmer series.

Answer / Solution

Transitions to n=2, visible lines

22. [MCQ] Hydrogen spectrum is?

Answer / Solution

Line spectrum

23. [Structured] Why discrete lines?

Answer / Solution

Quantised energy levels

24. [Long Answer] Explain Balmer series.

Answer / Solution

Transitions to n=2, visible lines

25. [MCQ] Hydrogen spectrum is?

Answer / Solution

Line spectrum

26. [Structured] Why discrete lines?

Answer / Solution

Quantised energy levels

27. [Long Answer] Explain Balmer series.

Answer / Solution

Transitions to n=2, visible lines

28. [MCQ] Hydrogen spectrum is?

Answer / Solution

Line spectrum

29. [Structured] Why discrete lines?

Answer / Solution

Quantised energy levels

30. [Long Answer] Explain Balmer series.

Answer / Solution

Transitions to n=2, visible lines

18

A-Level / British Curriculum

Cambridge and Edexcel — 40 questions.

1. [Cambridge] State Rydberg equation.

Answer / Solution

1/λ=R(1/n₁²−1/n₂²)

2. [Edexcel] Calculate λ for 2→1.

Answer / Solution

121.6 nm

3. [Cambridge] Explain absorption spectrum.

Answer / Solution

Atoms absorb photons matching transition energies.

4. [Edexcel] Name infrared series.

Answer / Solution

Paschen, Brackett, Pfund

5. [Cambridge] State Rydberg equation.

Answer / Solution

1/λ=R(1/n₁²−1/n₂²)

6. [Edexcel] Calculate λ for 2→1.

Answer / Solution

121.6 nm

7. [Cambridge] Explain absorption spectrum.

Answer / Solution

Atoms absorb photons matching transition energies.

8. [Edexcel] Name infrared series.

Answer / Solution

Paschen, Brackett, Pfund

9. [Cambridge] State Rydberg equation.

Answer / Solution

1/λ=R(1/n₁²−1/n₂²)

10. [Edexcel] Calculate λ for 2→1.

Answer / Solution

121.6 nm

11. [Cambridge] Explain absorption spectrum.

Answer / Solution

Atoms absorb photons matching transition energies.

12. [Edexcel] Name infrared series.

Answer / Solution

Paschen, Brackett, Pfund

13. [Cambridge] State Rydberg equation.

Answer / Solution

1/λ=R(1/n₁²−1/n₂²)

14. [Edexcel] Calculate λ for 2→1.

Answer / Solution

121.6 nm

15. [Cambridge] Explain absorption spectrum.

Answer / Solution

Atoms absorb photons matching transition energies.

16. [Edexcel] Name infrared series.

Answer / Solution

Paschen, Brackett, Pfund

17. [Cambridge] State Rydberg equation.

Answer / Solution

1/λ=R(1/n₁²−1/n₂²)

18. [Edexcel] Calculate λ for 2→1.

Answer / Solution

121.6 nm

19. [Cambridge] Explain absorption spectrum.

Answer / Solution

Atoms absorb photons matching transition energies.

20. [Edexcel] Name infrared series.

Answer / Solution

Paschen, Brackett, Pfund

21. [Cambridge] State Rydberg equation.

Answer / Solution

1/λ=R(1/n₁²−1/n₂²)

22. [Edexcel] Calculate λ for 2→1.

Answer / Solution

121.6 nm

23. [Cambridge] Explain absorption spectrum.

Answer / Solution

Atoms absorb photons matching transition energies.

24. [Edexcel] Name infrared series.

Answer / Solution

Paschen, Brackett, Pfund

25. [Cambridge] State Rydberg equation.

Answer / Solution

1/λ=R(1/n₁²−1/n₂²)

26. [Edexcel] Calculate λ for 2→1.

Answer / Solution

121.6 nm

27. [Cambridge] Explain absorption spectrum.

Answer / Solution

Atoms absorb photons matching transition energies.

28. [Edexcel] Name infrared series.

Answer / Solution

Paschen, Brackett, Pfund

29. [Cambridge] State Rydberg equation.

Answer / Solution

1/λ=R(1/n₁²−1/n₂²)

30. [Edexcel] Calculate λ for 2→1.

Answer / Solution

121.6 nm

31. [Cambridge] Explain absorption spectrum.

Answer / Solution

Atoms absorb photons matching transition energies.

32. [Edexcel] Name infrared series.

Answer / Solution

Paschen, Brackett, Pfund

33. [Cambridge] State Rydberg equation.

Answer / Solution

1/λ=R(1/n₁²−1/n₂²)

34. [Edexcel] Calculate λ for 2→1.

Answer / Solution

121.6 nm

35. [Cambridge] Explain absorption spectrum.

Answer / Solution

Atoms absorb photons matching transition energies.

36. [Edexcel] Name infrared series.

Answer / Solution

Paschen, Brackett, Pfund

37. [Cambridge] State Rydberg equation.

Answer / Solution

1/λ=R(1/n₁²−1/n₂²)

38. [Edexcel] Calculate λ for 2→1.

Answer / Solution

121.6 nm

39. [Cambridge] Explain absorption spectrum.

Answer / Solution

Atoms absorb photons matching transition energies.

40. [Edexcel] Name infrared series.

Answer / Solution

Paschen, Brackett, Pfund

19

Assertion Reason Questions

50 assertion-reason questions with explanations.

1. Assertion: Hydrogen produces line spectrum. Reason: Energy levels are discrete.

Correct Option

Both true; Reason explains Assertion.

Explanation

The assertion concerns hydrogen spectral behaviour; the reason links to quantised atomic structure and the Rydberg relation.

2. Assertion: Balmer series is visible. Reason: Transitions end at n=2.

Correct Option

Both true; Reason explains Assertion.

Explanation

The assertion concerns hydrogen spectral behaviour; the reason links to quantised atomic structure and the Rydberg relation.

3. Assertion: Lyman series is ultraviolet. Reason: Transitions end at n=1 with large ΔE.

Correct Option

Both true; Reason explains Assertion.

Explanation

The assertion concerns hydrogen spectral behaviour; the reason links to quantised atomic structure and the Rydberg relation.

4. Assertion: Absorption lines coincide with emission lines. Reason: Same transitions involve same energy difference.

Correct Option

Both true; Reason explains Assertion.

Explanation

The assertion concerns hydrogen spectral behaviour; the reason links to quantised atomic structure and the Rydberg relation.

5. Assertion: Rydberg formula applies to all atoms. Reason: It was derived for hydrogen.

Correct Option

Assertion false; Reason true.

Explanation

The assertion concerns hydrogen spectral behaviour; the reason links to quantised atomic structure and the Rydberg relation.

6. Assertion: Continuous spectrum has all wavelengths. Reason: It is produced by hot solids.

Correct Option

Both true; Reason does not fully explain Assertion.

Explanation

The assertion concerns hydrogen spectral behaviour; the reason links to quantised atomic structure and the Rydberg relation.

7. Assertion: Paschen series is infrared. Reason: Transitions to n=3 have smaller energy gaps.

Correct Option

Both true; Reason explains Assertion.

Explanation

The assertion concerns hydrogen spectral behaviour; the reason links to quantised atomic structure and the Rydberg relation.

8. Assertion: Photon energy equals energy difference. Reason: hν=|E₂−E₁|.

Correct Option

Both true; Reason explains Assertion.

Explanation

The assertion concerns hydrogen spectral behaviour; the reason links to quantised atomic structure and the Rydberg relation.

9. Assertion: Hydrogen produces line spectrum. Reason: Energy levels are discrete.

Correct Option

Both true; Reason explains Assertion.

Explanation

The assertion concerns hydrogen spectral behaviour; the reason links to quantised atomic structure and the Rydberg relation.

10. Assertion: Balmer series is visible. Reason: Transitions end at n=2.

Correct Option

Both true; Reason explains Assertion.

Explanation

The assertion concerns hydrogen spectral behaviour; the reason links to quantised atomic structure and the Rydberg relation.

11. Assertion: Lyman series is ultraviolet. Reason: Transitions end at n=1 with large ΔE.

Correct Option

Both true; Reason explains Assertion.

Explanation

The assertion concerns hydrogen spectral behaviour; the reason links to quantised atomic structure and the Rydberg relation.

12. Assertion: Absorption lines coincide with emission lines. Reason: Same transitions involve same energy difference.

Correct Option

Both true; Reason explains Assertion.

Explanation

The assertion concerns hydrogen spectral behaviour; the reason links to quantised atomic structure and the Rydberg relation.

13. Assertion: Rydberg formula applies to all atoms. Reason: It was derived for hydrogen.

Correct Option

Assertion false; Reason true.

Explanation

The assertion concerns hydrogen spectral behaviour; the reason links to quantised atomic structure and the Rydberg relation.

14. Assertion: Continuous spectrum has all wavelengths. Reason: It is produced by hot solids.

Correct Option

Both true; Reason does not fully explain Assertion.

Explanation

The assertion concerns hydrogen spectral behaviour; the reason links to quantised atomic structure and the Rydberg relation.

15. Assertion: Paschen series is infrared. Reason: Transitions to n=3 have smaller energy gaps.

Correct Option

Both true; Reason explains Assertion.

Explanation

The assertion concerns hydrogen spectral behaviour; the reason links to quantised atomic structure and the Rydberg relation.

16. Assertion: Photon energy equals energy difference. Reason: hν=|E₂−E₁|.

Correct Option

Both true; Reason explains Assertion.

Explanation

The assertion concerns hydrogen spectral behaviour; the reason links to quantised atomic structure and the Rydberg relation.

17. Assertion: Hydrogen produces line spectrum. Reason: Energy levels are discrete.

Correct Option

Both true; Reason explains Assertion.

Explanation

The assertion concerns hydrogen spectral behaviour; the reason links to quantised atomic structure and the Rydberg relation.

18. Assertion: Balmer series is visible. Reason: Transitions end at n=2.

Correct Option

Both true; Reason explains Assertion.

Explanation

The assertion concerns hydrogen spectral behaviour; the reason links to quantised atomic structure and the Rydberg relation.

19. Assertion: Lyman series is ultraviolet. Reason: Transitions end at n=1 with large ΔE.

Correct Option

Both true; Reason explains Assertion.

Explanation

The assertion concerns hydrogen spectral behaviour; the reason links to quantised atomic structure and the Rydberg relation.

20. Assertion: Absorption lines coincide with emission lines. Reason: Same transitions involve same energy difference.

Correct Option

Both true; Reason explains Assertion.

Explanation

The assertion concerns hydrogen spectral behaviour; the reason links to quantised atomic structure and the Rydberg relation.

21. Assertion: Rydberg formula applies to all atoms. Reason: It was derived for hydrogen.

Correct Option

Assertion false; Reason true.

Explanation

The assertion concerns hydrogen spectral behaviour; the reason links to quantised atomic structure and the Rydberg relation.

22. Assertion: Continuous spectrum has all wavelengths. Reason: It is produced by hot solids.

Correct Option

Both true; Reason does not fully explain Assertion.

Explanation

The assertion concerns hydrogen spectral behaviour; the reason links to quantised atomic structure and the Rydberg relation.

23. Assertion: Paschen series is infrared. Reason: Transitions to n=3 have smaller energy gaps.

Correct Option

Both true; Reason explains Assertion.

Explanation

The assertion concerns hydrogen spectral behaviour; the reason links to quantised atomic structure and the Rydberg relation.

24. Assertion: Photon energy equals energy difference. Reason: hν=|E₂−E₁|.

Correct Option

Both true; Reason explains Assertion.

Explanation

The assertion concerns hydrogen spectral behaviour; the reason links to quantised atomic structure and the Rydberg relation.

25. Assertion: Hydrogen produces line spectrum. Reason: Energy levels are discrete.

Correct Option

Both true; Reason explains Assertion.

Explanation

The assertion concerns hydrogen spectral behaviour; the reason links to quantised atomic structure and the Rydberg relation.

26. Assertion: Balmer series is visible. Reason: Transitions end at n=2.

Correct Option

Both true; Reason explains Assertion.

Explanation

The assertion concerns hydrogen spectral behaviour; the reason links to quantised atomic structure and the Rydberg relation.

27. Assertion: Lyman series is ultraviolet. Reason: Transitions end at n=1 with large ΔE.

Correct Option

Both true; Reason explains Assertion.

Explanation

The assertion concerns hydrogen spectral behaviour; the reason links to quantised atomic structure and the Rydberg relation.

28. Assertion: Absorption lines coincide with emission lines. Reason: Same transitions involve same energy difference.

Correct Option

Both true; Reason explains Assertion.

Explanation

The assertion concerns hydrogen spectral behaviour; the reason links to quantised atomic structure and the Rydberg relation.

29. Assertion: Rydberg formula applies to all atoms. Reason: It was derived for hydrogen.

Correct Option

Assertion false; Reason true.

Explanation

The assertion concerns hydrogen spectral behaviour; the reason links to quantised atomic structure and the Rydberg relation.

30. Assertion: Continuous spectrum has all wavelengths. Reason: It is produced by hot solids.

Correct Option

Both true; Reason does not fully explain Assertion.

Explanation

The assertion concerns hydrogen spectral behaviour; the reason links to quantised atomic structure and the Rydberg relation.

31. Assertion: Paschen series is infrared. Reason: Transitions to n=3 have smaller energy gaps.

Correct Option

Both true; Reason explains Assertion.

Explanation

The assertion concerns hydrogen spectral behaviour; the reason links to quantised atomic structure and the Rydberg relation.

32. Assertion: Photon energy equals energy difference. Reason: hν=|E₂−E₁|.

Correct Option

Both true; Reason explains Assertion.

Explanation

The assertion concerns hydrogen spectral behaviour; the reason links to quantised atomic structure and the Rydberg relation.

33. Assertion: Hydrogen produces line spectrum. Reason: Energy levels are discrete.

Correct Option

Both true; Reason explains Assertion.

Explanation

The assertion concerns hydrogen spectral behaviour; the reason links to quantised atomic structure and the Rydberg relation.

34. Assertion: Balmer series is visible. Reason: Transitions end at n=2.

Correct Option

Both true; Reason explains Assertion.

Explanation

The assertion concerns hydrogen spectral behaviour; the reason links to quantised atomic structure and the Rydberg relation.

35. Assertion: Lyman series is ultraviolet. Reason: Transitions end at n=1 with large ΔE.

Correct Option

Both true; Reason explains Assertion.

Explanation

The assertion concerns hydrogen spectral behaviour; the reason links to quantised atomic structure and the Rydberg relation.

36. Assertion: Absorption lines coincide with emission lines. Reason: Same transitions involve same energy difference.

Correct Option

Both true; Reason explains Assertion.

Explanation

The assertion concerns hydrogen spectral behaviour; the reason links to quantised atomic structure and the Rydberg relation.

37. Assertion: Rydberg formula applies to all atoms. Reason: It was derived for hydrogen.

Correct Option

Assertion false; Reason true.

Explanation

The assertion concerns hydrogen spectral behaviour; the reason links to quantised atomic structure and the Rydberg relation.

38. Assertion: Continuous spectrum has all wavelengths. Reason: It is produced by hot solids.

Correct Option

Both true; Reason does not fully explain Assertion.

Explanation

The assertion concerns hydrogen spectral behaviour; the reason links to quantised atomic structure and the Rydberg relation.

39. Assertion: Paschen series is infrared. Reason: Transitions to n=3 have smaller energy gaps.

Correct Option

Both true; Reason explains Assertion.

Explanation

The assertion concerns hydrogen spectral behaviour; the reason links to quantised atomic structure and the Rydberg relation.

40. Assertion: Photon energy equals energy difference. Reason: hν=|E₂−E₁|.

Correct Option

Both true; Reason explains Assertion.

Explanation

The assertion concerns hydrogen spectral behaviour; the reason links to quantised atomic structure and the Rydberg relation.

41. Assertion: Hydrogen produces line spectrum. Reason: Energy levels are discrete.

Correct Option

Both true; Reason explains Assertion.

Explanation

The assertion concerns hydrogen spectral behaviour; the reason links to quantised atomic structure and the Rydberg relation.

42. Assertion: Balmer series is visible. Reason: Transitions end at n=2.

Correct Option

Both true; Reason explains Assertion.

Explanation

The assertion concerns hydrogen spectral behaviour; the reason links to quantised atomic structure and the Rydberg relation.

43. Assertion: Lyman series is ultraviolet. Reason: Transitions end at n=1 with large ΔE.

Correct Option

Both true; Reason explains Assertion.

Explanation

The assertion concerns hydrogen spectral behaviour; the reason links to quantised atomic structure and the Rydberg relation.

44. Assertion: Absorption lines coincide with emission lines. Reason: Same transitions involve same energy difference.

Correct Option

Both true; Reason explains Assertion.

Explanation

The assertion concerns hydrogen spectral behaviour; the reason links to quantised atomic structure and the Rydberg relation.

45. Assertion: Rydberg formula applies to all atoms. Reason: It was derived for hydrogen.

Correct Option

Assertion false; Reason true.

Explanation

The assertion concerns hydrogen spectral behaviour; the reason links to quantised atomic structure and the Rydberg relation.

46. Assertion: Continuous spectrum has all wavelengths. Reason: It is produced by hot solids.

Correct Option

Both true; Reason does not fully explain Assertion.

Explanation

The assertion concerns hydrogen spectral behaviour; the reason links to quantised atomic structure and the Rydberg relation.

47. Assertion: Paschen series is infrared. Reason: Transitions to n=3 have smaller energy gaps.

Correct Option

Both true; Reason explains Assertion.

Explanation

The assertion concerns hydrogen spectral behaviour; the reason links to quantised atomic structure and the Rydberg relation.

48. Assertion: Photon energy equals energy difference. Reason: hν=|E₂−E₁|.

Correct Option

Both true; Reason explains Assertion.

Explanation

The assertion concerns hydrogen spectral behaviour; the reason links to quantised atomic structure and the Rydberg relation.

49. Assertion: Hydrogen produces line spectrum. Reason: Energy levels are discrete.

Correct Option

Both true; Reason explains Assertion.

Explanation

The assertion concerns hydrogen spectral behaviour; the reason links to quantised atomic structure and the Rydberg relation.

50. Assertion: Balmer series is visible. Reason: Transitions end at n=2.

Correct Option

Both true; Reason explains Assertion.

Explanation

The assertion concerns hydrogen spectral behaviour; the reason links to quantised atomic structure and the Rydberg relation.

20

Case Study Questions

25 case studies — 4 questions each with detailed solutions.

Case Study 1: Hydrogen Spectrum in discharge tube

Passage: A physics laboratory uses a hydrogen discharge tube and a spectrometer. When the gas is excited electrically, discrete coloured lines appear. Astronomers use the same principle to identify hydrogen in distant stars by matching observed wavelengths with the Rydberg formula 1/λ = R(1/n₁² − 1/n₂²).

Q1. Name the series for transitions ending at n=2. Answer: Balmer series.

Q2. Why are the lines discrete? Answer: Because only certain energy differences are allowed between quantised levels.

Q3. Calculate approximate wavelength for n=3→2. Answer: 656 nm (Hα line).

Q4. What region does the Lyman series lie in? Answer: Ultraviolet region.

Case Study 2: Balmer Series in astronomy

Passage: A physics laboratory uses a hydrogen discharge tube and a spectrometer. When the gas is excited electrically, discrete coloured lines appear. Astronomers use the same principle to identify hydrogen in distant stars by matching observed wavelengths with the Rydberg formula 1/λ = R(1/n₁² − 1/n₂²).

Q1. Name the series for transitions ending at n=2. Answer: Balmer series.

Q2. Why are the lines discrete? Answer: Because only certain energy differences are allowed between quantised levels.

Q3. Calculate approximate wavelength for n=3→2. Answer: 656 nm (Hα line).

Q4. What region does the Lyman series lie in? Answer: Ultraviolet region.

Case Study 3: Rydberg Formula application

Passage: A physics laboratory uses a hydrogen discharge tube and a spectrometer. When the gas is excited electrically, discrete coloured lines appear. Astronomers use the same principle to identify hydrogen in distant stars by matching observed wavelengths with the Rydberg formula 1/λ = R(1/n₁² − 1/n₂²).

Q1. Name the series for transitions ending at n=2. Answer: Balmer series.

Q2. Why are the lines discrete? Answer: Because only certain energy differences are allowed between quantised levels.

Q3. Calculate approximate wavelength for n=3→2. Answer: 656 nm (Hα line).

Q4. What region does the Lyman series lie in? Answer: Ultraviolet region.

Case Study 4: Lyman Series in UV spectroscopy

Passage: A physics laboratory uses a hydrogen discharge tube and a spectrometer. When the gas is excited electrically, discrete coloured lines appear. Astronomers use the same principle to identify hydrogen in distant stars by matching observed wavelengths with the Rydberg formula 1/λ = R(1/n₁² − 1/n₂²).

Q1. Name the series for transitions ending at n=2. Answer: Balmer series.

Q2. Why are the lines discrete? Answer: Because only certain energy differences are allowed between quantised levels.

Q3. Calculate approximate wavelength for n=3→2. Answer: 656 nm (Hα line).

Q4. What region does the Lyman series lie in? Answer: Ultraviolet region.

Case Study 5: Absorption in stellar atmospheres

Passage: A physics laboratory uses a hydrogen discharge tube and a spectrometer. When the gas is excited electrically, discrete coloured lines appear. Astronomers use the same principle to identify hydrogen in distant stars by matching observed wavelengths with the Rydberg formula 1/λ = R(1/n₁² − 1/n₂²).

Q1. Name the series for transitions ending at n=2. Answer: Balmer series.

Q2. Why are the lines discrete? Answer: Because only certain energy differences are allowed between quantised levels.

Q3. Calculate approximate wavelength for n=3→2. Answer: 656 nm (Hα line).

Q4. What region does the Lyman series lie in? Answer: Ultraviolet region.

Case Study 6: Spectral line identification

Passage: A physics laboratory uses a hydrogen discharge tube and a spectrometer. When the gas is excited electrically, discrete coloured lines appear. Astronomers use the same principle to identify hydrogen in distant stars by matching observed wavelengths with the Rydberg formula 1/λ = R(1/n₁² − 1/n₂²).

Q1. Name the series for transitions ending at n=2. Answer: Balmer series.

Q2. Why are the lines discrete? Answer: Because only certain energy differences are allowed between quantised levels.

Q3. Calculate approximate wavelength for n=3→2. Answer: 656 nm (Hα line).

Q4. What region does the Lyman series lie in? Answer: Ultraviolet region.

Case Study 7: Energy level transitions

Passage: A physics laboratory uses a hydrogen discharge tube and a spectrometer. When the gas is excited electrically, discrete coloured lines appear. Astronomers use the same principle to identify hydrogen in distant stars by matching observed wavelengths with the Rydberg formula 1/λ = R(1/n₁² − 1/n₂²).

Q1. Name the series for transitions ending at n=2. Answer: Balmer series.

Q2. Why are the lines discrete? Answer: Because only certain energy differences are allowed between quantised levels.

Q3. Calculate approximate wavelength for n=3→2. Answer: 656 nm (Hα line).

Q4. What region does the Lyman series lie in? Answer: Ultraviolet region.

Case Study 8: Hα red line in hydrogen

Passage: A physics laboratory uses a hydrogen discharge tube and a spectrometer. When the gas is excited electrically, discrete coloured lines appear. Astronomers use the same principle to identify hydrogen in distant stars by matching observed wavelengths with the Rydberg formula 1/λ = R(1/n₁² − 1/n₂²).

Q1. Name the series for transitions ending at n=2. Answer: Balmer series.

Q2. Why are the lines discrete? Answer: Because only certain energy differences are allowed between quantised levels.

Q3. Calculate approximate wavelength for n=3→2. Answer: 656 nm (Hα line).

Q4. What region does the Lyman series lie in? Answer: Ultraviolet region.

Case Study 9: Series limit calculation

Passage: A physics laboratory uses a hydrogen discharge tube and a spectrometer. When the gas is excited electrically, discrete coloured lines appear. Astronomers use the same principle to identify hydrogen in distant stars by matching observed wavelengths with the Rydberg formula 1/λ = R(1/n₁² − 1/n₂²).

Q1. Name the series for transitions ending at n=2. Answer: Balmer series.

Q2. Why are the lines discrete? Answer: Because only certain energy differences are allowed between quantised levels.

Q3. Calculate approximate wavelength for n=3→2. Answer: 656 nm (Hα line).

Q4. What region does the Lyman series lie in? Answer: Ultraviolet region.

Case Study 10: NEET spectral series

Passage: A physics laboratory uses a hydrogen discharge tube and a spectrometer. When the gas is excited electrically, discrete coloured lines appear. Astronomers use the same principle to identify hydrogen in distant stars by matching observed wavelengths with the Rydberg formula 1/λ = R(1/n₁² − 1/n₂²).

Q1. Name the series for transitions ending at n=2. Answer: Balmer series.

Q2. Why are the lines discrete? Answer: Because only certain energy differences are allowed between quantised levels.

Q3. Calculate approximate wavelength for n=3→2. Answer: 656 nm (Hα line).

Q4. What region does the Lyman series lie in? Answer: Ultraviolet region.

Case Study 11: Hydrogen Spectrum in discharge tube

Passage: A physics laboratory uses a hydrogen discharge tube and a spectrometer. When the gas is excited electrically, discrete coloured lines appear. Astronomers use the same principle to identify hydrogen in distant stars by matching observed wavelengths with the Rydberg formula 1/λ = R(1/n₁² − 1/n₂²).

Q1. Name the series for transitions ending at n=2. Answer: Balmer series.

Q2. Why are the lines discrete? Answer: Because only certain energy differences are allowed between quantised levels.

Q3. Calculate approximate wavelength for n=3→2. Answer: 656 nm (Hα line).

Q4. What region does the Lyman series lie in? Answer: Ultraviolet region.

Case Study 12: Balmer Series in astronomy

Passage: A physics laboratory uses a hydrogen discharge tube and a spectrometer. When the gas is excited electrically, discrete coloured lines appear. Astronomers use the same principle to identify hydrogen in distant stars by matching observed wavelengths with the Rydberg formula 1/λ = R(1/n₁² − 1/n₂²).

Q1. Name the series for transitions ending at n=2. Answer: Balmer series.

Q2. Why are the lines discrete? Answer: Because only certain energy differences are allowed between quantised levels.

Q3. Calculate approximate wavelength for n=3→2. Answer: 656 nm (Hα line).

Q4. What region does the Lyman series lie in? Answer: Ultraviolet region.

Case Study 13: Rydberg Formula application

Passage: A physics laboratory uses a hydrogen discharge tube and a spectrometer. When the gas is excited electrically, discrete coloured lines appear. Astronomers use the same principle to identify hydrogen in distant stars by matching observed wavelengths with the Rydberg formula 1/λ = R(1/n₁² − 1/n₂²).

Q1. Name the series for transitions ending at n=2. Answer: Balmer series.

Q2. Why are the lines discrete? Answer: Because only certain energy differences are allowed between quantised levels.

Q3. Calculate approximate wavelength for n=3→2. Answer: 656 nm (Hα line).

Q4. What region does the Lyman series lie in? Answer: Ultraviolet region.

Case Study 14: Lyman Series in UV spectroscopy

Passage: A physics laboratory uses a hydrogen discharge tube and a spectrometer. When the gas is excited electrically, discrete coloured lines appear. Astronomers use the same principle to identify hydrogen in distant stars by matching observed wavelengths with the Rydberg formula 1/λ = R(1/n₁² − 1/n₂²).

Q1. Name the series for transitions ending at n=2. Answer: Balmer series.

Q2. Why are the lines discrete? Answer: Because only certain energy differences are allowed between quantised levels.

Q3. Calculate approximate wavelength for n=3→2. Answer: 656 nm (Hα line).

Q4. What region does the Lyman series lie in? Answer: Ultraviolet region.

Case Study 15: Absorption in stellar atmospheres

Passage: A physics laboratory uses a hydrogen discharge tube and a spectrometer. When the gas is excited electrically, discrete coloured lines appear. Astronomers use the same principle to identify hydrogen in distant stars by matching observed wavelengths with the Rydberg formula 1/λ = R(1/n₁² − 1/n₂²).

Q1. Name the series for transitions ending at n=2. Answer: Balmer series.

Q2. Why are the lines discrete? Answer: Because only certain energy differences are allowed between quantised levels.

Q3. Calculate approximate wavelength for n=3→2. Answer: 656 nm (Hα line).

Q4. What region does the Lyman series lie in? Answer: Ultraviolet region.

Case Study 16: Spectral line identification

Passage: A physics laboratory uses a hydrogen discharge tube and a spectrometer. When the gas is excited electrically, discrete coloured lines appear. Astronomers use the same principle to identify hydrogen in distant stars by matching observed wavelengths with the Rydberg formula 1/λ = R(1/n₁² − 1/n₂²).

Q1. Name the series for transitions ending at n=2. Answer: Balmer series.

Q2. Why are the lines discrete? Answer: Because only certain energy differences are allowed between quantised levels.

Q3. Calculate approximate wavelength for n=3→2. Answer: 656 nm (Hα line).

Q4. What region does the Lyman series lie in? Answer: Ultraviolet region.

Case Study 17: Energy level transitions

Passage: A physics laboratory uses a hydrogen discharge tube and a spectrometer. When the gas is excited electrically, discrete coloured lines appear. Astronomers use the same principle to identify hydrogen in distant stars by matching observed wavelengths with the Rydberg formula 1/λ = R(1/n₁² − 1/n₂²).

Q1. Name the series for transitions ending at n=2. Answer: Balmer series.

Q2. Why are the lines discrete? Answer: Because only certain energy differences are allowed between quantised levels.

Q3. Calculate approximate wavelength for n=3→2. Answer: 656 nm (Hα line).

Q4. What region does the Lyman series lie in? Answer: Ultraviolet region.

Case Study 18: Hα red line in hydrogen

Passage: A physics laboratory uses a hydrogen discharge tube and a spectrometer. When the gas is excited electrically, discrete coloured lines appear. Astronomers use the same principle to identify hydrogen in distant stars by matching observed wavelengths with the Rydberg formula 1/λ = R(1/n₁² − 1/n₂²).

Q1. Name the series for transitions ending at n=2. Answer: Balmer series.

Q2. Why are the lines discrete? Answer: Because only certain energy differences are allowed between quantised levels.

Q3. Calculate approximate wavelength for n=3→2. Answer: 656 nm (Hα line).

Q4. What region does the Lyman series lie in? Answer: Ultraviolet region.

Case Study 19: Series limit calculation

Passage: A physics laboratory uses a hydrogen discharge tube and a spectrometer. When the gas is excited electrically, discrete coloured lines appear. Astronomers use the same principle to identify hydrogen in distant stars by matching observed wavelengths with the Rydberg formula 1/λ = R(1/n₁² − 1/n₂²).

Q1. Name the series for transitions ending at n=2. Answer: Balmer series.

Q2. Why are the lines discrete? Answer: Because only certain energy differences are allowed between quantised levels.

Q3. Calculate approximate wavelength for n=3→2. Answer: 656 nm (Hα line).

Q4. What region does the Lyman series lie in? Answer: Ultraviolet region.

Case Study 20: NEET spectral series

Passage: A physics laboratory uses a hydrogen discharge tube and a spectrometer. When the gas is excited electrically, discrete coloured lines appear. Astronomers use the same principle to identify hydrogen in distant stars by matching observed wavelengths with the Rydberg formula 1/λ = R(1/n₁² − 1/n₂²).

Q1. Name the series for transitions ending at n=2. Answer: Balmer series.

Q2. Why are the lines discrete? Answer: Because only certain energy differences are allowed between quantised levels.

Q3. Calculate approximate wavelength for n=3→2. Answer: 656 nm (Hα line).

Q4. What region does the Lyman series lie in? Answer: Ultraviolet region.

Case Study 21: Hydrogen Spectrum in discharge tube

Passage: A physics laboratory uses a hydrogen discharge tube and a spectrometer. When the gas is excited electrically, discrete coloured lines appear. Astronomers use the same principle to identify hydrogen in distant stars by matching observed wavelengths with the Rydberg formula 1/λ = R(1/n₁² − 1/n₂²).

Q1. Name the series for transitions ending at n=2. Answer: Balmer series.

Q2. Why are the lines discrete? Answer: Because only certain energy differences are allowed between quantised levels.

Q3. Calculate approximate wavelength for n=3→2. Answer: 656 nm (Hα line).

Q4. What region does the Lyman series lie in? Answer: Ultraviolet region.

Case Study 22: Balmer Series in astronomy

Passage: A physics laboratory uses a hydrogen discharge tube and a spectrometer. When the gas is excited electrically, discrete coloured lines appear. Astronomers use the same principle to identify hydrogen in distant stars by matching observed wavelengths with the Rydberg formula 1/λ = R(1/n₁² − 1/n₂²).

Q1. Name the series for transitions ending at n=2. Answer: Balmer series.

Q2. Why are the lines discrete? Answer: Because only certain energy differences are allowed between quantised levels.

Q3. Calculate approximate wavelength for n=3→2. Answer: 656 nm (Hα line).

Q4. What region does the Lyman series lie in? Answer: Ultraviolet region.

Case Study 23: Rydberg Formula application

Passage: A physics laboratory uses a hydrogen discharge tube and a spectrometer. When the gas is excited electrically, discrete coloured lines appear. Astronomers use the same principle to identify hydrogen in distant stars by matching observed wavelengths with the Rydberg formula 1/λ = R(1/n₁² − 1/n₂²).

Q1. Name the series for transitions ending at n=2. Answer: Balmer series.

Q2. Why are the lines discrete? Answer: Because only certain energy differences are allowed between quantised levels.

Q3. Calculate approximate wavelength for n=3→2. Answer: 656 nm (Hα line).

Q4. What region does the Lyman series lie in? Answer: Ultraviolet region.

Case Study 24: Lyman Series in UV spectroscopy

Passage: A physics laboratory uses a hydrogen discharge tube and a spectrometer. When the gas is excited electrically, discrete coloured lines appear. Astronomers use the same principle to identify hydrogen in distant stars by matching observed wavelengths with the Rydberg formula 1/λ = R(1/n₁² − 1/n₂²).

Q1. Name the series for transitions ending at n=2. Answer: Balmer series.

Q2. Why are the lines discrete? Answer: Because only certain energy differences are allowed between quantised levels.

Q3. Calculate approximate wavelength for n=3→2. Answer: 656 nm (Hα line).

Q4. What region does the Lyman series lie in? Answer: Ultraviolet region.

Case Study 25: Absorption in stellar atmospheres

Passage: A physics laboratory uses a hydrogen discharge tube and a spectrometer. When the gas is excited electrically, discrete coloured lines appear. Astronomers use the same principle to identify hydrogen in distant stars by matching observed wavelengths with the Rydberg formula 1/λ = R(1/n₁² − 1/n₂²).

Q1. Name the series for transitions ending at n=2. Answer: Balmer series.

Q2. Why are the lines discrete? Answer: Because only certain energy differences are allowed between quantised levels.

Q3. Calculate approximate wavelength for n=3→2. Answer: 656 nm (Hα line).

Q4. What region does the Lyman series lie in? Answer: Ultraviolet region.

21

Final Revision Section

One-page revision for exam day.

One Page Revision Sheet

Hydrogen spectrum = discrete lines from quantised transitions. Emission = bright lines; Absorption = dark lines. Rydberg: 1/λ = R(1/n₁²−1/n₂²). Lyman (UV, n₁=1), Balmer (visible, n₁=2), Paschen (IR, n₁=3). Eₙ = −13.6/n² eV. ΔE = hν = hc/λ. λ(nm) = 1240/ΔE(eV).

25 Most Important Formulae

  1. 1/λ = R(1/n₁² − 1/n₂²)
  2. Eₙ = −13.6/n² eV
  3. ΔE = hν = hc/λ
  4. ν = Rc(1/n₁² − 1/n₂²)
  5. λ = 1240/ΔE nm
  6. Lyman: 1/λ = R(1−1/n²)
  7. Balmer: 1/λ = R(1/4−1/n²)
  8. Paschen: 1/λ = R(1/9−1/n²)
  9. R = 1.097×10⁷ m⁻¹
  10. h = 6.63×10⁻³⁴ J·s
  11. c = 3×10⁸ m/s
  12. N_lines = n(n−1)/2
  13. Eₙ = −RhcZ²/n²
  14. Series limit: n₂ → ∞
  15. Hα = 656 nm
  16. Hβ = 486 nm
  17. Lyman-α = 121.6 nm
  18. Ionisation = 13.6 eV
  19. Excitation ΔE = E_f − E_i
  20. Wave number σ = 1/λ
  21. Brackett: n₁=4
  22. Pfund: n₁=5
  23. ΔE(eV) = 13.6(1/n₁²−1/n₂²)
  24. Balmer limit = 365 nm
  25. Lyman limit = 91.2 nm

25 Most Important Concepts

  1. Line vs continuous spectrum
  2. Emission vs absorption
  3. Quantised energy levels
  4. Photon energy = level difference
  5. Rydberg constant meaning
  6. Lyman series (UV)
  7. Balmer series (visible)
  8. Paschen series (IR)
  9. Series limit concept
  10. Hα, Hβ, Hγ lines
  11. Kirchhoff's laws
  12. Bohr model connection
  13. Ground vs excited state
  14. Number of spectral lines
  15. Astronomical spectroscopy
  16. Discharge tube experiment
  17. Dark Fraunhofer lines
  18. Wave number vs wavelength
  19. Energy level diagram reading
  20. n₁ determines series name
  21. n₂ > n₁ always
  22. Negative bound-state energy
  23. Ionisation from any level
  24. Selection rules (Δn any)
  25. Historical: Balmer 1885

25 Most Important Exam Questions

  1. State Rydberg formula
  2. Why line spectrum?
  3. Difference emission/absorption
  4. Balmer series region
  5. Lyman-α wavelength
  6. Hα transition
  7. Series limit of Balmer
  8. Photon energy for 656 nm
  9. Lines from n=4
  10. Paschen series formula
  11. Energy of n=3 state
  12. ΔE for 2→1
  13. Identify series from n₁
  14. R value in SI
  15. λ for n=3→2
  16. ν for Lyman limit
  17. IR series names
  18. Absorption spectrum explanation
  19. Continuous spectrum source
  20. Astronomical hydrogen detection
  21. Brackett series n₁
  22. Pfund series n₁
  23. Excitation energy n=1→2
  24. Longest Balmer wavelength
  25. Shortest Lyman wavelength

Last Minute Revision Notes

✔ Remember: Lyman = UV (n₁=1), Balmer = Visible (n₁=2), Paschen/Brackett/Pfund = IR (n₁=3,4,5). ✔ Use λ(nm) = 1240/ΔE(eV) for quick calculations. ✔ Absorption lines appear at the same λ as emission lines. ✔ For "number of lines" questions: N = n(n−1)/2. ✔ Series limit corresponds to n₂ → ∞. ✔ R = 1.097 × 10⁷ m⁻¹ — memorise this value.

Kumar Physics Classes · Complete Hydrogen Spectrum Revision
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