Stationary Orbits
Electrons revolve only in selected orbits without radiating electromagnetic energy.
Bohr Model
of Atom
A complete exam-oriented study of Bohr's postulates, quantised orbits, radius, velocity, energy, ionisation and hydrogen spectral series.
CBSENEETJEE MainJEE AdvancedIB PhysicsIGCSEICSEA-Level
01
Bohr Model of Atom
Rutherford's nucleus combined with quantum rules for electrons.
Bohr retained Rutherford's small positive nucleus but proposed that electrons may revolve only in certain permitted circular orbits. While in a stationary orbit an electron does not radiate energy.
Each orbit has a definite radius, speed and energy. Radiation is emitted or absorbed only when an electron jumps between two permitted states.
Hydrogen-like species: The model applies exactly to one-electron systems such as H, He⁺ and Li²⁺.
02
Bohr's Postulates
Quantisation
Permitted orbits satisfy mvr=nh/2π, where n=1,2,3...
Quantum Jumps
A photon is emitted or absorbed when an electron changes state: hν=|E₂−E₁|.
03
Quantisation of Angular Momentum
mvr = nℏ = nh/2π
m: electron mass, v: orbital speed, r: orbit radius, n: principal quantum numberThe condition can be understood as a standing de Broglie wave: 2πr=nλ. Only an integral number of wavelengths can fit around a stable circular orbit.
04
Stable Orbits
The electrostatic force provides centripetal acceleration:
mv²/r = (1/4πε₀)(Ze²/r²)
Together with angular-momentum quantisation, this selects discrete radii and energies. An electron in one of these stationary states does not continuously lose energy.
05
Radius of nth Orbit
rₙ = n²a₀/Z
a₀=5.29×10⁻¹¹ m is the Bohr radiusRadius increases as n² and decreases with nuclear charge Z. For hydrogen, r₁=a₀ and r₂=4a₀.
Derivation Result
Combining Coulomb force with mvr=nh/2π gives:
rₙ = ε₀h²n²/(πmZe²)
06
Velocity of Electron
vₙ = Z × 2.18×10⁶/n m s⁻¹
For a fixed Z, electron speed decreases as 1/n. For the ground-state hydrogen atom, v₁≈2.18×10⁶ m s⁻¹.
07
Energy of Electron
Kinetic Energy
K = +13.6Z²/n² eV
Potential Energy
U = −27.2Z²/n² eV
Total Energy
Eₙ = −13.6Z²/n² eV
08
Ionisation Energy
Ionisation takes the electron from a bound state to n=∞, where E=0.
Ionisation energy from nth state = 13.6Z²/n² eV
For ground-state hydrogen: 13.6 eVIonisation potential of hydrogen in its ground state is 13.6 V.
09
Emission and Absorption of Spectral Lines
A downward transition emits a photon; an upward transition requires absorption of exactly the energy difference.
ΔE = hν = hc/λ
Because energies are quantised, only selected photon wavelengths occur, producing a line spectrum.
10
Hydrogen Spectral Series
1/λ = RZ²(1/n₁² − 1/n₂²)
n₂>n₁; R=1.097×10⁷ m⁻¹11
Limitations of Bohr Model
Works mainly for one-electron atoms.
Cannot explain fine structure or relative line intensities.
Cannot explain Zeeman and Stark effects completely.
Assumes definite circular paths, conflicting with uncertainty principle.
Does not incorporate electron spin or full wave mechanics.
Fails for complex multi-electron spectra.
12
Complete Formula Sheet
mvr=nh/2π
rₙ=n²a₀/Z
vₙ=Z(2.18×10⁶)/n
Eₙ=−13.6Z²/n² eV
ΔE=hν=hc/λ
1/λ=RZ²(1/n₁²−1/n₂²)
13
Solved Numericals
Radius, velocity, energy, excitation, ionisation, wavelengths and series identification.
14
Exam Practice & PYQs
Toggle solutions for national and international curricula.
15
Assertion-Reason Questions
16
Case Study Questions
17
Extra Complete Question Bank Added
Additional CBSE, NEET, JEE, IB, IGCSE, ICSE, A-Level, assertion-reason and case-study questions with answers.
CBSE Board Important Questions
1. State Bohr's first postulate.
Answer / Solution
Electrons revolve only in certain permitted stationary orbits without radiating energy.
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2. Write Bohr's quantisation condition.
Answer / Solution
mvr = nh/2π, where n = 1, 2, 3, ...
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3. Why does hydrogen show line spectrum?
Answer / Solution
Because electrons can exist only in discrete energy levels and emit photons of definite energy during transitions.
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4. What is ionisation energy of hydrogen atom in ground state?
Answer / Solution
13.6 eV.
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5. Which hydrogen series lies in visible region?
Answer / Solution
Balmer series, because transitions terminate at n = 2.
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6. Define excitation energy.
Answer / Solution
Energy required to raise an electron from lower energy level to a higher bound energy level.
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7. Write the energy of electron in nth orbit of hydrogen.
Answer / Solution
E_n = -13.6/n² eV.
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8. Why is Bohr model not valid for helium atom?
Answer / Solution
Helium has two electrons, so electron-electron interaction is present; Bohr model works exactly only for one-electron atoms.
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9. What is the radius of first Bohr orbit?
Answer / Solution
a0 = 5.29 × 10^-11 m.
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10. Name the spectral series formed by transition to n=1.
Answer / Solution
Lyman series, in ultraviolet region.
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NEET Level Questions
1. The energy of electron in n=2 orbit of hydrogen is?
Answer / Solution
E2 = -13.6/4 = -3.4 eV.
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2. Energy required to excite hydrogen from n=1 to n=2 is?
Answer / Solution
ΔE = -3.4 - (-13.6) = 10.2 eV.
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3. Radius of third Bohr orbit of hydrogen is?
Answer / Solution
r3 = 9a0 = 9 × 5.29 × 10^-11 m.
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4. If electron jumps from n=3 to n=2, which series is obtained?
Answer / Solution
Balmer series.
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5. Ionisation energy from n=2 state of hydrogen is?
Answer / Solution
13.6/4 = 3.4 eV.
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6. For He+ ion, ground state energy is?
Answer / Solution
E1 = -13.6Z² = -13.6 × 4 = -54.4 eV.
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7. For Li2+ ion, radius of first orbit is?
Answer / Solution
r1 = a0/Z = a0/3.
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8. Velocity of electron in second orbit of hydrogen is?
Answer / Solution
v2 = 2.18×10^6/2 = 1.09×10^6 m/s.
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9. The shortest wavelength of Lyman series corresponds to which transition?
Answer / Solution
n = ∞ to n = 1.
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10. The longest wavelength of Balmer series corresponds to which transition?
Answer / Solution
n = 3 to n = 2.
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11. Which transition gives maximum energy photon: 2→1, 3→2, 4→3, 5→4?
Answer / Solution
2→1 gives maximum energy because energy gap is greatest among these.
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12. The angular momentum in third orbit is?
Answer / Solution
L = 3h/2π.
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13. If Z increases, Bohr radius for first orbit changes how?
Answer / Solution
It decreases as 1/Z.
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14. Hydrogen atom in n=3 state has energy?
Answer / Solution
E3 = -13.6/9 = -1.51 eV.
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15. Wavelength emitted for transition 2→1 has energy?
Answer / Solution
ΔE = 10.2 eV; wavelength = 1240/10.2 ≈ 121.6 nm.
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16. Paschen series lies in which region?
Answer / Solution
Infrared region.
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17. For hydrogen-like atom, energy is proportional to?
Answer / Solution
E_n ∝ -Z²/n².
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18. Number of possible spectral lines from n=4 to lower states is?
Answer / Solution
N = n(n-1)/2 = 4×3/2 = 6.
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19. For transition from n=5 to n=2, series is?
Answer / Solution
Balmer series.
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20. The limit of Balmer series corresponds to transition?
Answer / Solution
n = ∞ to n = 2.
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JEE Main Type Questions
1. Find the wavelength of photon emitted when hydrogen jumps from n=3 to n=2.
Answer / Solution
ΔE = 13.6(1/2² - 1/3²)=1.889 eV, λ=1240/1.889≈656 nm.
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2. A hydrogen atom absorbs 12.09 eV from ground state. Find final n.
Answer / Solution
Energy at final state = -13.6+12.09=-1.51 eV = -13.6/n², so n=3.
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3. How many spectral lines are possible when electron de-excites from n=5?
Answer / Solution
N = 5×4/2 = 10.
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4. Find ionisation potential of He+ from ground state.
Answer / Solution
Ionisation energy = 54.4 eV, so ionisation potential = 54.4 V.
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5. Find ratio of radii of second orbit of H and first orbit of He+.
Answer / Solution
r(H,n=2)=4a0; r(He+,n=1)=a0/2; ratio=8.
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6. Find ratio of velocities of electron in H n=1 and He+ n=2.
Answer / Solution
v ∝ Z/n. H:1/1=1, He+:2/2=1, ratio=1.
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7. Energy difference between n=4 and n=2 in hydrogen is?
Answer / Solution
ΔE=13.6(1/4-1/16)=2.55 eV.
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8. Find frequency for n=2 to n=1 transition in hydrogen.
Answer / Solution
ΔE=10.2 eV=10.2×1.6×10^-19 J; ν=ΔE/h≈2.47×10^15 Hz.
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9. The first line of Lyman series has wavelength about?
Answer / Solution
Transition 2→1; λ≈121.6 nm.
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10. The Hα line corresponds to which transition?
Answer / Solution
n=3 to n=2.
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11. Find de-excitation photon energy for He+ from n=2 to n=1.
Answer / Solution
ΔE=13.6×4(1-1/4)=40.8 eV.
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12. Calculate angular momentum of electron in n=4 orbit.
Answer / Solution
L=4h/2π=2h/π.
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13. In hydrogen, electron speed in first orbit is v. What is speed in fourth orbit?
Answer / Solution
v/4.
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14. In He+, radius of n=2 orbit equals?
Answer / Solution
r=4a0/2=2a0.
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15. What is energy of Li2+ in n=3 state?
Answer / Solution
E=-13.6×9/9=-13.6 eV.
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16. A photon of 13.6 eV is incident on ground-state hydrogen. What happens?
Answer / Solution
It can ionise the atom, electron just escapes with zero kinetic energy.
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17. A photon of 15 eV ionises ground-state hydrogen. Kinetic energy of electron?
Answer / Solution
15 - 13.6 = 1.4 eV.
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18. Find series limit wavelength of Lyman series.
Answer / Solution
ΔE=13.6 eV; λ=1240/13.6≈91.2 nm.
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19. Find series limit wavelength of Balmer series.
Answer / Solution
ΔE=13.6/4=3.4 eV; λ=1240/3.4≈365 nm.
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20. For hydrogen, E4/E2 equals?
Answer / Solution
(-13.6/16)/(-13.6/4)=1/4.
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JEE Advanced Conceptual Questions
1. Why is angular momentum quantisation related to standing waves?
Answer / Solution
For a stable orbit, electron wave must join smoothly after one round: 2πr=nλ. With λ=h/mv, this gives mvr=nh/2π.
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2. Why does Bohr model fail for multi-electron atoms?
Answer / Solution
It ignores electron-electron repulsion and cannot solve coupled motion exactly.
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3. Explain why energy of electron is negative in a bound state.
Answer / Solution
Zero energy is chosen at infinite separation; bound electron has less energy than a free electron at infinity.
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4. If nuclear charge increases, what happens to spectral line energy for same n transition?
Answer / Solution
Transition energy increases as Z².
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5. Why are lines closer at high n?
Answer / Solution
Energy levels vary as -1/n², so separation decreases as n becomes large.
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6. Can an electron absorb any photon energy in hydrogen?
Answer / Solution
No. Bound-bound absorption requires exactly matching energy gap; larger sufficient photon may ionise.
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7. Why is Balmer series visible but Lyman is ultraviolet?
Answer / Solution
Transitions to n=2 have smaller energy than transitions to n=1, so longer wavelengths lie in visible region.
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8. Compare ionisation from n=1 and n=3 in hydrogen.
Answer / Solution
From n=3 it is 13.6/9 eV, much smaller than 13.6 eV from ground state.
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9. What is physical meaning of series limit?
Answer / Solution
It corresponds to transition from n=∞ to a fixed lower level, giving maximum photon energy in that series.
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10. Why did Bohr introduce stationary states?
Answer / Solution
To avoid classical collapse of the Rutherford atom and explain discrete spectra.
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IB / IGCSE / A-Level Questions
1. Describe how Bohr model explains emission spectra.
Answer / Solution
Electrons emit photons only during transitions between quantised levels; each transition gives a definite wavelength.
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2. Calculate photon energy of wavelength 656 nm.
Answer / Solution
E=1240/656≈1.89 eV.
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3. What is meant by ground state?
Answer / Solution
Lowest energy state of an atom.
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4. What is meant by excited state?
Answer / Solution
A state with energy higher than ground state but still bound.
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5. Why does a gas discharge tube emit discrete colours?
Answer / Solution
Atoms emit photons of specific wavelengths due to transitions between discrete energy levels.
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6. What happens when an electron absorbs a photon with exactly the right energy?
Answer / Solution
It moves to a higher allowed energy level.
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7. What happens if photon energy is less than excitation energy?
Answer / Solution
It is not absorbed for that transition.
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8. Explain ionisation using energy levels.
Answer / Solution
Ionisation raises electron to E=0 level, corresponding to complete removal from atom.
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9. State Rydberg equation for hydrogen.
Answer / Solution
1/λ = R(1/n1² - 1/n2²).
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10. Why is hydrogen spectrum useful in astronomy?
Answer / Solution
Spectral lines identify hydrogen and reveal physical properties and motion of stars.
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Assertion Reason Practice
1. Assertion: Bohr orbits are called stationary states. Reason: Electron in these orbits does not radiate energy.
Answer / Solution
Both true; Reason correctly explains Assertion.
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2. Assertion: Lyman series is ultraviolet. Reason: It terminates at n=1.
Answer / Solution
Both true; Reason correctly explains Assertion.
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3. Assertion: Ionisation energy from n=2 is less than from n=1. Reason: Energy at n=2 is closer to zero.
Answer / Solution
Both true; Reason correctly explains Assertion.
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4. Assertion: Radius of orbit increases with n. Reason: r_n is proportional to n².
Answer / Solution
Both true; Reason correctly explains Assertion.
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5. Assertion: Bohr model explains all atomic spectra. Reason: It includes electron-electron interaction exactly.
Answer / Solution
Assertion false; Reason false.
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Case Study Questions
1. Case: Hydrogen atom is excited to n=4. How many maximum lines can be emitted?
Answer / Solution
Maximum lines = n(n-1)/2 = 6.
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2. Case: A Balmer photon is observed. What is the final level of transition?
Answer / Solution
n=2.
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3. Case: He+ emits photon from 2 to 1. Compare energy with H 2 to 1.
Answer / Solution
He+ energy is Z²=4 times hydrogen, so 40.8 eV.
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4. Case: Hydrogen absorbs 10.2 eV. What is final state?
Answer / Solution
From ground state, final state is n=2.
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5. Case: A line has wavelength 121.6 nm. Identify transition.
Answer / Solution
Lyman-alpha transition, n=2 to n=1.
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