PhysicsCBSE Class 12 · Atoms

Bohr Model of the Hydrogen Atom Class 12: Postulates, Radius and Energy

Updated 15 October 2026 · Fact-checked

Bohr's model says the electron in hydrogen moves only in certain stationary orbits without radiating, where its angular momentum is nh/2π, and emits or absorbs a photon only when it jumps between orbits. This gives orbit radius rₙ = 0.529n² Å, speed vₙ ≈ 2.18 × 10⁶/n m/s and energy Eₙ = −13.6/n² eV.

Understand Bohr Model of the Hydrogen Atom

Rutherford's nuclear atom could not explain why atoms are stable or why they give line spectra. In 1913 Bohr kept the nucleus and the orbiting electron, but added three bold rules.

Bohr's postulates.
1. Stationary orbits: an electron can revolve in certain stable orbits without radiating energy, in spite of its acceleration. Each such orbit has a definite total energy.
2. Quantisation of angular momentum: these orbits are the ones in which the angular momentum is a whole-number multiple of $\frac{h}{2 π}$ : $m v_{n} r_{n} = \frac{nh}{2 π}$ , n = 1, 2, 3, … (n is the principal quantum number).
3. Frequency condition: an electron jumping from an orbit of energy $E_{i}$ to a lower one of energy $E_{f}$ emits a photon with $h ν = E_{i} - E_{f}$ ; absorbing a photon of exactly this energy lifts it the other way.

Radius, speed and energy. The electrostatic attraction provides the centripetal force: $\frac{m v _{n}^{2}}{r _{n}} = \frac{1}{4 π ε _{0}} \frac{e ^{2}}{r _{n}^{2}}$ . Combining with the quantisation condition gives (expressions to remember; the derivation of the radius is not required for 2026-27):

Radius: $r_{n} = \frac{n ^{2} h ^{2} ε _{0}}{π m e ^{2}} = n^{2} a_{0}$ , with the Bohr radius $a_{0} \approx 0.529 \times 1 0^{- 10}$ m. So r ∝ n².
Speed: $v_{n} = \frac{e ^{2}}{2 ε _{0} nh} \approx \frac{2.18 \times 1 0 ^{6}}{n}$ m/s. So v ∝ 1/n; the ground-state speed is about c/137.
Energy: from the force balance, the kinetic energy is $K = \frac{1}{2} m v^{2} = \frac{e ^{2}}{8 π ε _{0} r}$ and the potential energy is $U = - \frac{e ^{2}}{4 π ε _{0} r}$ , so the total is $E = K + U = - \frac{e ^{2}}{8 π ε _{0} r}$ . Putting in $r_{n}$ : $E_{n} = - \frac{m e ^{4}}{8 ε _{0}^{2} h ^{2} n ^{2}} = - \frac{13.6}{n ^{2}}$ eV.

Note the useful relations $K = - E$ and $U = 2 E$ .

Meaning of the negative energy. The zero of energy is an electron at rest far from the nucleus. A negative total energy means the electron is bound. The ground state (n = 1) has E = −13.6 eV, so 13.6 eV is needed to ionise a hydrogen atom from its ground state. As n increases, E rises towards 0, the orbits get larger, and the electron moves more slowly. The levels crowd together near E = 0.

Other trends. The orbital period $T_{n} = \frac{2 π r _{n}}{v _{n}} \propto n^{3}$ , so the frequency of revolution ∝ 1/n³. Larger n means higher total energy but lower kinetic energy.

Limitations. Bohr's model works for hydrogen (and one-electron ions) but fails for atoms with two or more electrons, cannot explain the relative intensities of spectral lines or their fine structure, and mixes classical orbits with quantum rules without explaining why angular momentum should be quantised. Quantum mechanics later replaced the idea of definite orbits.

Bohr Model of the Hydrogen Atom: formulas and key facts

Quantisation of angular momentum: $m v_{n} r_{n} = \frac{nh}{2 π}$; n = 1, 2, 3, …
Frequency condition: $h ν = E_{i} - E_{f}$; Photon emitted or absorbed in a jump.
Radius of nth orbit: $r_{n} = \frac{n ^{2} h ^{2} ε _{0}}{π m e ^{2}} = 0.529 n^{2} \overset{˚}{A}$; Expression only; r ∝ n².
Speed in nth orbit: $v_{n} = \frac{e ^{2}}{2 ε _{0} nh} \approx \frac{2.18 \times 1 0 ^{6}}{n} m/s$; v ∝ 1/n.
Energy of nth orbit: $E_{n} = - \frac{m e ^{4}}{8 ε _{0}^{2} h ^{2} n ^{2}} = - \frac{13.6}{n ^{2}} eV$; E ∝ −1/n².
Kinetic and potential energy: $K = - E_{n}, U = 2 E_{n}$; For any orbit.
Period of revolution: $T_{n} = \frac{2 π r _{n}}{v _{n}} \propto n^{3}$; Frequency of revolution ∝ 1/n³.

How to solve Bohr Model of the Hydrogen Atom questions

Use this method for Bohr model numericals and ratios.

1Write the relevant n-dependence first: r ∝ n², v ∝ 1/n, E ∝ −1/n², T ∝ n³, L ∝ n.
2For actual values, use rₙ = 0.529n² Å, vₙ = 2.18 × 10⁶/n m/s and Eₙ = −13.6/n² eV.
3For kinetic and potential energy, use K = −E and U = 2E.
4For transitions, ΔE = Eᵢ − E_f (in eV), and photon wavelength λ ≈ 1240/ΔE nm.
5For angular momentum, L = nh/2π (h/2π ≈ 1.055 × 10⁻³⁴ J s).
6Keep signs: total and potential energies are negative; kinetic energy is positive.

Bohr Model of the Hydrogen Atom in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Size and binding of high orbits

Suppose a hydrogen atom in a very dilute gas in space has its electron in the orbit n = 100.

(a) Find the radius of this orbit. (b) Find the electron's energy and the energy needed to free it. (c) Why would such an atom be easily ionised?

Try it first, then show the solutionSolution

$r = 0.529 \times 1 0^{- 10} \times 10 0^{2} \approx 5.3 \times 1 0^{- 7}$ m, about half a micrometre, 10 000 times larger than a normal hydrogen atom.
$E = - \frac{13.6}{1 0 ^{4}} \approx - 1.36 \times 1 0^{- 3}$ eV; about 1.4 meV frees it.
Its binding energy is tiny, so a weak collision or a low-energy photon is enough to ionise it.

Answer: (a) about 5.3 × 10⁻⁷ m; (b) about −1.36 meV, needing about 1.36 meV; (c) it is very weakly bound.

Memory hook: High n: huge orbit, tiny binding.

Ionising excited atoms

Imagine hydrogen atoms that have already been excited to n = 2. You try to ionise them with red light (656 nm), violet light (400 nm) and ultraviolet light (300 nm).

(a) How much energy is needed to ionise an atom from n = 2? (b) Which of the three lights can do it? Use E(eV) ≈ 1240/λ(nm).

Try it first, then show the solutionSolution

From n = 2, E = −13.6/4 = −3.4 eV, so 3.4 eV is needed.
Photon energies: red ≈ 1.89 eV, violet ≈ 3.1 eV, ultraviolet ≈ 4.13 eV.
Only the 300 nm ultraviolet photons (4.13 eV > 3.4 eV) can ionise; the extra 0.73 eV becomes kinetic energy of the freed electron.

Answer: (a) 3.4 eV; (b) only the 300 nm ultraviolet light.

Memory hook: To leave level n, pay 13.6/n² eV.

Tricks to remember Bohr Model of the Hydrogen Atom

Short memory aids. Say each one aloud once, then use it on a question the same day.

r up as n², v down as n, E up towards 0 as 1/n²
Bigger orbit, slower electron, less tightly bound.
Use it when: Ratio questions.
0.529, 2.18, 13.6
Bohr radius (Å), ground-state speed (×10⁶ m/s), ground-state energy (eV).
Use it when: Quick values.
K = −E, U = 2E
Kinetic energy is minus the total; potential is twice the total.
Use it when: Energy split questions.
Angular momentum in steps of h/2π
L = nh/2π: first rung h/2π, second 2h/2π, …
Use it when: Quantisation condition.

Bohr Model of the Hydrogen Atom: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Radius and energy of an orbit · 2 marks · Very short answerBoard level

Find the radius and total energy of the electron in the third orbit (n = 3) of a hydrogen atom.

Try it first, then show the solutionSolution

$r_{3} = 0.529 \times 9 \approx 4.76$ Å $= 4.76 \times 1 0^{- 10}$ m. (1)
$E_{3} = - \frac{13.6}{9} \approx - 1.51$ eV. (1)

Answer: 4.76 Å and −1.51 eV.

Example 2: Bohr's postulates · 3 marks · Short answerBoard level

State Bohr's three postulates for the hydrogen atom. Which postulate explains the stability of the atom, and which explains line spectra?

Try it first, then show the solutionSolution

(1) Electrons revolve in certain stationary orbits without radiating energy. (2) In these orbits the angular momentum is mvr = nh/2π. (3) A photon of energy hν = Eᵢ − E_f is emitted (or absorbed) when the electron jumps between orbits. (2)
Postulate 1 explains stability (no continuous radiation); postulate 3, with discrete energy levels, explains line spectra (only definite photon energies). (1)

Answer: Stationary orbits; mvr = nh/2π; hν = Eᵢ − E_f. The first explains stability and the third, with discrete levels, explains line spectra.

Example 3: Total, kinetic and potential energy · 5 marks · Long answerHarder board level

(a) Starting from the condition that the electrostatic force provides the centripetal force, show that the total energy of the electron in a Bohr orbit of radius r is $E = - \frac{e ^{2}}{8 π ε _{0} r}$ . (3)
(b) For n = 2 in hydrogen, find the kinetic, potential and total energies. (2)

Try it first, then show the solutionSolution

(a) $\frac{m v ^{2}}{r} = \frac{e ^{2}}{4 π ε _{0} r ^{2}}$ gives $K = \frac{1}{2} m v^{2} = \frac{e ^{2}}{8 π ε _{0} r}$ . (1)
Potential energy of the electron-proton pair: $U = - \frac{e ^{2}}{4 π ε _{0} r}$ . (1)
$E = K + U = - \frac{e ^{2}}{8 π ε _{0} r}$ ; hence K = −E and U = 2E. (1)
(b) $E_{2} = - \frac{13.6}{4} = - 3.4$ eV. (1)
K = +3.4 eV, U = −6.8 eV. (1)

Answer: (a) E = −e²/(8πε₀r); (b) K = 3.4 eV, U = −6.8 eV, E = −3.4 eV.

Example 4: Speed and period of revolution · 3 marks · Short answerHarder board level

Find (a) the speed of the electron in the first and third orbits of hydrogen, and (b) the period and frequency of revolution in the first orbit.

Try it first, then show the solutionSolution

(a) $v_{1} \approx 2.18 \times 1 0^{6}$ m/s; $v_{3} = \frac{v _{1}}{3} \approx 7.27 \times 1 0^{5}$ m/s. (1)
(b) $T_{1} = \frac{2 π r _{1}}{v _{1}} = \frac{2 π \times 0.529 \times 1 0 ^{- 10}}{2.18 \times 1 0 ^{6}} \approx 1.52 \times 1 0^{- 16}$ s. (1)
Frequency $= \frac{1}{T _{1}} \approx 6.6 \times 1 0^{15}$ Hz. (1)

Answer: (a) about 2.18 × 10⁶ m/s and 7.27 × 10⁵ m/s; (b) about 1.52 × 10⁻¹⁶ s and 6.6 × 10¹⁵ Hz.

Example 5: Checking the quantisation and an excitation · 5 marks · Long answerHOTS / competency

(a) Using $r_{2}$ and $v_{2}$ for hydrogen, show numerically that the angular momentum of the electron in n = 2 equals 2h/2π. (2)
(b) An electron is raised from n = 1 to n = 3. Find the energy absorbed and the factors by which its radius, speed and kinetic energy change. (3)

Try it first, then show the solutionSolution

(a) $r_{2} = 4 \times 0.529 \times 1 0^{- 10} = 2.116 \times 1 0^{- 10}$ m; $v_{2} = 1.09 \times 1 0^{6}$ m/s. (1)
$m v_{2} r_{2} = 9.1 \times 1 0^{- 31} \times 1.09 \times 1 0^{6} \times 2.116 \times 1 0^{- 10} \approx 2.1 \times 1 0^{- 34}$ J s, and $\frac{2 h}{2 π} = \frac{6.63 \times 1 0 ^{- 34}}{π} \approx 2.11 \times 1 0^{- 34}$ J s. They agree. (1)
(b) Energy absorbed $= E_{3} - E_{1} = - 1.51 - (- 13.6) \approx 12.09$ eV. (1)
Radius becomes 9 times; speed becomes 1/3. (1)
Kinetic energy (= −E) falls from 13.6 eV to 1.51 eV, i.e. becomes 1/9 of its value. (1)

Answer: (a) about 2.1 × 10⁻³⁴ J s = 2h/2π; (b) 12.09 eV; radius × 9, speed ÷ 3, kinetic energy ÷ 9.

Example 6: Comparing orbits · 4 marks · Case-basedHarder board level

Read the passage and answer.

A hydrogen atom's electron is in the orbit n = 4.
(a) By what factor is its orbit radius larger than in the ground state? (1)
(b) By what factor is its speed smaller? (1)
(c) What fraction of the ground-state binding energy does it have? (1)
(d) By what factor is its period of revolution longer? (1)

Try it first, then show the solutionSolution

(a) r ∝ n²: 16 times. (1)
(b) v ∝ 1/n: 4 times smaller. (1)
(c) |E| ∝ 1/n²: 1/16 (0.85 eV against 13.6 eV). (1)
(d) T ∝ n³: 64 times longer. (1)

Answer: (a) 16; (b) 4; (c) 1/16; (d) 64.

Where students lose marks in Bohr Model of the Hydrogen Atom

The slips examiners see most often, and what to write instead.

Mistake: Writing the energy as +13.6/n² eV.
Dropping the sign.
Do this: Bound-state energies are negative: Eₙ = −13.6/n² eV.
Usually costs: 1 mark.
Mistake: Saying kinetic energy increases with n.
Confusing total and kinetic energy.
Do this: K = −E = 13.6/n² eV decreases as n increases.
Usually costs: 1 mark.
Mistake: Deriving the radius at length.
Older syllabi included it.
Do this: For 2026-27 quote rₙ = n²h²ε₀/(πme²); spend time on applications.
Usually costs: Time.
Mistake: Using r ∝ n or v ∝ 1/n² in ratio questions.
Mixing up powers.
Do this: Remember r ∝ n², v ∝ 1/n, E ∝ 1/n², T ∝ n³.
Usually costs: The answer mark.

Exam tips

The Bohr model was set in every board paper of 2023-2026 in our bank (for example 2023 55/5/1 Q30(b), 2024 55/1/1 Q27, 2025 55/7/1 Q27 and 2026 55/5/1 Q26(b)).
Write the three postulates as three numbered lines; each earns a mark.
Show the force balance and quantisation condition before quoting rₙ, vₙ and Eₙ.
Use K = −E, U = 2E to answer energy-split questions in one line.
Mention limitations (only one-electron atoms, no intensities or fine structure) when asked to evaluate the model.

Practice questions on Bohr Model of the Hydrogen Atom

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
In the Bohr model, the radius of the nth orbit of hydrogen is proportional to
(a) n (b) n² (c) 1/n (d) 1/n²
Hint
rₙ = 0.529n² Å.
Answer
(b) n².
Q2 · 1 mark · Very short answerBoard level
State Bohr's quantisation condition for angular momentum.
Hint
Multiples of h/2π.
Answer
The angular momentum of the electron in an allowed orbit is an integral multiple of h/2π: mvr = nh/2π.
Q3 · 2 marks · Very short answerHarder board level
Find the ratio of (a) the radii and (b) the energies (magnitudes) of the electron in orbits n = 2 and n = 3 of hydrogen.
Hint
r ∝ n², |E| ∝ 1/n².
Answer
(a) r₂ : r₃ = 4 : 9; (b) |E₂| : |E₃| = 9 : 4.
Q4 · 1 mark · Very short answerHarder board level
How much energy is needed to ionise a hydrogen atom whose electron is in the n = 3 orbit?
Hint
Bring E to 0.
Answer
13.6/9 ≈ 1.51 eV.
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): As the electron in a hydrogen atom moves to higher orbits, its total energy increases while its kinetic energy decreases.
Reason (R): In a Bohr orbit the kinetic energy equals the negative of the total energy.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
E = −13.6/n².
Answer
(a). E rises towards 0, so K = −E falls.
Q6 · 1 mark · Very short answerHOTS / competency
Find the angular momentum of the electron in the third orbit of hydrogen.
Hint
L = nh/2π.
Answer
L = 3h/2π = 3 × 6.63 × 10⁻³⁴/(2π) ≈ 3.16 × 10⁻³⁴ J s.
Q7 · 2 marks · Very short answerHOTS / competency
Find the ratio of the frequencies of revolution of the electron in the first and second orbits of hydrogen.
Hint
f = v/(2πr) ∝ 1/n³.
Answer
f ∝ v/r ∝ (1/n)/n² = 1/n³, so f₁ : f₂ = 8 : 1.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Bohr radius?Show answer
About 0.529 × 10⁻¹⁰ m.
2.Energy of the nth orbit of hydrogen?Show answer
−13.6/n² eV.
3.Speed in the nth orbit?Show answer
About 2.18 × 10⁶/n m/s.
4.Relation between K, U and E?Show answer
K = −E, U = 2E.
5.Ionisation energy of hydrogen from the ground state?Show answer
13.6 eV.
6.How does the period of revolution depend on n?Show answer
T ∝ n³.

Bohr Model of the Hydrogen Atom: frequently asked questions

Why doesn't the electron radiate in a stationary orbit?

Bohr simply postulated it. Classical physics says an accelerating charge radiates, but Bohr proposed that in the allowed orbits it does not; quantum mechanics later justified the existence of stable states.

Why is the energy of the electron negative?

Energy is measured from zero for an electron at rest far from the nucleus. A bound electron has less energy than that, so its total energy is negative; the magnitude is the energy needed to free it.

Is the derivation of the radius required?

No. For 2026-27, CBSE asks for the expressions for the radius, velocity and energy of the electron in the nth orbit; the NCERT book keeps only the expression for the radius.

What are the limitations of Bohr's model?

It works only for one-electron atoms, cannot explain the intensities or fine structure of spectral lines, and does not explain why angular momentum is quantised.

What is the ionisation energy of hydrogen?

13.6 eV: the energy needed to remove the electron from the ground state (n = 1) to infinity.