Q.The gravitational attraction between electron and proton in a hydrogen atom is weaker than the coulomb attraction by a factor of about 10−40. An alternative way of looking at this fact is to estimate the radius of the first Bohr orbit of a hydrogen atom if the electron and proton were bound by gravitational attraction. You will find the answer interesting.
Concept understanding — Bohr Model Quantization
Why does an electron not spiral into the nucleus?
Imagine you're pushing a child on a swing. If you push at random moments, the swing jerks and slows down. But if you push exactly in rhythm with the swing's natural motion, each push adds energy smoothly and the swing goes higher and higher. The swing "prefers" to move at specific frequencies — its natural modes.
An electron orbiting a nucleus is similar, but with a crucial twist from the quantum world. Classical physics says an accelerating charge (like an electron going in a circle) must continuously radiate energy. If that were true, the electron would lose energy, spiral into the nucleus, and atoms would collapse in a flash of light. But atoms are stable. So something is fundamentally different.
The radical idea: allowed orbits only
Niels Bohr proposed in 1913 that the electron cannot occupy just any orbit. It can only exist in certain stationary states — orbits where it does not radiate energy. These are like the swing's natural frequencies, but for an electron.
The key condition that picks out these special orbits is called quantization of angular momentum.
L=n2πh,n=1,2,3,…
Here L is the orbital angular momentum of the electron, h is Planck's constant, and n is a positive integer called the principal quantum number.
What this means physically
Angular momentum for a circular orbit is L=mvr, where m is the electron mass, v its speed, and r the orbit radius. So the quantization condition becomes:
mvr=n2πh
This is not a formula you derive — it is a postulate, a rule that nature follows. Bohr had no deeper explanation for why this rule works; he simply noticed it gave the right answers for hydrogen's spectrum.
The quantity 2πh appears so often that it has its own symbol: ℏ (h-bar). So the condition is often written as L=nℏ.
What it predicts
Combining this quantization with Newton's law for circular motion (centripetal force = Coulomb attraction) gives:
- Radius of the nth orbit: rn=n2a0, where a0=0.529A˚ is the Bohr radius (the smallest orbit, n=1).
- Energy of the nth orbit: En=−n213.6eV
The negative sign means the electron is bound to the nucleus. As n increases, the orbit gets larger and the energy becomes less negative (closer to zero).
The key insight for exams
Bohr quantization is not a derivation — it is a condition you apply. When you see a problem about hydrogen-like atoms (one electron), you:
- Write mvr=nℏ
- Write the force balance: rmv2=r2kZe2 (for nuclear charge Ze)
- Solve for r and v in terms of n
Bohr's model works perfectly only for single-electron systems: H, He+, Li2+, etc. It fails for multi-electron atoms because it ignores electron-electron repulsion and the wave nature of electrons.
Why "quantization"?
The word comes from the Latin quantus — "how much." In classical physics, angular momentum can take any value. In Bohr's atom, it comes only in discrete packets (quanta) of size ℏ. This is the first hint that at the atomic scale, nature is not continuous but granular.
The electron does not spiral because it cannot lose energy gradually — it can only jump from one allowed orbit to another, emitting or absorbing a photon of exactly the right energy. Between these jumps, it simply exists in a stationary state, defying classical expectations.
Bohr's quantization of angular momentum is one of the defining postulates covered in the NCERT Class 12 Physics Atoms chapter, and students frequently search for "Bohr model quantization condition and derivation" or "Bohr's model important questions" while preparing for CBSE boards and JEE Main/NEET. This concept is also a common launching point for numerical problems on orbital radius and energy levels of hydrogen-like atoms tested across competitive exams.
Why this formula?
Why Angular Momentum is Quantised in the Bohr Model
The Bohr model's most famous result — that angular momentum comes only in integer multiples of 2πh — is not an arbitrary assumption. It follows directly from a single, elegant idea: the electron's wave must close on itself.
The core problem Bohr faced
By 1913, physicists knew two things that seemed contradictory:
- Rutherford's nuclear model showed electrons orbiting the nucleus.
- Maxwell's equations predicted that any accelerating charge (like an orbiting electron) must radiate energy, spiral inward, and collapse in about 10−11 seconds.
Atoms are stable. Something was missing.
Bohr's breakthrough was to combine the newly discovered quantum idea (Planck's constant h) with classical mechanics, but only for allowed orbits. The key constraint came from thinking of the electron not as a tiny planet, but as a standing wave.
The de Broglie wavelength argument (the cleanest derivation)
A few years after Bohr, de Broglie proposed that every moving particle has a wavelength:
λ=ph=mvh
For an electron in a circular orbit of radius r, the circumference is 2πr. For the wave to be stable — not cancelling itself out — the circumference must contain an integer number of wavelengths:
2πr=nλ,n=1,2,3,…
Substitute λ=h/(mv):
2πr=n⋅mvh
Rearrange:
mvr=n⋅2πh
That's it. The left side mvr is the angular momentum L. So:
L=nℏ,where ℏ=2πh
This is not a separate postulate — it is a consequence of requiring the electron wave to be a standing wave. If the wave doesn't close on itself, it interferes destructively and the orbit cannot exist.
Why this fixes the energy levels
Once angular momentum is quantised, the rest follows from classical physics. For a circular orbit, the centripetal force is provided by the Coulomb attraction:
rmv2=4πϵ01r2e2
Combine this with mvr=nℏ and solve for r and E:
rn=me24πϵ0ℏ2⋅n2=a0n2
En=−8ϵ02h2me4⋅n21=−n213.6 eV
A common mistake is to think Bohr derived the quantization rule from first principles. He didn't — he postulated it. The de Broglie standing-wave argument came later and provides the physical reason for the postulate, but it is still a postulate in the full quantum theory.
The deeper reason: it's not really about orbits
The Bohr model is ultimately wrong — electrons don't orbit in neat circles. But the quantization of angular momentum survives in the full quantum mechanical treatment (Schrödinger equation) as the condition that the wavefunction must be single-valued. For the hydrogen atom, the angular momentum quantum number l can take values 0,1,2,…,n−1, and the magnitude is l(l+1)ℏ, not nℏ.
Yet the Bohr model's key insight — that only certain discrete states are allowed — remains the foundation of atomic physics. The formula L=nℏ is the simplest example of a quantum number, and it correctly predicts the hydrogen spectrum to within fine-structure corrections.
The Bohr quantization condition L=nℏ is a boundary condition on the electron wave, not a dynamical law. It says: for the electron to exist in a stable state, its wave must fit perfectly around the nucleus. This is the same principle that governs standing waves on a string or in an organ pipe — only certain wavelengths survive.
The Bohr-model derivation for the orbit radius never actually depends on the force being electrical -- only on it being an inverse-square central force -- so the same derivation goes through with the Coulomb force constant ke2 replaced by the gravitational analogue Gmemp.
r1≈1.2×1029 m -- vastly larger than the size of the observable universe, showing just how much weaker gravity is than the Coulomb force at atomic scales.
Replacing the Coulomb force ke2/r2 with the gravitational force Gmemp/r2 in the Bohr derivation gives a 'gravitational Bohr radius' of about 1.2×1029 m for n=1 -- enormously larger than an atom, or even the observable universe.
Step 1 -- Redo the Bohr derivation with gravity as the central force.
In the ordinary Bohr model, the centripetal force balance is
rmv2=r2ke2,k=4πε01
and angular momentum quantisation gives mvr=n2πh. Combining these (exactly as in the text's derivation of the Bohr radius) gives
rn=4π2mke2n2h2
Nothing about this derivation actually used the fact that the force was electrical -- only that it was a 1/r2 attractive force between the electron (mass me) and a much heavier fixed centre. So if the electron and proton were instead bound purely by gravity, we simply replace the Coulomb coupling ke2 by the gravitational coupling Gmemp:
rngrav=4π2me(Gmemp)n2h2=4π2Gme2mpn2h2
Step 2 -- Evaluate for n=1.
Using h=6.63×10−34 J s, G=6.67×10−11 N m2kg−2, me=9.11×10−31 kg, mp=1.67×10−27 kg:
r1grav=4π2Gme2mph2≈1.2×1029 m
Step 3 -- Put the number in perspective.
1.2×1029 m is about a thousand times larger than the radius of the observable universe (∼4×1026 m). This dramatically illustrates the exercise's opening fact -- gravity is weaker than the Coulomb attraction between an electron and proton by a factor of about 10−40 -- an atom held together by gravity alone, at the same quantum number, would be unimaginably larger than anything that actually exists.
r1grav≈1.2×1029 m
Showing the 12 most recent of 14 on this concept.
- GUJCET 2026Set x1 markMCQQ.The radius of the innermost electron orbit of a hydrogen atom is 5.3×10−11 m. What is the radius of the n=3 orbit? (A) 1.59×10−10 m (B) 1.06×10−10 m (C) 1.43×10−9 m (D) 4.77×10−10 m
›Reveal solutionSolution
rn=n2r1⇒r3=9(5.3×10−11)=4.77×10−10 m.
In the Bohr model the orbit radius scales as the square of the principal quantum number:
rn=n2r1
With r1=5.3×10−11 m and n=3:
r3=32×5.3×10−11=9×5.3×10−11=4.77×10−10 m
✓Final answerOption (D) 4.77×10−10 m
ANSWER: (D)
- GUJCET 2025Set 031 markMCQQ.According to Bohr's model, the orbital angular momentum of electrons in third excited state is ______ [h=6.63×10−34 Js] (A) 4.2×10−34 kg m2s−1 (B) 12.350×10−34 kg m2s−1 (C) 1.625×10−26 erg-s (D) 6.63×10−34 Js
›Reveal solutionSolution
Third excited state means n=4; Bohr angular momentum L=2πnh.
Ground n=1, so third excited ⇒n=4.
L=2π4h=2π4×6.63×10−34=6.28326.52×10−34≈4.2×10−34 kg m2s−1.
✓Final answer(A) 4.2×10−34 kg m2s−1
ANSWER: (A)
- GSEB Higher Secondary Certificate (HSC) Examination 2025Set ANNUAL1 markMCQQ.The ratio of radius of third and second orbits of a Hydrogen atom is ______.(a) 2/3(b) 4/9(c) 3/2(d) 9/4
›Reveal solutionSolution
In the Bohr model, the radius of the nth orbit scales as rn ∝ n².
r3/r2 = 3²/2² = 9/4.
✓Final answer(d) 9/4.
- GUJCET 2024Set 131 markMCQQ.The ratio of radius for second and third orbit of hydrogen atom is ________. (A) 4:9 (B) 3:2 (C) 9:4 (D) 2:3
›Reveal solutionSolution
Bohr orbit radius rn∝n2.
Steps. For hydrogen rn∝n2, so
r3r2=3222=94.
✓Final answerOption (A) 4:9
ANSWER: (A)
- GSEB Higher Secondary Certificate (HSC) Examination 2024Set ANNUAL1 markMCQQ.If the radius of hydrogen atom in its first orbit is a0, then its radius in third excited state is ___.(a) 3a0(b) 4a0(c) 9a0(d) 16a0
›Reveal solutionSolution
Bohr radius formula: rn = n²a0. Ground state is n=1; first excited n=2; second excited n=3; third excited state is n=4.
r4 = 4² a0 = 16a0.
✓Final answer(d) 16a0.
- GSEB Higher Secondary Certificate (HSC) Examination 2024Set ANNUAL1 markMCQQ.In accordance with the Bohr's model, the quantum number that characterises the earth's revolution around the sun in an orbit of radius 1.5 x 10^11 m with orbit speed 3 x 10^4 m/s is ___. (Mass of earth is 6.0 x 10^24 kg, h = 6.625 x 10^-34 Js)(a) 2.6 x 10^72(b) 2.6 x 10^74(c) 2.6 x 10^39(d) 2.6 x 10^73
›Reveal solutionSolution
Bohr's angular momentum quantization: L = Mvr = n(h/2π), so n = 2πMvr/h.
M = 6.0 × 10²⁴ kg, v = 3 × 10⁴ m/s, r = 1.5 × 10¹¹ m, h = 6.625 × 10⁻³⁴ Js.
Mvr = (6.0 × 10²⁴)(3 × 10⁴)(1.5 × 10¹¹) = 2.7 × 10⁴⁰
2π × 2.7 × 10⁴⁰ = 1.696 × 10⁴¹
n = 1.696 × 10⁴¹ / 6.625 × 10⁻³⁴ ≈ 2.56 × 10⁷⁴ ≈ 2.6 × 10⁷⁴.
✓Final answer(b) 2.6 × 10⁷⁴.
- GUJCET 2023Set 091 markMCQQ.In hydrogen atom an electron makes a transition from 5th orbit to 3rd orbit. The change in the angular momentum for this electron is ______. (A) πh (B) 2πh (C) π3h (D) π5h
›Reveal solutionSolution
[!TLDR]
Using L=2πnh, the change in angular momentum from the 5th to the 3rd orbit is πh.
Concept
In Bohr's model of the hydrogen atom, the orbital angular momentum is quantised: L=2πnh, where n is the orbit number.
Solution
For the two orbits:
L5=2π5h,L3=2π3h.
The magnitude of the change is
ΔL=L5−L3=2π(5−3)h=2π2h=πh.
[!ANSWER]
(A) πh
- GSEB Higher Secondary Certificate (HSC) Examination 2023Set ANNUAL1 markMCQQ.What is the angular momentum of electron of Be^+3 ion in n = 5 orbit?(a) 3.3 x 10^-34 Js(b) 6.6 x 10^-34 Js(c) 5.3 x 10^-34 Js(d) 1.3 x 10^-34 Js
›Reveal solutionSolution
Bohr's quantisation gives L = n h/(2*pi) for any hydrogen-like ion; for n = 5, L = 5 x 1.055 x 10^-34 = 5.3 x 10^-34 Js.
Bohr's postulate for angular momentum applies to any single-electron (hydrogen-like) system such as Be^3+:
L = n h/(2*pi) = n (h-bar).
It depends only on n, not on the nuclear charge Z. For n = 5:
L = 5 x (6.63 x 10^-34)/(2*pi) = 5 x 1.055 x 10^-34 = 5.27 x 10^-34 approximately 5.3 x 10^-34 Js.
✓Final answer(c) 5.3 x 10^-34 Js.
- GUJCET 2022Set 171 markMCQQ.The radius of the innermost electron orbit of a hydrogen atom is 5.3×10−11 m. What are the radii of the n=3 orbit? (A) 4.12×10−10 m (B) 4.77×10−10 m (C) 2.12×10−10 m (D) 2.24×10−10 m
›Reveal solutionSolution
Bohr radius scales as n2: rn=r1n2.
Concept. In the Bohr model the orbit radius is rn=r1n2, with r1=5.3×10−11 m.
Steps.
- r3=5.3×10−11×32=5.3×10−11×9.
- r3=4.77×10−10 m.
✓Final answer(B) 4.77×10−10 m
ANSWER: (B)
- GUJCET 2022Set 171 markMCQQ.In accordance with the Bohr's model, find the quantum number that characterises the earth's revolution around the sun in an orbit of radius 1.5×1011 m with orbital speed 3×104 m/s. (Mass of earth = 6×1024 kg, h=6.625×10−34 J.s.) (A) 3.6×1074 (B) 1.6×1074 (C) 2.6×1074 (D) 4.6×1074
›Reveal solutionSolution
Bohr quantisation of angular momentum: mvr=2πnh.
Steps.
- mvr=(6×1024)(3×104)(1.5×1011)=2.7×1040 kg m2/s.
- n=h2πmvr=6.625×10−346.283×2.7×1040.
- n≈2.6×1074.
✓Final answer(C) 2.6×1074
ANSWER: (C)
- GUJCET 2021Set 151 markMCQQ.What is the shortest wavelength present in the Balmer series of spectral line? [Where R is Rydberg constant] (A) R1 (B) R3 (C) R2 (D) R4
›Reveal solutionSolution
The Balmer series limit (n to infinity, to n=2) gives the shortest wavelength.
Concept. λ1=R(221−∞1)=4R.
Solution. λ=R4.
✓Final answer(D) R4
ANSWER: (D)
- GUJCET 2021Set 151 markMCQQ.The radius of the innermost electron orbit of a hydrogen atom is 5.3×10−11 m. What are the radii of the n=4 orbit? (A) 2.12×10−10 m (B) 8.48×10−10 m (C) 4.24×10−10 m (D) 10.6×10−10 m
›Reveal solutionSolution
Bohr radii scale as n^2: r_n = n^2 r1.
Concept. rn=n2r1.
Solution. r4=42×5.3×10−11=16×5.3×10−11=8.48×10−10 m.
✓Final answer(B) 8.48×10−10 m
ANSWER: (B)
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