Q.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 =6.0×1024 kg.)
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🔒 Start your 14-day free trial to unlock the full solution →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 …
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 …
Concept: Bohr Model Quantization — Bohr proposed that angular momentum in allowed orbits is an integer multiple of 2πh.
Reasoning:
- The angular momentum of Earth in its orbit is L=mvr.
- Bohr’s quantization condition: L=n2πh, where n is the quantum number.
- Equate and solve for n:
n=h2πmvr
Calculation:
n=6.63×10−342π×(6.0×1024)×(3×104)×(1.5×1011) …
Bohr’s angular momentum quantization condition mvr=nℏ is applied to Earth’s orbit. Plugging in the given values gives n≈2.6×1074, an astronomically large quantum number — showing that classical physics emerges from quantum mechanics for macroscopic systems.
The Bohr model was originally proposed for the hydrogen atom, where the electron’s angular momentum around the nucleus is quantized in integer multiples of ℏ=h/2π. The key insight is that this quantization condition — mvr=nℏ — is not limited to atoms. It can be applied to any orbiting system, including Earth around the Sun. The result tells us how “quantum” the orbit is: a small n means the system is truly quantum, while a huge n (like here) means the orbit behaves classically, because the spacing between adjacent quantum levels becomes vanishingly small.
Let’s work through the numbers step by step.
- Write down the quantization condition. Bohr’s postulate for angular momentum is:
mvr=nℏ
where m is the mass of the orbiting body (Earth), v is its orbital speed, r is the orbital radius, n is the quantum number (an integer), and ℏ=2πh with h=6.626×10−34 J⋅s.
-
Identify the given values.
- r=1.5×1011 m
- v=3×104 m/s
- m=6.0×1024 kg
- h=6.626×10−34 J⋅s, so ℏ=2πh≈1.0546×10−34 J⋅s
-
Calculate the left-hand side: Earth’s angular momentum.
L=mvr=(6.0×1024)×(3×104)×(1.5×1011)
Multiply stepwise:
6.0×3=18, and 1024×104=1028, so mv=18×1028=1.8×1029 kg⋅m/s.
Then L=(1.8×1029)×(1.5×1011)=2.7×1040 kg⋅m2/s.
- Solve for n. From L=nℏ, we have: …
Method: Angular Momentum Quantization (Bohr's Postulate)
Bohr's model says that stable orbits are those where the angular momentum of the revolving body is an integer multiple of 2πh.
Step 1: Write the quantization condition
For any orbiting body in Bohr's model:
mvr=n2πh
where m is mass, v is orbital speed, r is orbit radius, h is Planck's constant, and n is the quantum number (a positive integer).
Step 2: Identify the given values
- m=6.0×1024 kg
- v=3×104 m/s
- r=1.5×1011 m
- h=6.63×10−34 J⋅s (standard value)
Step 3: Solve for n
Rearrange the quantization condition:
n=h2πmvr
Step 4: Substitute and compute
First compute the numerator:
2πmvr=2×3.14×(6.0×1024)×(3×104)×(1.5×1011)
Multiply stepwise:
- 6.0×1024×3×104=18×1028
- 18×1028×1.5×1011=27×1039 …
Common Mistakes in the Bohr Model Quantization Problem
This question asks you to treat the Earth-Sun system as if it obeyed Bohr's angular momentum quantization condition — a purely hypothetical exercise, since Bohr's model applies to atoms, not planetary orbits. The calculation itself is straightforward, but students make several predictable errors.
Mistake 1: Forgetting the n must be an integer
The most fundamental error is to compute a value for n and not check whether it makes physical sense. The Bohr condition is:
mvr=n2πh
Plugging in the numbers:
n=h2πmvr=6.63×10−342π(6.0×1024)(3×104)(1.5×1011)
The numerator is roughly 1.7×1041, and dividing by 6.63×10−34 gives n≈2.6×1074.
How to avoid: Always note that n must be a positive integer. Here it is an astronomically large integer — that's fine conceptually, but if your calculation gave a non-integer like 2.6×1074, you've made an arithmetic error. The actual value is indeed an integer (approximately 2.55×1074), but the point is that Bohr quantization is meaningless for macroscopic orbits.
Mistake 2: Using the wrong value of Planck's constant
Students sometimes use h=6.63×10−34 correctly but then accidentally use h=6.63×10−34 in the denominator when the formula has h/2π. Or they use ℏ=h/2π but forget the factor of 2π entirely.
The Bohr condition is mvr=nℏ, where ℏ=h/2π. If you use h directly, you must write mvr=nh/2π, not mvr=nh.
How to avoid: Write the formula explicitly before substituting numbers. Either use mvr=n2πh or mvr=nℏ — but be consistent.
Mistake 3: Unit mismatch or missing conversions
All quantities must be in SI units. The radius is given in metres, speed in m/s, mass in kg — so no conversion is needed here. But students sometimes treat the radius as 1.5×1011 cm or the speed as 3×104 km/h without converting.
How to avoid: Before substituting, quickly verify each quantity is in SI base units. If any is not, convert first.
Mistake 4: Arithmetic errors with large exponents …
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: …
- 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. …
- 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.
…
- 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 …
- 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 …
- 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
…
- 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: …
- 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).
…
- 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. …
- 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. …
- 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. …
- 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. …
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