Q.(a) Using the Bohr's model calculate the speed of the electron in a hydrogen atom in the n=1,2, and 3 levels.
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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 — the electron's angular momentum is quantised, mevnrn=n2πh, and the Coulomb force supplies the centripetal force.
- Speed of the electron
Solving the force-balance and quantisation equations together gives vn=2ε0nhe2, which falls as 1/n. Substituting the constants gives v1≈2.19×106 m/s, so:
v1≈2.19×106 m/s,v2=2v1≈1.09×106 m/s,v3=3v1≈7.29×105 m/s
- Orbital period Using rn=n2a0 (with a0≈5.29×10−11 m) and Tn=vn2πrn: …
Bohr's model quantises angular momentum, which gives the electron's speed as vn=2ε0nhe2. For hydrogen, v1≈2.19×106 m/s, v2≈1.09×106 m/s, v3≈7.29×105 m/s. The orbital period Tn=vn2πrn then gives T1≈1.52×10−16 s, T2≈1.22×10−15 s, T3≈4.10×10−15 s.
Why Bohr's model works for this
Bohr's model combines classical circular motion with one quantum condition: the electron's angular momentum is an integer multiple of 2πh. The Coulomb force provides the centripetal force, and quantising the angular momentum ties the speed v to the orbit radius r — solving the two together gives both in terms of n alone.
Step-by-step calculation
1. The two governing equations
For an electron of mass me and charge −e orbiting a proton (charge +e) in a circular orbit of radius r with speed v:
- Coulomb force = centripetal force:
4πε01r2e2=rmev2
- Bohr's quantisation of angular momentum:
mevr=n2πh,n=1,2,3,…
2. Solve for the speed vn
Eliminating r between these two equations gives:
vn=2ε0nhe2
The speed falls as 1/n — higher orbits mean slower electrons.
3. Substitute the constants
Using e=1.602×10−19 C, ε0=8.854×10−12 F/m, h=6.626×10−34 J⋅s:
2ε0he2=2×8.854×10−12×6.626×10−34(1.602×10−19)2≈2.19×106 m/s
Since vn=v1/n:
- v1≈2.19×106 m/s
- v2=v1/2≈1.09×106 m/s
- v3=v1/3≈7.29×105 m/s
v1 is close to c/137 — the fine-structure constant α=2ε0hce2≈1371 appears naturally here. This is why relativistic corrections to the hydrogen atom are small.
4. Find the orbital radius rn
From the two governing equations, rn=n2a0, where a0=πmee2ε0h2≈5.29×10−11 m is the Bohr radius:
- r1=5.29×10−11 m …
Method: Bohr's Quantization of Angular Momentum
This problem uses the Bohr quantization condition — the idea that angular momentum comes in discrete packets — combined with the Coulomb force providing the centripetal acceleration.
Step 1: Write the two governing equations
Quantization of angular momentum (Bohr's postulate):
mevr=n2πh,n=1,2,3,…
Coulomb force = centripetal force (for a hydrogen nucleus with charge +e):
4πε01r2e2=rmev2
Where:
- me=9.11×10−31 kg
- e=1.60×10−19 C
- h=6.63×10−34 J⋅s
- ε0=8.85×10−12 C2/N⋅m2
Step 2: Solve for speed v in terms of n
From the quantization condition: r=2πmevnh
Substitute into the force equation:
4πε01(2πmevnh)2e2=2πmevnhmev2
This simplifies to:
4πε01n2h2e2⋅4π2me2v2=nh2πmev2
Cancel v2 (non-zero) and rearrange:
ε0n2h2e2me⋅π=nh2πmev
Cancel π and me:
ε0n2h2e2=nh2v
Multiply both sides by nh:
ε0nhe2=2v
vn=2ε0nhe2
This is the speed of the electron in the nth Bohr orbit.
Step 3: Calculate v1, v2, v3
First compute the constant factor:
2ε0he2=2(8.85×10−12)(6.63×10−34)(1.60×10−19)2
Numerator: 2.56×10−38
Denominator: 2×8.85×10−12×6.63×10−34=1.173×10−44
So the constant =1.173×10−442.56×10−38=2.18×106 m/s
Therefore:
vn=n2.18×106 m/s
| n | vn (m/s) |
|---|---|
| 1 | 2.18×106 |
| 2 | 1.09×106 |
| 3 | 7.27×105 |
v1≈c/137, the fine-structure constant times c. This is a famous result — the electron in the ground state moves at about 1% of the speed of light.
Part (b): Orbital period
Method: Period T=speedcircumference=v2πr
We need r for each n. From the quantization condition:
rn=2πmevnnh=2πmenh⋅e22ε0nh=πmee2ε0n2h2
rn=πmee2ε0n2h2
This is the Bohr radius a0=5.29×10−11 m when n=1. …
Common Mistakes in Bohr Model Calculations
Students often lose marks on this exact problem because they rush through the algebra or misapply the quantization condition. Let me walk through the most frequent errors and how to fix each.
Mistake 1: Using the wrong formula for velocity
Many students try to derive velocity from mvr=2πnh alone, forgetting that the Coulomb force provides the centripetal force. They end up with an expression that still contains r, which they don't know yet.
How to avoid: Always start from the force balance equation:
rmv2=r2ke2
This gives v2=mrke2. Then combine with the quantization condition mvr=nℏ (where ℏ=h/2π) to eliminate r. You get:
v=nℏke2
This is the clean, direct formula. Memorise it — it saves time and prevents algebra errors.
vn=nℏke2=nh2πke2
Mistake 2: Plugging in numbers with inconsistent units
Students use k=9×109 (SI), e=1.6×10−19 C, but then use h=6.63×10−34 J·s — all correct — but forget that ℏ=h/2π, not h itself. This off-by-a-factor-of-2π error is extremely common.
How to avoid: Write ℏ explicitly as h/2π in your formula before substituting numbers. For n=1:
v1=h2πke2
Now substitute: k=9×109, e=1.6×10−19, h=6.63×10−34.
v1=6.63×10−342π(9×109)(1.6×10−19)2
Calculate stepwise: e2=2.56×10−38, so numerator = 2π×9×109×2.56×10−38=2π×2.304×10−28≈1.447×10−27. Divide by 6.63×10−34 to get v1≈2.18×106 m/s.
A quick check: the answer should be about 2.2×106 m/s for n=1. If you get something like 1.4×107 or 3.4×105, you've likely used h instead of ℏ or vice versa.
Mistake 3: Forgetting that v∝1/n
Once you have v1, students sometimes recalculate everything from scratch for n=2 and n=3, wasting time and inviting arithmetic errors.
How to avoid: From the formula vn=nℏke2, it's clear that vn=v1/n. So:
- v2=v1/2≈1.09×106 m/s
- v3=v1/3≈7.27×105 m/s
No need to redo the full substitution.
Mistake 4: Confusing orbital period with frequency
For part (b), students often write T=v2πr but then use the wrong r or forget that r also depends on n.
How to avoid: First, recall that rn=n2a0, where a0=mke2ℏ2≈5.29×10−11 m is the Bohr radius. Then:
Tn=vn2πrn=v1/n2π(n2a0)=v12πa0⋅n3
So Tn∝n3. Calculate T1 once, then multiply by n3 for higher levels.
For n=1:
T1=2.18×1062π(5.29×10−11)≈1.52×10−16 s …
Showing the 12 most recent of 29 on this concept.
- CBSE 2026Set 55/2/11 markMCQQ.In Bohr model of hydrogen atom, for large values of n, the distance between the consecutive orbits is proportional to (A) n (B) n (C) n2 (D) n3
›Reveal solutionSolution
In the Bohr model, the orbital radius scales as rn∝n2. For large n, the gap between consecutive orbits Δr=rn+1−rn behaves like 2n+1, which is proportional to n — so the answer is (B).
The Bohr model gives a beautifully simple picture of the hydrogen atom: electrons orbit the nucleus in fixed circular paths, with quantised angular momentum. The radius of the n-th orbit is
rn=πme2n2h2ε0or, more compactly,rn=a0n2,
where a0≈0.529A˚ is the Bohr radius. So the radius grows as n2.
The question asks about the distance between consecutive orbits — that is, the difference rn+1−rn — for large n. Many students instinctively think this difference is constant, or that it grows like n2 because the radii themselves do. But a difference between two quadratic terms behaves differently from either term alone. Let’s work it out.
- Write the radii for two neighbouring orbits:
rn=a0n2,rn+1=a0(n+1)2.
- The gap between them is
Δr=rn+1−rn=a0[(n+1)2−n2].
- Expand (n+1)2=n2+2n+1. Then
Δr=a0(n2+2n+1−n2)=a0(2n+1).
- For large n, the constant 1 becomes negligible compared to 2n. So
Δr≈2a0n.
Thus the spacing between consecutive orbits is proportional to n itself — not n2, not n, not n3. …
- CBSE 2026Set A1 markMCQQ.The radius of the lowest Bohr's orbit in hydrogen atom is r0. The radius of Bohr's second orbit is (A) r0 (B) 2r0 (C) 4r0 (D) r0/2
›Reveal solutionSolution
Bohr radii scale as n², so r₂ = 4 r₀.
In Bohr's model of hydrogen the radius of the n-th orbit is:
rn=n2r0
…
- CBSE 2026Set ANNUAL1 markMCQQ.If the first Bohr radius of hydrogen atom be R, then the radius of the third orbit is(a) 9R(b) R/3(c) 3R(d) R/9
›Reveal solutionSolution
Bohr radius of the nth orbit scales as n2, so the third orbit's radius is 9R.
In the Bohr model, the radius of the nth orbit of the hydrogen atom is
rn=n2r1
…
- CBSE 2026Set ANNUAL1 markMCQQ.Case study: Bohr's model addressed the instability of the Rutherford model by introducing quantization. The model is based on three postulates, which successfully explained the discrete line spectrum of hydrogen. According to the model, the radius of the nth stationary orbit is r_n ∝ n², and the total energy is E_n = −13.6 eV / n², when an electron jumps from a higher energy level (E_i) to a lower one (E_f), a photon of energy hν = E_i − E_f is emitted. Transitions ending at the n = 1 level form the Lyman series. According to Bohr's second postulate, which quantity is quantized?(a) Energy of electron(b) Orbital angular momentum(c) Linear momentum(d) Frequency of revolution
›Reveal solutionSolution
Bohr's second postulate quantizes the electron's orbital angular momentum in integer multiples of h/2π.
Bohr's second postulate states that an electron can revolve only in those orbits for which its orbital angular momentum is an integral multiple of ℏ=h/2π: …
- CBSE 2025Set ANNUAL1 markMCQQ.According to Bohr's hypothesis the following physical quantity is quantised(a) angular momentum(b) angular velocity(c) potential energy(d) momentum
›Reveal solutionSolution
Bohr's key postulate (beyond classical mechanics) was that only orbits where the electron's angular momentum is an integer multiple of h/2π are allowed.
Bohr's second postulate states that the electron can revolve only in those orbits for which its orbital angular momentum is an integral multiple of h/2π:
L=mvr=2πnh,n=1,2,3,…
…
- CBSE 2025Set ANNUAL1 markMCQQ.The radius of Bohr's stable orbit for hydrogen is r. The radius of Bohr's second orbit is(a) r(b) r/2(c) 2r(d) 4r
›Reveal solutionSolution
In the Bohr model, the radius of the nth orbit of hydrogen scales as n^2, so if the (first, ground-state) orbit has radius r, the second orbit has radius 4r.
Bohr's model gives the radius of the nth stationary orbit of hydrogen as:
r_n = n^2 r_1
where r_1 is the radius of the first (n=1) orbit. Taking the given "stable orbit" radius r to be r_1, the second orbit (n=2) has radius:
…
- CBSE 2025Set ANNUAL1 markQ.If radius of first electron orbit of hydrogen is r0, radius of second electron orbit of hydrogen is ______.
›Reveal solutionSolution
In the Bohr model, the radius of the nth orbit of hydrogen scales as rn=n2r0, where r0 is the first-orbit (Bohr) radius.
Bohr's model gives the radius of the nth stationary orbit of the hydrogen atom as
rn=n2r0 …
- CBSE 2024Set ANNUAL1 markQ.If the radius of first orbit of hydrogen atom is 0.5 x 10^-10 m, then the radius of its second orbit will be __________ m.
›Reveal solutionSolution
Bohr's model gives orbit radius proportional to n², so the second orbit's radius is 2² = 4 times the first orbit's radius.
In Bohr's model of the hydrogen atom, the radius of the nth orbit is:
rn=n2r1
…
- CBSE 2024Set ANNUAL1 markQ.What is Bohr's quantisation condition for the angular momentum of an electron in the second orbit ?
›Reveal solutionSolution
Bohr's second postulate: angular momentum is quantised as L=2πnh; for the second orbit, n=2.
Bohr's quantisation condition states that an electron can revolve only in those orbits for which its orbital angular momentum is an integral multiple of h/2π: …
- CBSE 2023Set 55/1/11 markMCQQ.The radius of the nth orbit in Bohr model of hydrogen atom is proportional to :(a) n21(b) n1(c) n2(d) n
›Reveal solutionSolution
In the Bohr model, the electron's orbit radius grows with the principal quantum number because higher orbits require more angular momentum and lower electrostatic attraction. The radius is proportional to n2.
The Bohr model treats the hydrogen atom as a miniature solar system where the electron orbits the nucleus in circular paths. But unlike planets, the electron can only occupy certain allowed orbits, determined by quantum conditions. The question asks how the orbital radius scales with the quantum number n.
The key insight is that two forces govern the electron's motion: the electrostatic attraction pulling it inward and the requirement that its angular momentum be quantized. Let me show you how these constraints lead to the radius formula.
The physics behind the orbit
For a stable circular orbit, the centripetal force must equal the electrostatic force:
rmv2=r2ke2
where m is the electron mass, v its speed, k is Coulomb's constant, and e the electron charge. This gives us one equation relating r and v.
Bohr's quantum condition provides the second equation. He postulated that angular momentum is quantized:
mvr=nℏ
where ℏ=2πh and n=1,2,3,… is the principal quantum number.
Deriving the radius dependence
- From the angular momentum condition, solve for v:
v=mrnℏ
- Substitute this into the force balance equation:
rm(mrnℏ)2=r2ke2
- Simplify the left side:
m2r3m⋅n2ℏ2=mr3n2ℏ2=r2ke2
- Multiply both sides by r3:
mn2ℏ2=ke2r
- Solve for r: r=mke2n2ℏ2 …
- CBSE 2023Set ANNUAL1 markQ.Write the mathematical form of Bohr's postulate regarding angular momentum of electron in atom. (Write the answer only)
›Reveal solutionSolution
Bohr's second postulate states that the angular momentum of the electron in a stationary orbit is an integral multiple of h/2π.
Bohr postulated that an electron can revolve only in those orbits for which its orbital angular momentum is quantized:
L=mvr=2πnh,n=1,2,3,…
…
- CBSE 2023Set ANNUAL1 markMCQQ.The Bohr model of atoms(1) assumes that the angular momentum of electrons is quantized(2) uses Einstein's photoelectric equation(3) predicts continuous emission spectra for atoms(4) predicts the same emission spectra for all types of atoms
›Reveal solutionSolution
Bohr's central postulate was that electrons can only occupy orbits where angular momentum is an integer multiple of h/2π.
Bohr postulated that an electron revolves only in those orbits for which its angular momentum is quantized: L=mvr=2πnh, n=1,2,3,…. This quantization condition (not option b, which is unrelated to Bohr's model; and not options c/d, since Bohr's model correctly predicts DISCRETE, …
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