Q.According to the classical electromagnetic theory, calculate the initial frequency of the light emitted by the electron revolving around a proton in hydrogen atom.
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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 …
The key idea is that in classical electromagnetic theory, an accelerating charge radiates energy at the same frequency as its orbital motion. For the hydrogen atom, the electron's orbital frequency is found from the centripetal force condition and the Bohr radius.
Step 1: Orbital frequency from force balance
The centripetal force is provided by the Coulomb attraction:
rmv2=4πε01r2e2
The orbital frequency is f=2πrv.
Step 2: Express v and r
From the force equation, v2=4πε0mre2.
Using the Bohr radius r=a0=me24πε0ℏ2≈5.29×10−11 m, we get:
The classical electromagnetic theory predicts that an electron orbiting a proton radiates energy continuously, causing its orbit to shrink and the emitted light frequency to increase. The initial frequency of the emitted light equals the orbital frequency of the electron in its ground state, which is approximately 6.6×1015Hz.
Why Classical Theory Fails — and What It Predicts
The question asks you to step into the shoes of a 19th-century physicist, before quantum mechanics. According to classical electrodynamics, an accelerating charge radiates electromagnetic waves. An electron orbiting a proton is constantly accelerating (centripetal acceleration), so it must continuously lose energy by emitting light.
This is a disaster for the classical model: as the electron loses energy, it spirals into the nucleus, and the frequency of the emitted light changes continuously. But the problem asks for the initial frequency — the frequency of light emitted at the very start, when the electron is in its smallest stable orbit (the Bohr radius).
The key insight: the frequency of the emitted light equals the orbital frequency of the electron, because the electron's circular motion generates a wave at that same frequency.
For an electron in a circular orbit of radius r with speed v, the orbital frequency is:
f=2πrv
Step-by-Step Calculation
1. Set up the force balance for a hydrogen atom
The electron (charge −e) orbits a proton (charge +e) at a distance r. The Coulomb force provides the centripetal acceleration:
4πϵ01r2e2=rmev2
where me=9.11×10−31kg, e=1.60×10−19C, and ϵ0=8.85×10−12C2/N⋅m2.
2. Solve for the orbital speed v
From the force equation:
v2=4πϵ01mere2
So:
v=4πϵ0mere2
3. Use the Bohr radius for the initial orbit
The smallest stable orbit in the classical sense corresponds to the Bohr radius (the ground state radius in quantum mechanics, which classical theory cannot derive — but we use it as the starting point):
r=a0=5.29×10−11m
You can derive the Bohr radius from the quantization condition mevr=nℏ with n=1, but the problem assumes you know it. In exams, a0=0.529A˚ is a standard constant.
4. Calculate the orbital frequency
First, find v:
v=(9.11×10−31kg)(5.29×10−11m)(9.00×109N⋅m2/C2)(1.60×10−19C)2
Let's compute step by step:
- Numerator: (9.00×109)(2.56×10−38)=2.304×10−28
- Denominator: (9.11×10−31)(5.29×10−11)=4.82×10−41 …
Method: Bohr's Frequency Condition (Quantum Jump Model)
This problem is a trick — classical electromagnetic theory cannot correctly predict the frequency of light emitted by a hydrogen atom. Classical physics says an accelerating electron radiates continuously, spiralling into the nucleus. The actual discrete spectrum comes from quantum mechanics. However, the exam often expects you to use Bohr's frequency condition, which bridges classical orbit ideas with quantum jumps.
Classical theory alone gives a continuous spectrum, not a single initial frequency. The method below uses Bohr's model, which is semi-classical and is the standard approach for such problems in Indian exams.
Steps
Step 1: Recall Bohr's frequency condition
When an electron jumps from a higher orbit (n2) to a lower orbit (n1), the frequency of emitted light is:
f=hEn2−En1
where h=6.63×10−34 J⋅s is Planck's constant.
Step 2: Write the energy of the electron in the n-th orbit of hydrogen
From Bohr's model:
En=−n213.6 eV
Convert to joules if needed: 1 eV=1.6×10−19 J.
Step 3: Identify the "initial" transition
The phrase "initial frequency" usually means the first emission line of the Lyman series (highest energy jump): from n=2 to n=1.
Step 4: Calculate the energy difference
E2−E1=(−413.6)−(−13.6)=−3.4+13.6=10.2 eV
In joules:
ΔE=10.2×1.6×10−19=1.632×10−18 J …
Common Mistakes on This Question (Photon Energy & Classical Hydrogen)
This question is a classic trap — it asks you to use classical electromagnetic theory to calculate something that classical theory cannot correctly describe. The very act of doing the calculation reveals why classical physics fails for the atom. Here are the mistakes students make most often.
Mistake 1: Forgetting that classical theory predicts a continuous spectrum, not a single frequency
Students often try to calculate one "initial frequency" and stop there. But classical electrodynamics says an accelerating charge radiates at the instantaneous orbital frequency of the electron. Since the electron spirals inward as it loses energy, this frequency changes continuously — there is no single answer.
How to avoid: Recognise that the question is asking for the frequency at the start, when the electron is in its ground-state orbit (Bohr radius a0). You must calculate the orbital frequency from the centripetal force condition, then state that this is the initial frequency of the emitted radiation — and that it will increase as the electron spirals in.
Mistake 2: Using the wrong radius or velocity
Some students plug in r=0.529A˚ without deriving it, or they use the Bohr radius but forget it comes from quantisation. Others use r=10−10m as a guess.
How to avoid: Derive the orbital radius from the Coulomb force and circular motion:
rmv2=r2ke2
This gives v=mrke2. But you still need r. In classical theory, there is no fixed r — so you must choose the ground-state Bohr radius a0=0.529×10−10m as the starting point. State this assumption clearly.
Mistake 3: Confusing orbital frequency with photon frequency
Students sometimes calculate the orbital period T and then say the photon frequency is f=1/T. That is correct for classical radiation — the emitted wave has the same frequency as the orbital motion. But then they stop, not realising this is the initial frequency only.
How to avoid: After finding f=2πrv, explicitly note: "According to classical theory, the radiation frequency equals the orbital frequency at that instant."
Mistake 4: Arithmetic errors in the final calculation
The numbers are messy: k=9×109, e=1.6×10−19, m=9.1×10−31, a0=5.29×10−11. Students often misplace exponents or forget to square e.
How to avoid: Work step by step with symbols first, then substitute once. The cleanest path:
- From rmv2=r2ke2, get v=mrke2.
- Orbital frequency: f=2πrv=2π1mr3ke2.
- Substitute r=a0 and compute.
f=2π1ma03ke2
Plugging in:
f=2π1(9.1×10−31)(5.29×10−11)3(9×109)(1.6×10−19)2
Compute the denominator inside the square root first: ma03=9.1×10−31×1.48×10−31≈1.35×10−61. Then numerator: ke2=9×109×2.56×10−38=2.30×10−28. The ratio is about 1.70×1033, square root gives 4.12×1016, divided by 2π gives f≈6.6×1015Hz. …
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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