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 25 on this concept.
- AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQQ.An electron in the ground state of hydrogen atom is revolving in anti-clockwise direction in a circular orbit. The orbital magnetic moment of the electron is given by (A) 2πh (B) 2πmeh (C) 4πmeh (D) 4πmh
›Reveal solutionSolution
This tests the gyromagnetic-ratio relation between orbital magnetic moment and angular momentum for the ground-state hydrogen electron; the answer is 4πmeh.
Concept and Intuition
A charge moving in a circular orbit constitutes a tiny current loop, and any current loop has a magnetic moment μ=IA. This can always be re-expressed in terms of the particle's orbital angular momentum L via the classical gyromagnetic ratio μL=2meL, valid regardless of orbit size or speed, as long as we know L. For hydrogen's ground state, the Bohr model fixes L=2πh (i.e., n=1, L=nℏ), so combining the two relations gives the magnetic moment directly.
Step-by-Step Solution
- Orbital magnetic moment in terms of angular momentum: μL=2meL (standard result for a charge −e orbiting; magnitude used here). …
- AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQQ.The energy of an electron in Bohr's hydrogen atom is −3.4 eV. The angular momentum of the electron is (A) π2h (B) 2πh (C) πh (D) 4πh
›Reveal solutionSolution
Identify the orbit number from the given energy, then use Bohr's quantization rule L=nh/2π. E=−3.4 eV corresponds to n=2, giving L=h/π.
Concept and Intuition
In the Bohr model, an electron in the n-th orbit of hydrogen has energy En=−13.6/n2 eV, and its orbital angular momentum is quantized as Ln=nℏ=2πnh — this quantization condition is the postulate that let Bohr explain the discrete hydrogen spectrum.
Step-by-Step Solution
- Given En=−3.4 eV. Set −n213.6=−3.4⇒n2=3.413.6=4⇒n=2. …
- AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQQ.If the ratio of the time periods of the electrons revolving in the first and nth orbits of Hydrogen atom is 1 : 64, then the angular momentum of the electron in the nth excited state of Hydrogen atom is (h – Planck's constant) (A) π3.5h (B) π5h (C) π2.5h (D) π2h
›Reveal solutionSolution
T∝n3 gives orbit n=4; the 'nth (fourth) excited state' means principal number 5, so L=5⋅2πh=π2.5h.
Concept and Intuition
In Bohr's model r∝n2 and v∝1/n, so the period T=v2πr∝n3. The period ratio pins the orbit number. The angular momentum is quantised as L=n2πh.
Step-by-Step Solution
- TnT1=n313=641⇒n3=64⇒n=4.
- The 'nth excited state' with n=4 means the fourth excited state; counting ground =1, 1st excited =2, …, 4th excited corresponds to principal quantum number 5. …
- AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQQ.Suppose an electron is attracted towards the origin by a force rK, where K is a constant and r is the distance of the electron from the origin. By applying Bohr model to this system, the radius of the nth orbit of the electron is found to be rn, and the kinetic energy of the electron to be Tn, then which of the following is true? (A) Tn is independent of n, rn∝n (B) Tn∝n1, rn∝n (C) Tn∝n1, rn∝n2 (D) Tn∝n21, rn∝n2
›Reveal solutionSolution
An unusual force law F=K/r makes the orbital speed the same in every Bohr orbit, so kinetic energy doesn't depend on n at all, while quantized angular momentum then forces the radius to grow linearly with n. Answer: (A).
Concept and Intuition
Bohr's model has two ingredients regardless of the force law: (i) the given force supplies the centripetal force for circular motion, and (ii) angular momentum is quantized, mvrn=nℏ. Normally (Coulomb force ∝1/r2) both v and r depend on n in specific ways; here the unusual 1/r force makes the speed itself independent of r (hence of n), which is the key simplifying feature of this problem.
Step-by-Step Solution
- Centripetal condition: rmv2=rK⇒mv2=K, i.e. v=K/m — a constant, the same for every orbit (independent of r or n).
- Kinetic energy: Tn=21mv2=2K — a constant, independent of n.
- Bohr's angular-momentum quantization: mvrn=nℏ. …
- AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQQ.An electron has an angular momentum of 90ℏ J−S while orbiting with a linear velocity of π×105 ms−1 then the radius of the orbit is (mass of electron =9×10−3 Kg and Planks Constant =6.6×10−34 J s) (A) 66×10−15 m (B) 33×10−15 m (C) 66×10−10 m (D) 33×10−8 m
›Reveal solutionSolution
Angular momentum of a circulating particle is L=mvr, so the orbit radius is r=L/(mv), with L given as a multiple of ℏ=h/2π. Answer: ≈3.3×10−8 m.
Concept and Intuition
For a particle moving in a circular orbit with speed v at radius r, its angular momentum about the center is simply L=mvr (mass times linear momentum times the lever arm, which here is the radius itself since velocity is tangential/perpendicular to the radius vector). Given the angular momentum in units of the reduced Planck constant (L=nℏ), we can invert this relation to solve directly for the radius.
Step-by-Step Solution
- Angular momentum: L=nℏ=n2πh, with n=90, h=6.6×10−34 Js. L=90×2π6.6×10−34=90×1.05×10−34≈9.45×10−33 Js.
- Since L=mvr, the radius is r=mvL.
- Momentum: mv=(9×10−31)×(π×105)≈2.83×10−25 kgm/s. …
- AP EAPCET 2026Set ap-2026-05-19-AN1 markMCQQ.The ratio of areas of 2nd and 3rd Bohr's orbits in a doubly ionized Lithium atom is (A) 16 : 81 (B) 4 : 5 (C) 4 : 9 (D) 2 : 3
›Reveal solutionSolution
Bohr orbit radius scales as n2, so orbit area scales as n4; for n=2 and n=3 this gives an area ratio of 16:81.
Concept and Intuition
In the Bohr model, rn=Zn2a0. Since Z (here, for doubly-ionised lithium, Z=3) is the same for both orbits being compared, it cancels out in any ratio of radii for the same atom/ion. The area of a circular orbit scales as the square of the radius, so it scales as n4.
Step-by-Step Solution
- rn∝n2/Z; since Z is fixed (same ion, Z=3), rn∝n2.
- Area An=πrn2∝n4.
- Ratio of areas of 2nd and 3rd orbits: A3A2=3424=8116. …
- AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQQ.An electron is moving in an orbit of hydrogen atom in which there can be a maximum of six transitions. Another electron is moving in an another orbit of hydrogen atom in which there can be maximum of three transitions. The ratio of the velocity of electrons in these two orbits is (A) 3/4 (B) 5/4 (C) 2/1 (D) 1/2
›Reveal solutionSolution
Maximum-transitions count fixes the principal quantum numbers as n=4 and n=3; since vn∝1/n, their speed ratio is 3/4.
Concept and Intuition
In the Bohr model, an electron in level n can make a transition to any of the (n−1) lower levels, and the total number of distinct spectral lines obtainable from levels 1 to n (or equivalently, the number of possible transitions starting from the topmost level n when all electrons are in it) is (2n)=2n(n−1). Once we know n for each orbit, we use the Bohr result that orbital speed vn=nv1∝n1 — higher orbits move slower.
Step-by-Step Solution
- First orbit: 2n(n−1)=6⇒n(n−1)=12⇒n=4 (since 4×3=12).
- Second orbit: 2n(n−1)=3⇒n(n−1)=6⇒n=3 (since 3×2=6).
- Bohr orbital speed: vn∝n1, so v4∝41 and v3∝31. …
- AP EAPCET 2025Set ap-2025-05-19-AN1 markMCQQ.If the angular momenta of electrons in two orbits of hydrogen atom are πh and π1.5h, then the ratio of velocities of electrons in these two orbits is (h - Planck's constant) (A) 3 : 2 (B) 3 : 4 (C) 9 : 4 (D) 1 : 3
›Reveal solutionSolution
Converting the given angular momenta to orbit numbers (n1=2,n2=3) and using vn∝1/n gives v1:v2=3:2.
Concept and Intuition
Bohr's quantization condition states the angular momentum of an electron in the n-th orbit is Ln=n2πh. Also, from the Bohr model, the orbital speed is vn=2ε0nhZe2, i.e. vn∝n1 — electrons in higher (larger) orbits move slower. So once we identify which orbit numbers correspond to the given angular momenta, the velocity ratio follows immediately by inverting the n ratio.
Step-by-Step Solution
- First orbit: L1=πh=n12πh⟹n1=2.
- Second orbit: L2=π1.5h=n22πh⟹n2=3. …
- AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQQ.The ratio of the time periods of the revolution of the electrons in the second and third excited states of hydrogen atom is (A) 9:16 (B) 27:64 (C) 4:9 (D) 8:27
›Reveal solutionSolution
Using Bohr's model, the orbital period scales as n3; for n=3 (second excited) and n=4 (third excited) the ratio is 27:64.
Concept and Intuition
In the Bohr model, orbit radius grows as rn∝n2 while orbital speed falls as vn∝1/n. The time period is Tn=vn2πrn, so combining these gives Tn∝n2×n=n3 — outer orbits take dramatically longer to complete one revolution.
Step-by-Step Solution
- Identify the principal quantum numbers: ground state is n=1, so first excited =n=2, second excited =n=3, third excited =n=4.
- Since Tn∝n3: T4T3=4333=6427. …
- AP EAPCET 2024Set ap-2024-05-17-AN1 markMCQQ.If the angular momentum of an electron in the second orbit of hydrogen atom is J, then the angular momentum of an electron in the third excited state of hydrogen atom is (A) 2 J (B) 3 J (C) 4 J (D) 6 J
›Reveal solutionSolution
Bohr's angular momentum quantization (Ln∝n) turns the given L2=J into L4=2J for the third excited state (n=4).
Concept and Intuition
Bohr's model postulates that the angular momentum of an electron in the n-th orbit is quantized as Ln=2πnh — it simply scales linearly with the orbit number n. The key subtlety here is counting orbits correctly: the ground state is n=1, so the "first excited state" is n=2, the "second excited state" is n=3, and the "third excited state" is n=4.
Step-by-Step Solution
- Given: angular momentum in the second orbit (n=2) is J. So L2=2π2h=J, giving 2πh=2J.
- Identify the third excited state: ground state n=1 → 1st excited n=2 → 2nd excited n=3 → 3rd excited n=4. …
- AP EAPCET 2024Set eng-2024-05-20-FN1 markMCQQ.μ – meson of charge 'e', mass 208 me moves in a circular orbit around a heavy nucleus having charge +3e. The quantum state 'n' for which the radius of the orbit is same as that of the first Bohr orbit for hydrogen atom is [approximately] (A) n≈20 (B) n≈25 (C) n≈28 (D) n≈29
›Reveal solutionSolution
The Bohr radius scales as n2/(Z⋅m); matching the muon's orbit radius (Z=3, m=208
electron masses) to hydrogen's first Bohr radius gives n2=624, i.e. n≈25.
Concept and Intuition
The Bohr model radius for a hydrogen-like system is rn=Zme2/(4πϵ0)n2ℏ2∝Zmn2 (for fixed fundamental constants), where m is the orbiting
particle's mass. A heavier, more charge-attracted particle (the muon, with m=208me around
Z=3) needs a much larger quantum number n to reach the same radius as the electron's smallest
hydrogen orbit.
Step-by-Step Solution
- Write rn(muon)=Zn2⋅mμmea0, where a0 is hydrogen's first Bohr radius (with n=1,Z=1,m=me).
- We want rn(muon)=a0 (matches the first Bohr orbit of hydrogen). …
- AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQQ.The speed of the electron is a hydrogen atom in the n=3 level is (Plank constant =6.6×10−34 Js) (A) 6.2×105 ms−1 (B) 3.7×105 ms−1 (C) 7.3×105 ms−1 (D) 1.6×105 ms−1
›Reveal solutionSolution
The Bohr-model orbital speed scales as 1/n; using the standard first-orbit speed of hydrogen, the n=3 speed comes out to about 7.3×105 m/s.
Concept and Intuition
In Bohr's model, angular momentum is quantized (mvr=nh/2π) and the Coulomb force provides centripetal force. Solving these together gives an orbital speed vn=2ε0nhe2=nv1 for hydrogen, where v1≈2.18×106 ms−1 is the speed in the ground state (n=1). Higher orbits therefore have progressively lower orbital speeds.
Step-by-Step Solution
- Use vn=v1/n with v1=2.18×106 ms−1 (standard hydrogen ground-state speed, consistent with e2/2ε0h using the given Planck constant).
- For n=3: v3=2.18×106/3=7.27×105 ms−1. …
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