When a hydrogen atom absorbs an energy of 10.2 eV, the change in its angular momentum is : (a) $4.14\times10^{-15}$ Js (b) $0.525\times10^{-34}$ Js (c) $1.05\times10^{-34}$ Js (d) $2.1\times10^{-34}$ Js

When a hydrogen atom absorbs an energy of 10.2 eV, the change in its angular momentum is : (a) 4.14×10⁻¹⁵ Js (

Question 98 of 127

Q. When a hydrogen atom absorbs an energy of 10.2 eV, the change in its angular momentum is :

(a)

4.14\times10^{-15}

(b)

0.525\times10^{-34}

(c)

1.05\times10^{-34}

(d)

2.1\times10^{-34}

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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 = n \frac{h}{2\pi}, \quad n = 1, 2, 3, \dots

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 $L = m v r$ , where $m$ is the electron mass, $v$ its speed, and $r$ the orbit radius. So the quantization condition becomes:

m v r = n \frac{h}{2\pi}

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.

Note

The quantity $\frac{h}{2\pi}$ appears so often that it has its own symbol: $\hbar$ (h-bar). So the condition is often written as $L = n\hbar$ .

What it predicts

Combining this quantization with Newton's law for circular motion (centripetal force = Coulomb attraction) gives:

Radius of the $n$ th orbit: $r_n = n^2 a_0$ , where $a_0 = 0.529 \, \text{Å}$ is the Bohr radius (the smallest orbit, $n=1$ ).
Energy of the $n$ th orbit: $E_n = -\frac{13.6 \, \text{eV}}{n^2}$

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 $mvr = n\hbar$
Write the force balance: $\frac{mv^2}{r} = \frac{kZe^2}{r^2}$ (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 $\frac{h}{2\pi}$ — 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:

\lambda = \frac{h}{p} = \frac{h}{mv}

For an electron in a circular orbit of radius $r$ , the circumference is $2\pi r$ . For the wave to be stable — not cancelling itself out — the circumference must contain an integer number of wavelengths:

2\pi r = n\lambda, \quad n = 1, 2, 3, \dots

Substitute $\lambda = h/(mv)$ :

2\pi r = n \cdot \frac{h}{mv}

Rearrange:

mvr = n \cdot \frac{h}{2\pi}

That's it. The left side $mvr$ is the angular momentum $L$ . So:

L = n\hbar, \quad \text{where } \hbar = \frac{h}{2\pi}

Note

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:

\frac{mv^2}{r} = \frac{1}{4\pi\epsilon_0} \frac{e^2}{r^2}

Combine this with $mvr = n\hbar$ and solve for $r$ and $E$ :

r_n = \frac{4\pi\epsilon_0 \hbar^2}{m e^2} \cdot n^2 = a_0 n^2

$E_n = -\frac{m e^4}{8\epsilon_0^2 h^2} \cdot \frac{1}{n^2} = -\frac{13.6 \text{ eV}}{n^2}$ …

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Absorbing 10.2 eV takes a hydrogen electron from $n=1$ to $n=2$ , and the resulting change in orbital angular momentum is $h/2\pi \approx 1.05\times10^{-34}$ Js.

The energy of the $n$ -th stationary state of the hydrogen atom is $E_n = -\dfrac{13.6}{n^2}$ eV. The ground state energy is $E_1=-13.6$ eV and the first excited state energy is $E_2=-3.4$ eV.

The energy absorbed for a transition from $n=1$ to $n=2$ is: $\Delta E = E_2-E_1 = -3.4-(-13.6) = 10.2$ eV

This exactly matches the given absorbed energy, confirming the transition is $n=1 \to n=2$ .

According to Bohr's quantisation condition, the orbital angular momentum in the $n$ -th orbit is $L_n = \dfrac{nh}{2\pi}$ , where $h$ is Planck's constant.

…

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 29 on this concept.

CBSE 2026Set 55/2/11 markMCQ
$n$
›Reveal solutionSolution
In the Bohr model, the orbital radius scales as $r_n \propto n^2$ . For large $n$ , the gap between consecutive orbits $\Delta r = r_{n+1} - r_n$ 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
$r_n = \frac{n^2 h^2 \varepsilon_0}{\pi m e^2} \quad \text{or, more compactly,} \quad r_n = a_0 n^2,$
where $a_0 \approx 0.529 \, \text{Å}$ is the Bohr radius. So the radius grows as $n^2$ .

The question asks about the distance between consecutive orbits — that is, the difference $r_{n+1} - r_n$ — for large $n$ . Many students instinctively think this difference is constant, or that it grows like $n^2$ 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:

$r_n = a_0 n^2, \qquad r_{n+1} = a_0 (n+1)^2.$

The gap between them is

$\Delta r = r_{n+1} - r_n = a_0 \left[ (n+1)^2 - n^2 \right].$

Expand $(n+1)^2 = n^2 + 2n + 1$ . Then

$\Delta r = a_0 \left( n^2 + 2n + 1 - n^2 \right) = a_0 (2n + 1).$

For large $n$ , the constant $1$ becomes negligible compared to $2n$ . So

$\Delta r \approx 2 a_0 n.$
Thus the spacing between consecutive orbits is proportional to $n$ itself — not $n^2$ , not $\sqrt{n}$ , not $n^3$ . …
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CBSE 2026Set A1 markMCQ
$Q. The radius of the lowest Bohr's orbit in hydrogen atom is r0. The radius of Bohr's second orbit is (A) r0 (B) 2r0 (C) 4r0 (D) r0/2$
›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:

$r_n=n^2 r_0$

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. If the first Bohr radius of hydrogen atom be R, then the radius of the third orbit is$
(a) 9R
(b) R/3
(c) 3R
(d) R/9
›Reveal solutionSolution

Bohr radius of the nth orbit scales as $n^2$ , so the third orbit's radius is $9R$ .

In the Bohr model, the radius of the $n$ th orbit of the hydrogen atom is

$r_n = n^2 r_1$

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. Case study: Bohr's model addressed the instability of the Rutherford model by introducing quantization. The model is based on three postulates, which successfully explained the discrete line spectrum of hydrogen. According to the model, the radius of the nth stationary orbit is r_n \propto n², and the total energy is E_n = -13.6 eV / n², when an electron jumps from a higher energy level (E_i) to a lower one (E_f), a photon of energy hν = E_i - E_f is emitted. Transitions ending at the n = 1 level form the Lyman series. According to Bohr's second postulate, which quantity is quantized?$
(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 $h/2\pi$ .

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 $\hbar = h/2\pi$ : …

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CBSE 2025Set ANNUAL1 markMCQ
$Q. According to Bohr's hypothesis the following physical quantity is quantised$
(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 $h/2\pi$ 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 $h/2\pi$ :

$L = mvr = \dfrac{nh}{2\pi}, \quad n = 1, 2, 3, \ldots$

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. The radius of Bohr's stable orbit for hydrogen is r. The radius of Bohr's second orbit is$
(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:

…

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CBSE 2025Set ANNUAL1 mark
$Q. If radius of first electron orbit of hydrogen is r0, radius of second electron orbit of hydrogen is ______.$
›Reveal solutionSolution

In the Bohr model, the radius of the nth orbit of hydrogen scales as $r_n = n^2 r_0$ , where $r_0$ is the first-orbit (Bohr) radius.

Bohr's model gives the radius of the nth stationary orbit of the hydrogen atom as

$r_n = n^2 r_0$ …

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CBSE 2024Set ANNUAL1 mark
$Q. If the radius of first orbit of hydrogen atom is 0.5 x 10^-10 m, then the radius of its second orbit will be __________ m.$
›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:

$r_n = n^2 r_1$

…

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CBSE 2024Set ANNUAL1 mark
$Q. What is Bohr's quantisation condition for the angular momentum of an electron in the second orbit ?$
›Reveal solutionSolution

Bohr's second postulate: angular momentum is quantised as $L=\dfrac{nh}{2\pi}$ ; 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 $h/2\pi$ : …

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CBSE 2023Set 55/1/11 markMCQ
$n^{\text{th}}$
(a) $\dfrac{1}{n^2}$
(b) $\dfrac{1}{n}$
(c) $n^2$
(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 $n^2$ .

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:
$\frac{mv^2}{r} = \frac{ke^2}{r^2}$
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:
$mvr = n\hbar$
where $\hbar = \frac{h}{2\pi}$ and $n = 1, 2, 3, \ldots$ is the principal quantum number.

Deriving the radius dependence

From the angular momentum condition, solve for $v$ :

$v = \frac{n\hbar}{mr}$

Substitute this into the force balance equation:

$\frac{m}{r}\left(\frac{n\hbar}{mr}\right)^2 = \frac{ke^2}{r^2}$

Simplify the left side:

$\frac{m \cdot n^2\hbar^2}{m^2r^3} = \frac{n^2\hbar^2}{mr^3} = \frac{ke^2}{r^2}$

Multiply both sides by $r^3$ :

$\frac{n^2\hbar^2}{m} = ke^2 r$

Solve for $r$ : $r = \frac{n^2\hbar^2}{mke^2}$ …
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CBSE 2023Set ANNUAL1 mark
$Q. Write the mathematical form of Bohr's postulate regarding angular momentum of electron in atom. (Write the answer only)$
›Reveal solutionSolution

Bohr's second postulate states that the angular momentum of the electron in a stationary orbit is an integral multiple of $h/2\pi$ .

Bohr postulated that an electron can revolve only in those orbits for which its orbital angular momentum is quantized:

$L = mvr = \dfrac{nh}{2\pi}, \quad n = 1, 2, 3, \ldots$

…

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CBSE 2023Set ANNUAL1 markMCQ
$Q. The Bohr model of atoms$
(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: $L=mvr=\dfrac{nh}{2\pi}$ , $n=1,2,3,\dots$ . 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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Q.When a hydrogen atom absorbs an energy of 10.2 eV, the change in its angular momentum is :

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