In a hydrogen atom, the electron revolving in the fourth orbit has angular momentum equal to (a) h (b) (h)/(π)

I Multiple Choice Questions · Q2

Q. In a hydrogen atom, the electron revolving in the fourth orbit has angular momentum equal to

(a)

h

(b)

\dfrac{h}{\pi}

(c)

\dfrac{4h}{\pi}

(d)

\dfrac{2h}{\pi}

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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 $n$

Watch out

Bohr's model works perfectly only for single-electron systems: H, He $^+$ , Li $^{2+}$ , etc. It fails for multi-electron atoms because it ignores electron-electron repulsion and the wave nature of electrons.

Why "quantization"?

The word comes from the Latin quantus — "how much." In classical physics, angular momentum can take any value. In Bohr's atom, it comes only in discrete packets (quanta) of size $\hbar$ . This is the first hint that at the atomic scale, nature is not continuous but granular.

The electron does not spiral because it cannot lose energy gradually — it can only jump from one allowed orbit to another, emitting or absorbing a photon of exactly the right energy. Between these jumps, it simply exists in a stationary state, defying classical expectations.

Bohr's quantization of angular momentum is one of the defining postulates covered in the NCERT Class 12 Physics Atoms chapter, and students frequently search for "Bohr model quantization condition and derivation" or "Bohr's model important questions" while preparing for CBSE boards and JEE Main/NEET. This concept is also a common launching point for numerical problems on orbital radius and energy levels of hydrogen-like atoms tested across competitive exams.

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}

Watch out

A common mistake is to think Bohr derived the quantization rule from first principles. He didn't — he postulated it. The de Broglie standing-wave argument came later and provides the physical reason for the postulate, but it is still a postulate in the full quantum theory.

The deeper reason: it's not really about orbits

The Bohr model is ultimately wrong — electrons don't orbit in neat circles. But the quantization of angular momentum survives in the full quantum mechanical treatment (Schrödinger equation) as the condition that the wavefunction must be single-valued. For the hydrogen atom, the angular momentum quantum number $l$ can take values $0, 1, 2, \dots, n-1$ , and the magnitude is $\sqrt{l(l+1)}\hbar$ , not $n\hbar$ .

Yet the Bohr model's key insight — that only certain discrete states are allowed — remains the foundation of atomic physics. The formula $L = n\hbar$ is the simplest example of a quantum number, and it correctly predicts the hydrogen spectrum to within fine-structure corrections.

Important

The Bohr quantization condition $L = n\hbar$ is a boundary condition on the electron wave, not a dynamical law. It says: for the electron to exist in a stable state, its wave must fit perfectly around the nucleus. This is the same principle that governs standing waves on a string or in an organ pipe — only certain wavelengths survive.

Step 1. Bohr's quantisation condition states the angular momentum of an electron in the $n$ th orbit is $l=n\dfrac{h}{2\pi}$ .

Step 2. For the fourth orbit, $n=4$ :

l=4\times\frac{h}{2\pi}=\frac{4h}{2\pi}=\frac{2h}{\pi}

Step 3. This matches option (d). Option (a) would be $n=2\pi$ (not an integer orbit label), option (b) is the $n=1$ (ground state) value, and option (c) $4h/\pi$ is twice too large (a common slip is forgetting the factor of $2$ in $2\pi$ in the denominator).

✓Final answer

\dfrac{2h}{\pi}

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Q.In a hydrogen atom, the electron revolving in the fourth orbit has angular momentum equal to

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