A set of atoms in an excited state decays. (a) in general to any of the states with lower energy. (b) into a l

NCERT Exemplar · Q7

Q. A set of atoms in an excited state decays.

(a) in general to any of the states with lower energy.

(b) into a lower state only when excited by an external electric field.

(d) to emit photons only when they collide.

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Concept understanding — Bohr Model Energy Levels

The Intuition: Why Can't an Electron Just Sit Anywhere?

Imagine you're rolling a marble on a staircase. The marble can rest on any step — step 1, step 2, step 3 — but it can never float halfway between two steps. The staircase forces the marble into specific, fixed positions.

That's the core idea of the Bohr model. Before Bohr, physicists thought electrons orbited the nucleus like planets around the sun — they could be at any distance, any energy. But experiments showed something strange: atoms only emit or absorb light at very specific colours (wavelengths), not a continuous rainbow. That meant electrons could only have certain, fixed energies — like the steps of a staircase.

Bohr's genius was to say: an electron in an atom cannot have any arbitrary energy. It can only occupy certain "allowed" energy levels. When it jumps from one level to another, it either absorbs or emits a photon of light whose energy exactly matches the difference between those levels.

The Precise Statement

In the Bohr model of the hydrogen atom (and hydrogen-like ions with one electron), the electron moves in circular orbits around the nucleus. But only those orbits are allowed where the electron's angular momentum is an integer multiple of $\frac{h}{2\pi}$ (where $h$ is Planck's constant).

This quantisation condition leads to a simple formula for the energy of the electron in the $n$ -th orbit:

E_n = -\frac{13.6 \text{ eV}}{n^2}

Here:

$n$ is the principal quantum number — a positive integer ( $n = 1, 2, 3, \dots$ )
$E_n$ is the energy of the electron in that level (in electronvolts)
The negative sign means the electron is bound to the nucleus — you need to add energy to free it

The lowest energy state ( $n=1$ ) is called the ground state. Its energy is $-13.6 \text{ eV}$ . The higher states ( $n=2, 3, 4, \dots$ ) are excited states — they have less negative (higher) energies.

Note

As $n$ increases, the energy levels get closer together. At $n = \infty$ , the energy becomes $0 \text{ eV}$ — the electron is completely free from the atom (ionisation).

What This Explains

When an electron jumps from a higher level ( $n_i$ ) to a lower level ( $n_f$ ), it emits a photon of energy:

\Delta E = E_{n_f} - E_{n_i} = 13.6 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \text{ eV}

This single formula predicts all the spectral lines of hydrogen — the Lyman series (jumps to $n=1$ ), Balmer series (to $n=2$ ), Paschen series (to $n=3$ ), and so on. Each series corresponds to a different "final step" on the staircase. …

Why this formula?

Why the Bohr Model Gives Those Energy Levels

The Bohr model is a beautiful piece of physics because it takes a simple, almost desperate idea — "electrons only exist in certain orbits" — and derives the entire hydrogen spectrum from it. The key is that Bohr didn't just assume the energy levels; he forced them to be consistent with classical physics in one specific way, then broke with it in another.

The Two Non-Negotiable Pieces

First, the electron moves in a circle around the proton. That's pure classical mechanics: the Coulomb attraction provides the centripetal force.

\frac{1}{4\pi\varepsilon_0} \frac{e^2}{r^2} = \frac{mv^2}{r}

This gives you a relation between speed $v$ and radius $r$ :

v^2 = \frac{e^2}{4\pi\varepsilon_0 m r}

Second, the total energy of the electron is the sum of its kinetic and potential energies. Potential energy for a Coulomb force is $-\frac{1}{4\pi\varepsilon_0}\frac{e^2}{r}$ (negative because the force is attractive, and we set zero at infinity).

E = \frac12 mv^2 - \frac{1}{4\pi\varepsilon_0}\frac{e^2}{r}

Substitute $v^2$ from above:

E = \frac12 \left( \frac{e^2}{4\pi\varepsilon_0 r} \right) - \frac{e^2}{4\pi\varepsilon_0 r} = -\frac{1}{2} \frac{e^2}{4\pi\varepsilon_0 r}

So far, nothing is quantised. Any radius $r$ gives a valid classical orbit, and the energy just follows from that radius. The problem is that a classical electron in a curved path radiates energy and spirals into the nucleus — atoms should collapse. Bohr needed a rule to pick out stable orbits.

The Quantisation Condition

Bohr's revolutionary step was to postulate that the angular momentum of the electron is quantised in units of $\hbar = h/2\pi$ :

m v r = n \hbar, \quad n = 1, 2, 3, \dots

Why this particular rule? Bohr later said it was the simplest way to get the right answer. But there's a deeper physical motivation: if you think of the electron as a wave (de Broglie's idea, which came a decade later), the condition that a standing wave fits exactly around the circumference $2\pi r = n \lambda$ gives $m v r = n \hbar$ directly. So the quantisation condition is really a wave condition imposed on a particle picture.

Note

The angular momentum quantisation is the only non-classical assumption in the Bohr model. Everything else follows from classical mechanics and electromagnetism.

Deriving the Allowed Radii and Energies

From $m v r = n \hbar$ , we get $v = n\hbar / (mr)$ . Substitute this into the centripetal force equation:

\frac{1}{4\pi\varepsilon_0} \frac{e^2}{r^2} = \frac{m}{r} \left( \frac{n\hbar}{mr} \right)^2 = \frac{n^2 \hbar^2}{m r^3}

Solve for $r$ :

r_n = \frac{4\pi\varepsilon_0 \hbar^2}{m e^2} \, n^2

The constant in front is the Bohr radius $a_0 \approx 0.529 \, \text{Å}$ . So the radii are $r_n = a_0 n^2$ .

Now plug $r_n$ back into the energy expression $E = -\frac12 \frac{e^2}{4\pi\varepsilon_0 r}$ :

E_n = -\frac12 \frac{e^2}{4\pi\varepsilon_0} \cdot \frac{1}{a_0 n^2} = -\frac{1}{2} \frac{e^2}{4\pi\varepsilon_0 a_0} \cdot \frac{1}{n^2}

Substitute $a_0 = \frac{4\pi\varepsilon_0 \hbar^2}{m e^2}$ :

E_n = -\frac{1}{2} \frac{e^2}{4\pi\varepsilon_0} \cdot \frac{m e^2}{4\pi\varepsilon_0 \hbar^2} \cdot \frac{1}{n^2} = -\frac{m e^4}{8 \varepsilon_0^2 h^2} \cdot \frac{1}{n^2}

E_n = -\frac{13.6 \text{ eV}}{n^2}

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The correct answer is (A) — an excited atom can decay spontaneously to any lower-energy state, subject to selection rules, without needing an external field, simultaneous emission, or collisions.

Why This Question Tests a Core Idea

The question is about spontaneous emission — the process by which an excited atom returns to a lower energy state by emitting a photon. This is a fundamental quantum process that happens on its own, without any external trigger. The key insight is that an excited state can decay to any lower state that is allowed by quantum selection rules, not just the ground state.

Let’s examine each option carefully.

Option (A): "in general to any of the states with lower energy"

This is correct. An atom in an excited state (say energy $E_n$ ) can transition to any lower state $E_m$ (where $m < n$ ) as long as the transition is allowed by conservation laws (especially angular momentum and parity). The emitted photon has energy $E_\gamma = E_n - E_m$ . For example, in hydrogen, the $n=3$ state can decay to $n=2$ (Balmer series) or directly to $n=1$ (Lyman series). The probability of each path is given by the Einstein A coefficient, which depends on the specific states involved.
Option (B): "into a lower state only when excited by an external electric field"

This describes stimulated emission, not spontaneous decay. In stimulated emission, an incoming photon of the right energy triggers the atom to emit an identical photon. But spontaneous emission happens without any external field — it is driven by the interaction of the atom with the vacuum fluctuations of the electromagnetic field. So (B) is false.
Option (C): "all together simultaneously into a lower state"

This would imply that all atoms in an excited state decay at the same instant. In reality, spontaneous emission is a random, probabilistic process. Each atom decays independently, with a characteristic mean lifetime $\tau$ . The number of atoms remaining in the excited state decays exponentially: $N(t) = N_0 e^{-t/\tau}$ . They do not all decay together — that would require a synchronized trigger, which is not present. So (C) is false.
Option (D): "to emit photons only when they collide" …

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Q.A set of atoms in an excited state decays.

Concept understanding — Bohr Model Energy Levels

The Intuition: Why Can't an Electron Just Sit Anywhere?

The Precise Statement

What This Explains

Why this formula?

Why the Bohr Model Gives Those Energy Levels

The Two Non-Negotiable Pieces

The Quantisation Condition

Deriving the Allowed Radii and Energies

Why This Question Tests a Core Idea

Method: Classifying "How Does an Excited Atom Decay" Statements

Steps

Q.A set of atoms in an excited state decays.

Why This Question Tests a Core Idea

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