Q.Let $E_n = -\dfrac{m e^4}{8 n^2 \varepsilon_0^2 h^2}$ be the energy of the $n$th level of H-atom. If all the H-atoms are in the ground state and radiation of frequency $(E_2 - E_1)/h$ falls on it,

The key idea is photon energy: a photon is absorbed only if its energy exactly matches the difference between two energy levels. Here the incident photon energy is $E_2 - E_1$, which is precisely the gap from $n=1$ to $n=2$.

Reasoning:

1. Ground state H-atoms can absorb a photon only if $h\nu = E_f - E_1$ for some final level $f$.
2. Given $\nu = (E_2 - E_1)/h$, the photon energy equals $E_2 - E_1$ exactly. …

The key here is to understand what happens when light of a specific frequency shines on a collection of hydrogen atoms. The energy of a photon is $E = h\nu$. The problem gives the frequency as $\nu = (E_2 - E_1)/h$, so the photon energy is exactly $E_2 - E_1$.

This is the textbook condition for resonant absorption. An electron in the ground state ($n=1$) can absorb this photon only if the photon’s energy precisely matches the gap to some higher energy level. Here, the gap to $n=2$ is exactly $E_2 - E_1$, so absorption is possible.

But does every atom absorb? No. Absorption is a quantum mechanical process — it’s probabilistic. When a beam of photons passes through a gas of atoms, each atom has a certain probability of absorbing a photon. Some atoms will absorb and get excited; others will not. So we never get “all atoms” excited unless the light is extremely intense and the interaction time is long enough — and even then, stimulated emission competes. The question’s phrasing “all atoms” is a trap.

Also, can an atom in the ground state absorb this photon and jump to $n=3$? No — the energy required for $n=1 \to n=3$ is $E_3 - E_1$, which is larger than $E_2 - E_1$. A single photon cannot supply that energy unless its frequency is higher. So option (D) is actually correct: no atom can go to $n=3$ with this photon.

Let’s go through the options step by step.

1. Photon energy matches exactly one transition.

   The photon energy is $h\nu = E_2 - E_1$. This is the precise energy difference between the ground state ($n=1$) and the first excited state ($n=2$). So absorption is energetically allowed.

2. Not all atoms absorb — absorption is probabilistic.

   When a beam of photons passes through a sample, each atom has a certain cross-section for absorption. Some atoms will absorb a photon and excite to $n=2$; others will not. The fraction that absorbs depends on the intensity and duration of exposure, but it is never “all” in a typical scenario unless specified otherwise. Option (C) says “all atoms will be excited” — that is false.

3. The $n=3$ state is unreachable with this photon.

   The energy needed to go from $n=1$ to $n=3$ is $E_3 - E_1$. Since $E_3$ is higher than $E_2$, this gap is larger than $E_2 - E_1$. A single photon of energy $E_2 - E_1$ cannot bridge that gap. So no atom can transition to $n=3$ via this photon. Option (D) is correct. …

Method: The Discrete-Resonance Decision Procedure

For any "photon of a given frequency shines on ground-state atoms" question, run through this fixed three-question decision procedure instead of reasoning freshly about each option.

Steps

Step 1: List every transition energy available from the initial state

From the ground state, the atom can only absorb a photon whose energy exactly equals $E_m - E_1$ for some integer $m$. Write out the first few: $E_2-E_1$, $E_3-E_1$, $E_4-E_1$, ... — these are the only "doors" the photon can walk through.

Step 2: Compare the given photon energy against every door, not just the obvious one

The photon here has energy exactly $E_2-E_1$ — so the $n=1\to2$ door is open. But check the other doors too: is $h\nu$ also equal to $E_3-E_1$? No — that gap is strictly larger. So the $n=3$ door stays firmly shut, regardless of how many photons or how intense the beam is, because resonant absorption is an exact-match condition, not an "energy is enough" condition.

Step 3: Apply the "all atoms" red-flag heuristic …