Three photons originating from an excited atomic-hydrogen sample...

The energy levels of a hydrogen atom are given by $E_n = -\dfrac{13.6}{n^2} \text{ eV}$.

The energies of the first few levels are:
$E_1 = -13.6 \text{ eV}$
$E_2 = -3.4 \text{ eV}$
$E_3 = -1.51 \text{ eV}$

The energy differences between these levels are:
$E_3 - E_1 = -1.51 - (-13.6) = 12.09 \text{ eV} \approx 12.1 \text{ eV}$
$E_2 - E_1 = -3.4 - (-13.6) = 10.2 \text{ eV}$
$E_3 - E_2 = -1.51 - (-3.4) = 1.89 \text{ eV} \approx 1.9 \text{ eV}$

The detected photons correspond to the transitions $n=3 \to n=1$ ($12.1 \text{ eV}$), $n=2 \to n=1$ ($10.2 \text{ eV}$), and $n=3 \to n=2$ ($1.9 \text{ eV}$).

A single hydrogen atom in the $n=3$ state can either transition directly to $n=1$ (emitting one photon of $12.1 \text{ eV}$) or transition to $n=2$ and then to $n=1$ (emitting two photons of $1.9 \text{ eV}$ and $10.2 \text{ eV}$). It cannot emit all three photons as it only has one electron that eventually reaches the ground state.

To observe all three photons, we need at least two atoms:
Case 1 (Two atoms): One atom undergoes the $3 \to 1$ transition, and the second atom undergoes the $3 \to 2 \to 1$ transitions.
Case 2 (Three atoms): One atom undergoes $3 \to 1$, another undergoes $3 \to 2$, and a third undergoes $2 \to 1$.

Therefore, the photons must have been emitted by either two atoms or three atoms.