What is the minimum energy that must be given to a $H$ atom in the ground state so that it can emit an $H_{\gamma}$ line in the Balmer series? If the angular momentum of the system is conserved,what would be the angular momentum of such an $H_{\gamma}$ photon?

(D) For the Balmer series of the atomic spectrum of hydrogen,the transition ends at $n=2$. The lines are defined as $H_{\alpha} (n=3 \to 2)$,$H_{\beta} (n=4 \to 2)$,$H_{\gamma} (n=5 \to 2)$,and so on.
To emit an $H_{\gamma}$ line,the electron must be excited from the ground state $(n=1)$ to the $n=5$ energy level.
The energy required is $\Delta E = E_{5} - E_{1}$.
Using $E_{n} = -\frac{13.6 \text{ eV}}{n^{2}}$,we get:
$\Delta E = -\frac{13.6}{5^{2}} - (-13.6) = 13.6 \left(1 - \frac{1}{25}\right) = 13.6 \times \frac{24}{25} = 13.056 \text{ eV}$.
The angular momentum of the emitted photon is equal to the change in the orbital angular momentum of the electron during the transition from $n=5$ to $n=2$.
Using Bohr's quantization condition $L = \frac{nh}{2\pi}$:
$L_{\text{photon}} = L_{5} - L_{2} = \frac{5h}{2\pi} - \frac{2h}{2\pi} = \frac{3h}{2\pi}$.
Substituting $h = 6.626 \times 10^{-34} \text{ Js}$:
$L_{\text{photon}} = \frac{3 \times 6.626 \times 10^{-34}}{2 \times 3.14159} \approx 3.165 \times 10^{-34} \text{ Js}$.