$A$ $12.5\; eV$ electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted?

(A) The energy of the electron beam is $12.5\; eV$. The ground state energy of hydrogen is $E_1 = -13.6\; eV$.
When bombarded,the hydrogen atom can absorb energy to reach an excited state $E_n = E_1 + 12.5\; eV = -13.6 + 12.5 = -1.1\; eV$.
Using the formula $E_n = -13.6 / n^2$,we find $n^2 = -13.6 / -1.1 \approx 12.36$,which implies $n \approx 3.5$. Thus,the electron can be excited up to the $n = 3$ level.
During de-excitation from $n = 3$,the possible transitions are $n = 3 \to 2$,$n = 2 \to 1$,and $n = 3 \to 1$.
$1$. For $n = 3 \to 1$ (Lyman series): $\frac{1}{\lambda} = R_y (\frac{1}{1^2} - \frac{1}{3^2}) = 1.097 \times 10^7 \times \frac{8}{9} \implies \lambda \approx 102.6\; nm$.
$2$. For $n = 2 \to 1$ (Lyman series): $\frac{1}{\lambda} = R_y (\frac{1}{1^2} - \frac{1}{2^2}) = 1.097 \times 10^7 \times \frac{3}{4} \implies \lambda \approx 121.6\; nm$.
$3$. For $n = 3 \to 2$ (Balmer series): $\frac{1}{\lambda} = R_y (\frac{1}{2^2} - \frac{1}{3^2}) = 1.097 \times 10^7 \times \frac{5}{36} \implies \lambda \approx 656.3\; nm$.
Thus,wavelengths from both the Lyman and Balmer series are emitted.