Answer
$ 3 \to 2 = 658 \, \text{nm} $
$ 3 \to 1 = 103 \, \text{nm} $
$ 2 \to 1= 122\, \text{nm} $
Work Step by Step
First, we need to determine the highest energy level allowed.
We know that
$$
E = e V
$$
where $ e $ is the electron charge and $ V $ is the potential difference.
In this case, $ V = 12.5 \, \text{V} $, so:
$$
E =\bf 12.5 \,\rm eV\tag 1
$$
We know that the energy levels of a hydrogen atom are given by
$$
E_n = - \frac{13.6}{n^2} \, \text{eV}
$$
where $ n $ is the principal quantum number.
So to find the highest energy level $ n $ that the hydrogen atom can be excited to, we need to find the value of $ n $ such that the energy difference between the ground state ( $ n = 1 $ ) and level $ n $ equals the energy gained by the electron (12.5 eV).
$$
E_n - E_1 = 12.5 \, \text{eV}
$$
$$
- \frac{13.6}{n^2} + 13.6 = 12.5
$$
Solving for $n$;
$$
\frac{13.6}{n^2} = 1.1
$$
$$
n^2 = \frac{13.6}{1.1} \approx 12.36
$$
$$
n \approx 3.5
$$
Since $ n $ must be an integer, the highest allowed energy level is
$$ n = \bf 3 $$.
Now we can see that the possible quantum-jump transitions are:
$ 3 \to 2 $, $ 3 \to 1 $, $ 2 \to 1 $
Recalling that the energy difference between two levels is given by:
$$
\Delta E = E_{n_i} - E_{n_f}=\dfrac{hc}{\lambda}
$$
So,
$$
\lambda = \dfrac{hc}{\Delta E }
$$
Transition $ 3 \to 2 $:
$$
\lambda_{3\to 2} =\dfrac{hc}{ E_3-E_2 } =\dfrac{(6.63\times 10^{-34})(3\times 10^8)}{\left(- \frac{13.6}{9} + \frac{13.6}{4} \right)(1.6\times 10^{-19})}
$$
$$
\lambda_{3\to 2} =\color{red}{\bf 658} \, \text{nm}
$$
Transition $ 3 \to 1 $:
$$
\lambda_{3\to 1} =\dfrac{hc}{ E_3-E_1} =\dfrac{(6.63\times 10^{-34})(3\times 10^8)}{\left(- \frac{13.6}{9} + 13.6 \right)(1.6\times 10^{-19})}
$$
$$
\lambda_{3\to 1} =\color{red}{\bf 103} \, \text{nm}
$$
Transition $ 2 \to 1 $:
$$
\lambda_{2\to 1} =\dfrac{hc}{ E_2-E_1} =\dfrac{(6.63\times 10^{-34})(3\times 10^8)}{\left(- \frac{13.6}{4} + 13.6 \right)(1.6\times 10^{-19})}
$$
$$
\lambda_{2\to 1} =\color{red}{\bf 122} \, \text{nm}
$$
