Answer
See the detailed answer below.
Work Step by Step
When a hydrogen atom collides with a free electron with 12.75 eV kinetic energy it absorbs this amount of energy that makes its electron excited and go for farther orbits. After a while, the electron goes back to its original orbit and emits a photon.
The author asks us to find all emitted photons' wavelengths in this process.
We know that the hydrogen atoms are initially on the ground state at $n=1$. Now we need to find its state after absorbing the 12.75 eV.
We know that
$$\Delta E=E_n-E_1$$
And hence,
$$E_n=\Delta E+E_1 $$
where $\Delta E$ here is the kinetic energy $KE$ of the free electron.
$$E_n=KE+E_1 $$
Plugging the known;
$$\dfrac{-13.6 }{n^2}=12.75+\dfrac{-13.6}{1^2}=-0.85$$
Solving for $n$;
$$n^2=\dfrac{-13.6 }{-0.85}$$
$$n =\sqrt{\dfrac{ 13.6 }{ 0.85}}=4$$
Now we know that the maximum state for the hydrogen atom when absorbing the 12.75 Ev electron is at $n=4$.
And hence, the emitted photons occur when the electron cascades back to the ground state from $n=4$ or $n=3$, or $n=2$.
$$\lambda=\dfrac{hc}{E_n-E_{n'}}$$
But remember to convert the energy to the units of joules from electronvolt.
$$\lambda =\dfrac{6.626\times10^{-34}\times3\times10^8}{\left(E_n-E_{n'}\right)\times1.6\times10^{-19}}$$
$$\boxed{\lambda=\dfrac{1.24\times10^{-6} }{ E_n-E_{n'}}}$$
$\Rightarrow$ From $n=4$ to $n=3$;
$$ \lambda=\dfrac{1.24\times10^{-6} }{ \dfrac{-13.6}{4^2}-\dfrac{-13.6}{3^2}} =\color{red}{\bf 1875.6 }\;\rm nm$$
$\Rightarrow$ From $n=4$ to $n=2$;
$$ \lambda=\dfrac{1.24\times10^{-6} }{ \dfrac{-13.6}{4^2}-\dfrac{-13.6}{2^2}} =\color{red}{\bf 486.3}\;\rm nm$$
$\Rightarrow$ From $n=4$ to $n=1$;
$$ \lambda=\dfrac{1.24\times10^{-6} }{ \dfrac{-13.6}{4^2}-\dfrac{-13.6}{1^2}} =\color{red}{\bf 97.25}\;\rm nm$$
$\Rightarrow$ From $n=3$ to $n=2$;
$$ \lambda=\dfrac{1.24\times10^{-6} }{ \dfrac{-13.6}{3^2}-\dfrac{-13.6}{2^2}} =\color{red}{\bf 656.5}\;\rm nm$$
$\Rightarrow$ From $n=3$ to $n=1$;
$$ \lambda=\dfrac{1.24\times10^{-6} }{ \dfrac{-13.6}{3^2}-\dfrac{-13.6}{1^2}} =\color{red}{\bf 102.6}\;\rm nm$$
$\Rightarrow$ From $n=2$ to $n=1$;
$$ \lambda=\dfrac{1.24\times10^{-6} }{ \dfrac{-13.6}{2^2}-\dfrac{-13.6}{1^2}} =\color{red}{\bf 121.6}\;\rm nm$$
