Answer
See the detailed answer below.
Work Step by Step
a) We sketched the energy level diagram of this unknown element, as you see in the figure below.
b) When the atom abosorbs an entry of 5.1 eV, it sends it to the third excited state at $n=4$ which has an energy of -6.4 eV, as we see in the figure below.
The eenrgyies of the emitted photons will be as follows:
- When the electron moves from $n=4$ to $n=3$.
$$E_p=\Delta E_e=E_4-E_3=-6.4-(-6.8)=\color{red}{\bf 0.4}\;\rm eV$$
- When the electron moves from $n=4$ to $n=2$.
$$E_p=\Delta E_e=E_4-E_2=-6.4-(-9)=\color{red}{\bf 2.6}\;\rm eV$$
- When the electron moves from $n=4$ to $n=1$.
$$E_p=\Delta E_e=E_4-E_1=-6.4-(-11.5)=\color{red}{\bf 5.1}\;\rm eV$$
- When the electron moves from $n=3$ to $n=2$.
$$E_p=\Delta E_e=E_3-E_2=-6.8-(-9)=\color{red}{\bf 2.2}\;\rm eV$$
- When the electron moves from $n=3$ to $n=1$.
$$E_p=\Delta E_e=E_3-E_1=-6.8-(-11.5)=\color{red}{\bf 4.7}\;\rm eV$$
- When the electron moves from $n=2$ to $n=1$.
$$E_p=\Delta E_e=E_2-E_1=-9-(-11.5)=\color{red}{\bf 2.5}\;\rm eV$$