Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We draw the atom’s energy levels diagram as shown below.
$$\color{blue}{\bf [b]}$$
Now we need to find the wavelengths that appear in the atom’s emission spectrum.
Recalling that
$$E=\dfrac{hc}{\lambda} $$
Hence,
$$\lambda=\dfrac{hc}{E} \tag 1$$
And we have here 3 levels of energy emitted, see the figure below:
$\bullet$ $E_{2\rightarrow 1}=4-0=\bf 4\;\rm eV$
$\bullet$ $E_{3\rightarrow 1}=6-0=\bf6\;\rm eV$
$\bullet$ $E_{3\rightarrow 2}=6-4=\bf2\;\rm eV$
Plug these into (1) to find the emitted wavelengths.
$$\lambda_{2\rightarrow 1}=\dfrac{hc}{E_{2\rightarrow 1}} =\dfrac{(4.14\times 10^{-15})(3\times 10^8)}{(4)}=\color{red}{\bf 311}\;\rm nm$$
$$\lambda_{3\rightarrow 1}=\dfrac{hc}{E_{3\rightarrow 1}} =\dfrac{(4.14\times 10^{-15})(3\times 10^8)}{(6)}=\color{red}{\bf 207}\;\rm nm$$
$$\lambda_{3\rightarrow 2}=\dfrac{hc}{E_{3\rightarrow 2}} =\dfrac{(4.14\times 10^{-15})(3\times 10^8)}{(2)}=\color{red}{\bf 621}\;\rm nm$$
$$\color{blue}{\bf [c]}$$
Now we need to find the wavelengths that appear in the atom’s absorption spectrum.
And we have here 2 levels of energy absorption:
$\bullet$ $E_{1\rightarrow 2}=4-0=\bf 4\;\rm eV$
$\bullet$ $E_{1\rightarrow 3}=6-0=\bf6\;\rm eV$
Plug these into (1) to find the absorption wavelengths.
$$\lambda_{1\rightarrow 2}=\dfrac{hc}{E_{1\rightarrow 2}} =\dfrac{(4.14\times 10^{-15})(3\times 10^8)}{(4)}=\color{red}{\bf 311}\;\rm nm$$
$$\lambda_{1\rightarrow 3}=\dfrac{hc}{E_{1\rightarrow 3}} =\dfrac{(4.14\times 10^{-15})(3\times 10^8)}{(6)}=\color{red}{\bf 207}\;\rm nm$$
