Answer
c. $3.0$ $eV$
Work Step by Step
The longest wavelength emitted by our hypothetical atom must correspond the the lowest energy transition. $(\lambda \sim \frac{1}{E})$
Now, the allowed transitions are conditioned upon, $\Delta l = 0$.
Note that for the $s$-$state$, $l = 0$, $p$-$state$, $l = 1$, and $d$-$state$, $l = 2$.
Therefore, transitions from, for example, $5p \to 4p$ are $\mathbf{not}$ allowed because $\Delta l =1$ in that case.
The $5p \to 3d$ transition is the allowed transition with the $\mathbf{lowest}$ energy and therefore longest wavelength. Note that the $5p \to 4s$ is allowed but the energy gap is larger than the $5p \to 3d$ transition so it does not satisfy the question.
To find the energy of the $5p \to 3d$ transition simply look at the figure provided. The $5p$ is at $4$ $eV$ and the $3d$ is at $1$ $eV$. So we subtract; $4$ $eV$ $-$ $1$ $eV$ $=$ $3$ $eV$ thus the answer is c.
