Answer
See the detailed answer below, But....
Note that in my version of the textbook, there is a mistake in the graph for this problem. The error involves the repetition of the term $6s$ at each energy level. You can simply ignore this and work with the correct term next to it. For example, $6s6p$ should actually be $6p$, and so on.
Work Step by Step
$$\color{blue}{\bf [a]}$$
To calculate the wavelength, we need to use the formula of
$$
\Delta E =E_{\rm emitted}= \frac{hc}{\lambda}
$$
So,
$$
\lambda = \frac{hc}{\Delta E}
$$
For each transition, the energy difference \( \Delta E \) is given by:
$$
\Delta E = E_{\text{higher}} - E_{\text{lower}}
$$
So
$$
\lambda = \frac{hc}{(E_{\text{higher}} - E_{\text{lower}})e}
$$
And to get it in nanometers,
$$
\lambda = \frac{hc}{(E_{\text{higher}} - E_{\text{lower}})e(10^{-9})}
$$
Plug the known
$$
\boxed{\lambda = \frac{(6.63\times 10^{-34})(3\times 10^8)}{(E_{\text{higher}} - E_{\text{lower}})(1.6\times 10^{-19})(10^{-9})}}
$$
Assuming that the allowed transitions must satisfy the selection rule $ \Delta l = \pm 1 $, the angular momentum quantum number must change by 1 during the transition.
See the table below.
\begin{array}{|c|c|c|}
\hline
\text{Transition} & \lambda (\text{nm}) \\
\hline
6p \rightarrow 6s & 185 \\\hline
6d \rightarrow 6p & 581 \\\hline
7s \rightarrow 6p & 1019\\\hline
7p \rightarrow 6s & 141 \\\hline
7p \rightarrow 7s & 1351 \\\hline
8s \rightarrow 6p & 493 \\\hline
8s \rightarrow 7p & 3271 \\\hline
8p \rightarrow 6s & 130 \\\hline
8p \rightarrow 7s & 772 \\\hline
8p \rightarrow 8s & 4010 \\\hline
8p \rightarrow 6d & 1802 \\\hline
\hline
\end{array}
$$\color{blue}{\bf [b]}$$
We are given that the 492-nm-wavelength blue emission line in the Hg spectrum corresponds to the transition $ 8s \to 6p $, as we see in the table above.
The energy of the $ 8s $ state is 9.22 eV. The problem asks us to find the minimum kinetic energy that an electron must have to excite the atom from the ground state to the $ 8s $ state.
The minimum kinetic energy $ K $ required to excite the atom is given by
$$
K = \frac{1}{2}mv^2
$$
The energy required to excite the atom to the $ 8s $ state is $ 9.22 \, \text{eV} $. Therefore, the minimum kinetic energy $ K $ of the electron must be at least 9.22 eV:
$$
\frac{1}{2}mv^2 \geq 9.22 (1.6\times 10^{-19})
$$
So the minimum speed is given by
$$
v_{\rm min} = \sqrt{\dfrac{2 (9.22) (1.6\times 10^{-19})}{m}}
$$
Plug the known;
$$
v_{\rm min} = \sqrt{\dfrac{2 (9.22) (1.6\times 10^{-19})}{9.11 \times 10^{-31}}}
$$
$$ v_{\rm min} =\color{red}{\bf 1.80 \times 10^6 }\;\rm m/s$$
