Answer
${\bf 1.13\times 10^{6}}\;\rm m/s$
Work Step by Step
We are given that the wavelength is 818 nm. The energy corresponding to this wavelength is given by
$$
E = \frac{hc}{\lambda}
$$
Plug the known;
$$
E =\dfrac{(6.63\times 10^{-34})(3\times 10^8)}{(818\times 10^{-9})}=\bf 2.43\times 10^{-19} \;\rm J$$
$$E=\bf 1.52\;\rm eV$$
We can see, from Figure 41.24, that the transition that obeys the selection rule $ \Delta l = 1 $ and has an energy difference of around 1.5 eV is the transition from $ 3p $ to $ 3d $.
The energy difference between $ E_{3d} $ and $ E_{3p} $ is:
$$
E_{3d} - E_{3p} = 3.620 \, \text{eV} - 2.104 \, \text{eV} = 1.516 \, \text{eV}
$$
This value is very close to the energy of the emitted photon ($ 1.52 \, \text{eV} $). This means that the atom was excited from the ground state to the $ 3d $ state.
So, the minimum kinetic energy $ K $ of the electron is given by:
$$
K =\frac{1}{2} mv^2= E_{3d}-E_0
$$
Plug the known from Figure 41.24;
$$
\frac{1}{2} mv^2\geq (3.62-0) (1.6\times 10^{-19})
$$
We know the energy must be at least $ 3.62 \, \text{eV} $.
Now, solve for the speed $ v $:
$$
v \geq\sqrt{ \dfrac{2(3.62-0) (1.6\times 10^{-19})}{m}}
$$
$$
v \geq\sqrt{ \dfrac{2(3.62-0) (1.6\times 10^{-19})}{(9.11\times 10^{-31})}}
$$
So the minimum speed is
$$v_{\rm min}\approx \color{red}{\bf 1.13\times 10^{6}}\;\rm m/s$$
