Explanation

(c) The energy of first excitation of sodium is
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ E = hv$=$\frac{hc}{\lambda}$
Where $h$ is Planck’s constants, $v$ is frequency, $c$ is speed of light and $\lambda$ is wavelength.
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ E = \frac{6.63\ \times 10^{- 34} \times 3\ \times 10^{8}}{5896\ \times 10^{- 10}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ E = \ 3.37\ \times 10 - 19\ J$
Also since $\ 1.6\ \times \ 10 - 19\ J\ = \ 1eV$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ E\ = \frac{3.37\ \times 10^{- 19}}{1.6\ \times 10^{- 19}}\ eV$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ E = \ 2.1\ eV$
Hence ,corresponding first excitation potential is 2.1 V.