Question

The wavelength of the first spectral line of sodium is $5896\mathring{A}$. The first excitation potential of sodium atom will be $\left(h=6.63\times 10^{-34}Js\right)$

• A. 4.2 V
• B. 3.5 V
• C. 2.1 V
• D. None of these

Answer 1

Answer: (C)

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