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 $E=h v=\frac{h c}{\lambda}$
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 $1.6 \times 10^{-19} J=1\, e V$
$\therefore E=\frac{3.37 \times 10^{-19}}{1.6 \times 10^{-19}} e V$
$E=2.1 \,eV$
Hence, corresponding first excitation potential is $2.1\, V$.