Question

The first excitation potential of an atom is V. The ionization potential of the atom $V_1$ is

• A. $\frac{5}{4}V$
• B. $\frac{4}{3}V$
• C. $\frac{2}{3}V$
• D. $\frac{1}{3}V$

Answer 1

The correct option is B $\frac{4}{3}V$

$E_{excitation}=E_{ionisation}(\frac{1}{n_1^2}-\frac{1}{n_2^2})$
For transition $n_1=1to{n}_2=2,$
$V_{excitation}=V_{ionisation}(\frac{1}{1}-\frac{1}{2^2})=V_{ionisation}\left(\frac{3}{4}\right)$
$\therefore V_{ionisation}=\frac{4}{3}{V}_{excitation}$