Question

If the ionization potential of an atom is $122.4V$, then its first excitation potential will be

• A. $71.8V$
• B. $91.8V$
• C. $51.8V$
• D. $101.8V$

Answer 1

The correct option is B $91.8V$

Ionization Potential $=122.4eV$
$\Rightarrow 122.4=13.6\times \frac{Z^2}{1^2}$ --- (1)

Then $1^{st}$ excitation potential is
$E=13.6\times \frac{Z^2}{1}[\frac{1}{1^2}-\frac{1}{2^2}]$
$E=13.6Z^2\times \frac{3}{4}$ --- (2)

Dividing (1) and (2) :
$E=91.8eV$