Question

If the ionisation 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 $91.8V$

First excitation potential is:
$E_2-E_1=I.E-\frac{I.E}{4}=\frac{3}{4}I.E=\frac{3\times 122.4}{4}=91.8V$