Question

If the ionisation potential of an atom is 122.4 V. The second excitation potential of the atom will be

• A. $108.8V$
• B. $8.8V$
• C. $208.8V$
• D. $308.8V$

Answer 1

The correct option is A $108.8V$

Ionisation potenital = $122.4=13.6\times \frac{Z^2}{1^2}$ --- (1)
For $2^{nd}$ excitation, $E=13.6\times Z^2\times [\frac{1}{1^2}-\frac{1}{3^2}]$ --- (2)
Dividing (1) and (2): $E=108.8V$