Question

In hydrogen atom the energy difference between the states $n=2$ and $n=3$ is $EeV$. The ionisation potential of $H$ atom is:

• A. $3.2E$
• B. $5.6E$
• C. $7.2E$
• D. $13.2E$

Answer 1

The correct option is C $7.2E$

Energy difference, $E=13.6[\frac{1}{(2{)}^2}-\frac{1}{(3{)}^2}]$

$⟹E=\frac{5}{9\times 4}\times 13.6$ $eV$

$⟹E=\left(\frac{5}{36}\right)\times 13.6$ $eV$

Ionization potential $=13.6$ $V=\frac{36}{4}E$ $=7.2E$