MCQ
One energy difference between the states $n = 2$ and $n = 3$ is $E\ eV$ , in hydrogen atom. The ionisation potential of $H$ atom is ............. $\mathrm{E}$
- A$3.2$
- B$5.6$
- ✓$7.2$
- D$13.2$
✓
Answer
Correct option: C.
$7.2$
c
Energy difference $=E=E_{o}\left[\frac{1}{(2)^{2}}-\frac{1}{(3)^{2}}\right]=\frac{5 E_{o}}{36}$
Energy difference $=E=E_{o}\left[\frac{1}{(2)^{2}}-\frac{1}{(3)^{2}}\right]=\frac{5 E_{o}}{36}$
$\Rightarrow E_{o}=7.2 E$
Ionization potential= $E^{1}=E_{o}\left[\frac{1}{(1)^{2}}-\frac{1}{\infty}\right]=E_{o}$
$\therefore E^{\prime}=7.2 E$
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