MCQ
The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is [ $a _0$ is Bohr radius]:
- A$\frac{h^2}{4 \pi^2 m a_0^2}$
- B$\frac{h^2}{16 \pi^2 ma _0^2}$
- ✓$\frac{h^2}{32 \pi^2 ma _0^2}$
- D$\frac{h^2}{64 \pi^2 ma _0^2}$
✓
Answer
Correct option: C.
$\frac{h^2}{32 \pi^2 ma _0^2}$
c
$m v\left(4 a_0\right)=\frac{h}{\pi} $
$m v\left(4 a_0\right)=\frac{h}{\pi} $
$\text { so, } v=\frac{h}{4 m \pi a_0}$
so $K E=\frac{1}{2} m v^2=\frac{1}{2} m \cdot \frac{h^2}{16 m^2 \pi^2 a_0^2}=\frac{h^2}{32 m \pi^2 a_0^2}$
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