MCQ
If in Bohr's atomic model, it is assumed that force between electron and proton varies inversely as $r^4$, energy of the system will be proportional to
- A$n^2$
- B$n^4$
- ✓$n^6$
- D$n^8$
✓
Answer
Correct option: C.
$n^6$
c
$\frac{m v^2}{r}=\frac{k}{r^4}$
$\frac{m v^2}{r}=\frac{k}{r^4}$
$V^2=\frac{k}{m r^3}$
we know
$m v r=\frac{n h}{2 x}$
$m^2 v^2 r^2=\frac{n^2 h^2}{4 \pi^2}$
$\frac{m^2 k r}{m r}=\frac{n^2 h^2}{4 x^2}$
$r=\frac{m k^3 \pi^2}{n^2 h^2}$
$k \cdot f=\frac{1}{2} m v^2=\frac{1}{2} m / \frac{k}{m r^3}$
$=\frac{1}{2} m^m \frac{K}{m^1 m^3}$
$E \alpha n^6$
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