Question

For a hypothetical hydrogen like atom, the potential energy of the system is given by $U\left(r\right)=\frac{-K{e}^2}{r^3}$, where $r$ is the distance between the two particles. If Bohr's model of quantization of angular momentum is applicable, then velocity of particle is given by:

• A. $v=\frac{n^2{h}^3}{K{e}^{2}8{\pi }^3{m}^2}$
• B. $v=\frac{n^3{h}^3}{8K{e}^2{\pi }^3{m}^2}$
• C. $v=\frac{n^3{h}^3}{24K{e}^2{\pi }^3{m}^2}$
• D. $v=\frac{n^2{h}^3}{24K{e}^2{\pi }^3{m}^2}$

Answer 1

The correct option is C $v=\frac{n^3{h}^3}{24K{e}^2{\pi }^3{m}^2}$

$\frac{d\left[U\right(r\left)\right]}{dr}=\frac{3K{e}^2}{r^4}\Rightarrow$ Magnitude of the force
$\frac{3K{e}^2}{r^4}=\frac{m{v}^2}{r}$
and we know, $mvr=\frac{nh}{2\pi }orr=\frac{nh}{2\pi m.v},$ (Bohr model)
$3K{e}^2\times \frac{8{\pi }^3{m}^3{v}^3}{n^3{h}^3}=m{v}^2,v=\frac{n^3{h}^3}{24K{e}^2{\pi }^3{m}^2}$