Question

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

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

Answer 1

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

$magnitudeoftheforce\Rightarrow \frac{d\left[U\right(r\left)\right]}{dr}=\frac{3k{e}^2}{r^4}$
$\frac{3k{e}^2}{r^4}=\frac{m{v}^2}{r}\left(i\right)$
we know that
$mvr=\frac{nh}{2\pi }\left(ii\right)$
By substituting value of $r$ from $\left(ii\right)$ to $\left(i\right)$ we get
$v=\frac{n^3{h}^3}{24K{e}^2{\pi }^3{m}^2}$