Question

For a hypothetical hydrogen like atom, the potential energy of the system is given by $\mathrm{U}(\mathrm{r})=\frac{-{\mathrm{Ke}}^2}{{\mathrm{r}}^3}$, 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

Answer: (C)

$\frac{\mathrm{d}[\mathrm{U}(\mathrm{r})]}{\mathrm{dr}}=\frac{3{\mathrm{Ke}}^2}{{\mathrm{r}}^4}$
$\Rightarrow$ Magnitude of the centripetal force $(=$ Electrostatic force)
$\therefore \frac{3K{e}^2}{r^4}=\frac{m{v}^2}{r}$
and we know mvr $=\frac{nh}{2\pi }$ or $r=\frac{nh}{2\pi m\cdot v}$
$3K{e}^2\times \frac{8{\pi }^3{m}^3{v}^3}{n^3{h}^3}=m{v}^2,v=\frac{n^3{h}^3}{24K{e}^2{\pi }^3{m}^2}$