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

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)
$\therefore \frac{3 Ke ^{2}}{ r ^{4}}=\frac{ mv ^{2}}{ r }$
and we know mvr $=\frac{n h}{2 \pi}$ or $r=\frac{n h}{2 \pi m \cdot v}$
$3 Ke ^{2} \times \frac{8 \pi^{3} m ^{3} v ^{3}}{ n ^{3} h ^{3}}= mv ^{2}, v=\frac{ n ^{3} h ^{3}}{24 Ke ^{2} \pi^{3} m ^{2}}$