Physics - Modern Physics Question with Solution | TestHub

The centripetal acceleration of an electron in a circular orbit is given by:

$a=\frac{v^2}{r}$

According to Bohr's quantization condition for angular momentum:

$mvr=\frac{nh}{2\pi }$

For the first Bohr's orbit, $n=1$:

$mvr=\frac{h}{2\pi }$

$v=\frac{h}{2\pi mr}$

Substituting the value of $v$ in the expression for acceleration:

$a=\frac{1}{r}{\left(\frac{h}{2\pi mr}\right)}^2$

$a=\frac{h^2}{4{\pi }^2{m}^2{r}^3}$

Since $4{\pi }^2$ is a constant, the acceleration is proportional to:

$a\propto \frac{h^2}{m^2{r}^3}$

Answer: $\frac{{\mathrm{h}}^2}{{\mathrm{m}}^2{\mathrm{r}}^3}$