Question

The ratio of the speed of an electron in the first Bohr orbit of hydrogen to the speed of light is equal to -

Here $e,h,k$ and $c$ have their usual meanings.

• A. $\frac{4\pi k{e}^2}{ch}$
• B. $\frac{\pi k{e}^2}{ch}$
• C. $\frac{3\pi k{e}^2}{ch}$
• D. $\frac{2\pi k{e}^2}{ch}$

Answer 1

The correct option is D $\frac{2\pi k{e}^2}{ch}$

For electron in first Bohr orbit,

$\frac{m{v}^2}{r}=\frac{k{e}^2}{r^2}$

$\Rightarrow mvr=\frac{k{e}^2}{v}...\left(1\right)$

Also,

$mvr=\frac{h}{2\pi }[\text{Here}n=1]...\left(2\right)$

From equations $\left(1\right)$ and $\left(2\right)$,

$\frac{h}{2\pi }=\frac{k{e}^2}{v}$

$\Rightarrow v=\frac{2\pi k{e}^2}{h}$

So,

$\frac{v}{c}=\frac{2\pi k{e}^2}{ch}$

Hence, option $\left(D\right)$ is the correct answer.