Question

The time of revolution of an electron, around a nucleus of charge $Ze$, in $n^{th}$ Bohr orbit is directly proportional to,

• A. $n$
• B. $\frac{n^3}{Z^2}$
• C. $\frac{n^2}{Z}$
• D. $\frac{Z}{n}$

Answer 1

The correct option is B $\frac{n^3}{Z^2}$

The time period of electron revolving in an orbit is given by,
$T=\frac{2\pi r_n}{v}----\left(1\right)$

Radius of $n^{\text{th}}$ orbit is, $r_n=\frac{{ϵ}_0{n}^2{h}^2}{\pi mZ{e}^2}$
Also, $v_n=\frac{Z{e}^2}{2{ϵ}_{0}nh}$

Putting the values of $r_n$ & $v_n$, in the equation of $r_n$, we get,

$T=2\pi \left(\frac{{ϵ}_0{n}^2{h}^2}{\pi mZ{e}^2}\right)\times \frac{2{ϵ}_{0}nh}{Z{e}^2}$
$\Rightarrow T=\frac{2\pi n^3{h}^3\times 2{ϵ}_0^2}{\pi m{Z}^2{e}^4}$

$\Rightarrow T\propto \frac{n^3}{Z^2}$

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

Why this question ? This question is based on the time period of electron in an orbit, in a Bohr's model.