Question

In Bohr's model of hydrogen atom, the ratio of the period of revolution of an electron in the first orbit to that in the second orbit, is,

• A. $1:2$
• B. $2:1$
• C. $1:4$
• D. $1:8$

Answer 1

The correct option is D $1:8$

The period of an electron in the $n^{\text{th}}$ orbit is,

$T_n=\frac{2\pi r_n}{v_n}$

Putting $r_n=\frac{{ϵ}_0{h}^2{n}^2}{\pi m{e}^2}\text{and}{v}_n=\frac{e^2}{2h{ϵ}_{0}n}$ and simplifying, we get,

$T_n=\frac{4h^2{ϵ}_0^2{n}^3}{m{e}^4}$

$\therefore T_n\propto n^3$

$\Rightarrow \frac{T_1}{T_2}=\frac{n_1^3}{n_2^3}={\left(\frac{1}{2}\right)}^3$

$\therefore \frac{T_1}{T_2}=\frac{1}{8}$

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

Why this Question? This question is based on Time period of the electron in $n^{th}$ Bohr's orbit.