Question

The electron in hydrogen atom makes a transition $n_1\to n_2$ where $n_1\text{and}{n}_2$ are principle quantum numbers of the two states. The time period of the electron in the initial state is eight times that of in the final state. The possible values of $n_1\text{and}{n}_2$ are,

• A. $n_1=4,n_2=2$
• B. $n_1=8,n_2=2$
• C. $n_1=8,n_2=1$
• D. $n_1=6,n_2=8$

Answer 1

The correct option is A $n_1=4,n_2=2$

The time period of an electron, revolving in circular orbit, is,

$T=\frac{4h^3{ϵ}_0^2{n}^3}{m{e}^4}$

$\Rightarrow T\propto n^3$

$\therefore \frac{T_1}{T_2}=\frac{n_1^3}{n_2^3}$

As, $T_2=8T_1\Rightarrow \frac{n_1^3}{n_2^3}=8$

$\therefore n_1=2n_2$

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