Question

An electron in a hydrogen atom makes a transition from $n=n_1$ to $n=n_2$. The time period of the electron in the initial state is eight times that in the final state. The possible values of $n_1$ 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. $n_1=6,n_2=3$

Answer 1

Answer: (A), (D)

The correct options are
A $n_1=4,n_2=2$
D $n_1=6,n_2=3$

The time period of revolution of an electron in an orbit is
$T=\frac{1}{\nu }=\frac{4{{ϵ}_0}^2{n}^{3}h}{m{e}^4}$
$\Rightarrow T\propto n^3$
Therefore, for the transition from $n_1$ to $n_2$ we get
$T_1\propto n_1^3$ and $T_2\propto n_2^3$
$\frac{T_1}{T_2}=\frac{n_1^3}{n_2^3}$
$8=(\frac{n_1}{n_2}{)}^3$
$\frac{n_1}{n_2}=2$

Therefore only the values of $n_1$ and $n_2$ having ratio 2 are possible for transition.

i.e. $n_1=4$ and $n_2=2$ and $n_1=6$ and $n_2=3$