Question

The electron in a hydrogen atom makes a transition $n_1\to n_2$ where $n_1\&n_2$ are the principal quantum numbers of the two states. Assume the Bohr's model to be valid the time period of the electron in the initial state is eight times that in the final state. The possible values of $n_1\&n_2$ are:

• A. $n_1=6,n_2=2$
• B. $n_1=8,n_2=2$
• C. $n_1=8,n_2=1$
• D. $n_1=6,n_2=3$

Answer 1

The correct option is D $n_1=6,n_2=3$

The expression for the time period of revolution of an electron in a hydrogen atom is given by

$T=\frac{4n^3{h}^3{\epsilon }_0^2}{mZ{e}^4}$

For same atom,

$\Rightarrow T\propto n^3$

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

Given that, $T_1=8T_2$

$\frac{T_1}{T_2}=\frac{8}{1}......\left(2\right)$

Comparing $\left(1\right)$ and $\left(2\right)$,

${\left(\frac{n_1}{n_2}\right)}^3=\frac{8}{1}$

$\Rightarrow \frac{n_1}{n_2}=\frac{2}{1}$

Only option $\left(\text{D}\right)$ satisfies the above condition,

Hence, option $\left(\text{D}\right)$ is correct.

Key idea: $T\propto n^3$