Question

An electron in hydrogen atom makes a transition $n_1 \rightarrow n_2$ where $n_1$, and $n_2$ are principal quantum numbers of the two states. Assuming 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

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

Answer 1

Answer: (C)

The time period $T$ of an electron in a Bohr orbit of principal quantum number $n$ is
$T =\frac{4 \varepsilon_{0}^{2} h^{3} n^{3}}{m e^{4}} $
$T \propto n^{3} $
$\therefore \frac{T_{1}}{T_{2}}=\frac{n_{1}^{3}}{n_{2}^{3}}$
As $T_{1}=8 T_{2}$, the above relation gives
$\left(\frac{n_{1}}{n_{2}}\right)^{3}=8 \text { or } n_{1}=2 n_{2}$
Thus the possible values of $n_{1}$ and $n_{2}$ are
$n_{1}=2, n_{2}=1 ;$
$n_{1}=4, n_{2}=2 ; n_{1}=6, n_{2}=3, $ and so on.