An electron in hydrogen atom makes a transition n_1 arrow n_2 where...

An electron in hydrogen atom makes a transition

n_1 \rightarrow n_2

where

n_1

are principle 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 values of

n_1

and

n_2

are:

Options

A $n_1=6$ and $n_2=2$
B $n_1=8$ and $n_2=1$
C $n_1=4$ and $n_2=2$
D $n_1=4$ and $n_2=2$

Correct answer

D. $n_1=4$ and $n_2=2$

Step-by-step solution

Time period of an electron in Bohr orbit is

\begin{aligned} & T=\frac{4 \varepsilon_0^2 h^3 n^3}{m e^4} \\ \Rightarrow T & \propto n^3 \\ \Rightarrow \frac{T_1}{T_2} & =\frac{n_1^3}{n_2^3} \end{aligned}

T_1=8 T_2

then

\begin{aligned} \Rightarrow & \left(\frac{8 T_2}{T_2}\right) & =\frac{n_1^3}{n_2^3} \\ \Rightarrow & \left(\frac{n_1}{n_2}\right)^3 & =\left(2^2\right)^3 \\ \Rightarrow & \frac{n_1}{n_2} & =2^2 \\ \Rightarrow & n_1 & =2 n_2 \end{aligned}

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