Question

An electron in a hydrogen atom makes a transition from $n=n_1$ and $n=n_2$ The time period of the electron in the initial state is eight times that in the final state. The possible values of $n_1\text{and}{n}_2$ are

• A. $n_1=4,n_2=2$
• B. $n_1=8,n_2=2$
• C. $n_1=8,n_2=1$
• D. $n_1=6,n_2=3$

Answer 1

Answer: (A), (D)

$n_1=6,n_2=3$

In the $n^{th}$ orbit:
let $r_n$ and $V_n$ speed of electron

Time period $T_n=\frac{2\pi r_n}{V_n}\propto \frac{r_n}{V_n}$
Now, $r_n\propto n^2\text{and}{V}_n\propto \frac{1}{n}$
$\frac{r_n}{V_n}\propto n^3\text{or}{T}_n\propto n^3$
Here $T_1=8T_2,$ we get
$8=(\frac{n_1}{n_2}{)}^3\Rightarrow \frac{n_1}{n}=2$
$\Rightarrow n_1=2n_2$
So the possible values of $n_1$ and $n_2$ are
$n_1=2$, $n_2=1$ and
$n_1=4;n_2=2$ and
$n_1=6;n_2=3$ and so on.