Question

The orbital electron of the hydrogen atom jumps from the ground state to a higher energy state and its orbital velocity is reduced to one third of its initial value. If the radius of the orbit in ground state is $r$, what is the radius of new orbit?

• A. $2r$
• B. $3r$
• C. $4r$
• D. $9r$

Answer 1

The correct option is D $9r$

According to question,
$v_3=\frac{v_1}{3}\Rightarrow \frac{v_1}{v_3}=3-----\left(1\right)$
where $v_3$ is velocity of electron in higher state and $v_1$ is velocity of electron in ground state.
Now, velocity of the electron is $v=\frac{e^{2}Z}{2h{ϵ}_{0}n}$
$\Rightarrow \frac{v_1}{v_3}=\frac{n_3}{n_1}-----\left(2\right)$

From $\left(1\right)\text{and}\left(2\right)$, we can write,
$3=\frac{n_3}{n_1}\Rightarrow n_3=3(\because n_1=1)$

Now, the radius of an electron in $n^{\text{th}}$ orbit is given as
$r_n=\frac{0.529\times n^2}{z}\text{in}\mathring{\text{A}}$

$\Rightarrow r_n\propto n^2$

$\Rightarrow \frac{r_1}{r_2}=\frac{n_1^2}{n_3^2}$

$\frac{r}{r_3}=\frac{1}{9}$

$\therefore r_3=9r$

Hence, $\left(D\right)$ is the correct answer.

Why this question ? This question is based on, velocity of an electron and radius of $n^{\text{th}}$ orbit, of Bohr's atomic model.