Question

In hydrogen atom, electron excites from ground state to higher energy state and its orbital velocity is reduced to $\frac{1}{3^{rd}}$ of its initial value. The radius of the orbit in the ground state is $R$. The radius of the orbit in that higher energy state is:

• A. $2R$
• B. $3R$
• C. $27R$
• D. $9R$

Answer 1

The correct option is D $9R$

Velocity of electron in $n^{th}$ orbit is inversely proportional to $n$.

$v_n\propto \frac{1}{n}$

Given:

$\frac{v_n}{v_1}=\frac{1}{3}$

$⟹n=3$

Radius of electron in $n^{th}$ orbit is proportional to $n^2$.

$r_n\propto n^2$

Hence, $r_3=9r_1=9R$