Question

In hydrogen atom, electron excites from ground state to higher energy state and its orbital velocity is reduced to $\left(\frac{1}{3}\right)rd$ 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. 9 R
• B. 2 R
• C. 3 R
• D. 27 R

Answer 1

Answer: (A)

Velocity of electron in $n ^{\text {th }}$ orbit is inversely proportional to $n$.
$v _{ n } \propto \frac{ 1 }{ n }$
Given:
$\frac{ v _{ n }}{ v _{ 1 }}=\frac{ 1 }{ 3 }$
$\Longrightarrow n = 3$
Radius of electron in
$n ^{\text {th }}$
orbit is proportional to
$n ^{2}$.
$r _{ n } \propto n ^{2}$
Hence, $r _{3}=9 r _{1}= 9 R$