Question

The velocity of an electron in the second orbit of sodium atom $(\text{atomic no.}=11)$, is $v$. The velocity of an electron in the fifth orbit of sodium atom will be,

• A. $v$
• B. $\frac{22}{5}v$
• C. $\frac{5}{2}v$
• D. $\frac{2}{5}v$

Answer 1

The correct option is D $\frac{2}{5}v$

Velocity of an electron in $n^{th}$ orbit of Bohr's model is given as,
$v_n=\frac{e^{2}Z}{2h{ϵ}_{0}n}$

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

Let $v_2$ & $v_5$ be the velocity of an electron in second and fifth orbit respectively.

$\Rightarrow \frac{v_2}{v_5}=\frac{n_5}{n_2}=\frac{5}{2}$

$v_5=\frac{2}{5}{v}_2$

$\therefore v_5=\frac{2}{5}v$

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

Why this question ? This question gives clear idea of velocity of electron in $n^{th}$ orbit in Bohr's hydrogen atom.