Question

The velocity of an electron in $2^{nd}$ orbit of double ionized sodium atom $(Z=11)$ is $v.$ The velocity of an electron in its fifth orbit will be

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

Answer 1

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

Given, $(Z=11)$

The orbital velocity of revolving electron in the $n^{th}$ orbit is,

$v_n=2.18\times {10}^6\left(\frac{Z}{n}\right){\text{ms}}^{-1}$

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

$\Rightarrow \frac{v_5}{v_2}=\frac{2}{5}[\because Z=\text{same}]$

$\Rightarrow v_5=\frac{2}{5}v[\because v_2=v]$

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

Why this question? Key concept: The orbital velocity of revolving electron in the $n^{th}$ orbit is, $v_n=2.18\times {10}^6\left(\frac{Z}{n}\right){\text{ms}}^{-1}$ $\Rightarrow v_n\propto \frac{Z}{n}$