Question

The magnitude of acceleration of the electron in the $n^{th}$ orbit of hydrogen atom is $a_H$ and that of singly ionized helium atom is $a_{He}$. The ratio $a_H:a_{He}$ is?

• A. $1:8$
• B. $1:4$
• C. $1:2$
• D. Dependent on n

Answer 1

The correct option is A $1:8$

Centripetal acceleration of electron in nth orbit $a_n=\frac{v_n^2}{r_n}$
Since, $v_n\propto \frac{Z}{n}$ and $r_n\propto \frac{n^2}{Z}$
$⟹a_n\propto \frac{Z^3}{n^3}$
So we get $\frac{a_H}{a_{He}}=\frac{Z_H^3}{Z_{He}^3}$
Given : $Z_{He}=2$ and $Z_H=1$
$⟹$ $\frac{a_H}{a_{He}}=\frac{1^3}{2^3}=\frac{1}{8}$