Question

Ratio of centripetal acceleration of electron in 2nd excited state of hydrogen atom to ground state in $H{e}^{+}$ ion is

• A. 9:4
• B. 1:648
• C. 8:31
• D. 3:2

Answer 1

The correct option is B 1:648

$v^2\propto E\propto \frac{z^2}{n^2}$
$r\propto \frac{n^2}{z}$

$a=\frac{v^2}{r}$
Therefore,
$a\propto \frac{\frac{z^2}{n^2}}{\frac{n^2}{z}}\propto \frac{z^3}{n^4}$
So,
$\frac{a_H}{a_{H{e}^{+}}}=\frac{z_H^3}{n_H^4}\frac{n_{He}^4}{z_{He}^3}=\frac{1}{3^4\times 2^3}=\frac{1}{648}$