Question

If the energy of a hydrogen atom in $n^{th}$ orbit is $E_n$ then energy in the $n^{th}$ orbit of a singly ionized helium atom will be.

• A. $4E_n$
• B. $E_n/4$
• C. $2E_n$
• D. $E_n/2$

Answer 1

The correct option is A $4E_n$

$E=\frac{-m{z}^2{e}^4}{8{\epsilon }_0{h}^2}\cdot \frac{1}{n^2}$
$Z_H=1$
$Z_{He}=2$
$\frac{E_H}{E_{He}}=\frac{Z_H^2}{Z_{He}^2}=\frac{1}{4}$
$E_H=\frac{Z_H^2}{Z_{He}^2}=\frac{2}{4}$
$E_{He}=4E_H$
$E_H=E_n$
$E_{He}=4E_n$.