Question

The energy of an electron in a hydrogen atom in the $n^{th}$ orbit is $E_n$. Then, the energy in the $n^{th}$ orbit of double 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 $4E_n$

For Hydrogen atom,
$E_n=-\frac{13.6}{n^2}$
For helium ion
$E^{\prime }=4\times \left(-\frac{13.6}{n^2}\right)E^{\prime }=4E_n$