Question

The energy of $H$ atom in $n^{th}$ orbit is $E_n$, then energy in $n^{th}$ orbit of singly ionised helium atom will be :

• A. $4E_n$
• B. $\frac{E_n}{4}$
• C. $2E_n$
• D. $\frac{E_n}{2}$

Answer 1

The correct option is A $4E_n$

The energy is proportional to $Z^2$.

The energy of $H$ atom in $n^{th}$ orbit is $E_n$ then energy in $n^{th}$ orbit of singly ionized helium atom will be :

$\frac{E_{He+}}{E_n}=\frac{Z_{He+}^2}{Z_{H+}^2}=\frac{2^2}{1^2}=4$

$\therefore E_{He+}=4En$

Hence the correct option is A.