Question

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

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

Answer 1

The correct option is A 4 $E_n$

We know,

$E_n=\frac{-2{\pi }^2{mK}^2{K}^2{Z}^2{e}^4}{n^2{h}^2}$

For $n^{th}$ orbit of singly ionized helium, Z=2

$\therefore E=4E_n$