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Q. The energy of hydrogen atom in $n^{\text {th }}$ orbit is $E_{n}$ then the energy in $n^{\text {th }}$ orbit of singly ionised helium atom will be

AIPMTAIPMT 2001Atoms

Solution:

The energy of the electron in the $n^{th}$ orbit is
$ E = E \propto Z^2 \frac {1}{n^2} $
Given, $ Z_H=1,Z_{He}=2 $
$ \therefore \frac {E_H}{E_{He}}= \frac {Z_H^2}{Z^2_{He}}= \frac {1}{4} $
$ \Rightarrow E_{He}=4 E_H $
Given, $ E_H=E_n $
$ \therefore E_{He}=4E_n $