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Q.
If $E$ is the energy of $n^{th}$ orbit of hydrogen atom the energy of $n^{th}$ orbit of $He$ atom will be
Atoms
Solution:
Since Bohr’s formula for energy in $n^{th}$ orbit is
$E_{n} = -\frac{me^{4}Z^{2}}{8\varepsilon_{0}^{2}n^{2}h^{2}}$
$\Rightarrow E_{n} \propto Z^{2} $
Now for hydrogen $\left(_{1}H^{2}\right), Z_{H}= 1, $and $E_H = E$ and for helium $\left(_{2}H^{4}\right), Z_{He}= 2$,
$\therefore \frac{E_{H}}{E_{He}} = \frac{\left(Z_{H}\right)^{2}}{\left(Z_{He}\right)^{2}} $
$\Rightarrow E_{He} = \left(\frac{Z_{He}}{Z_{H}}\right)^{2} E_{H}$
$ = \left(2\right)^{2} E = 4 E$