In the second orbit of hydrogen atom, the energy of an electron is ' $E$ '. In the third orbit of helium atom, the energy of the electron will be (atomic number of helium $=2$ )

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Aicharya · Accepted Answer

$\begin{aligned} & E \propto \frac{Z^2}{n^2} \ 	herefore \quad \frac{E_H}{E_{H e}} & =\frac{\left(Z^2ight)_H}{(n)_H^2} 	imes \frac{\left(n_{\mathrm{He}}ight)^2}{\left(Z^2ight)_{\mathrm{He}}} \ & =\frac{1}{2^2} 	imes \frac{3^2}{2^2}=\frac{9}{16} \ 	herefore \quad E_{H e} & =\frac{16}{9} E_H=\frac{16}{9} E\end{aligned}$

In the second orbit of hydrogen atom, the energy of an electron is ' $E$ '. In the third orbit of helium…

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