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The ratio of the total energy of the $2^{\text {nd }}$ orbit electron for the hydrogen atom $\left({ }_1 H^1\right)$ to that of helium ion $\left(H_e^{+}\right)\left[\left({ }_2^4 \mathrm{He}\right)\right]$ is
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$\frac{1}{4}$
Total energy of an electron in a Hydrogen like atom is given by,
$E=-13.6 \frac{Z^2}{n^2}$
$\frac{E_H}{E_{H e}}=\left(\frac{1}{2}\right)^2\left(\frac{2}{2}\right)^2=\frac{1}{4}$
$E=-13.6 \frac{Z^2}{n^2}$
$\frac{E_H}{E_{H e}}=\left(\frac{1}{2}\right)^2\left(\frac{2}{2}\right)^2=\frac{1}{4}$
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