The ratio of the total energy of the $2^{ ext {nd }}$ orbit electron for the hydrogen atom $\left({ }_1 H^1 ight)$ to that of helium ion $\left(H_e^{+} ight)\left[\left({ }_2^4 \mathrm{He} ight) ight]$ is

Question

The ratio of the total energy of the $2^{	ext {nd }}$ orbit electron for the hydrogen atom $\left({ }_1 H^1ight)$ to that of helium ion $\left(H_e^{+}ight)\left[\left({ }_2^4 \mathrm{He}ight)ight]$ is

Aicharya · Accepted Answer

Total energy of an electron in a Hydrogen like atom is given by, $E=-13.6 \frac{Z^2}{n^2}$ $\frac{E_H}{E_{H e}}=\left(\frac{1}{2}\right)^2\left(\frac{2}{2}\right)^2=\frac{1}{4}$

The ratio of the total energy of the $2^{\text {nd }}$ orbit electron for the hydrogen atom $\left({ }_1…

Solution