The ratio of the total energy of the 2^{text{ | vedclass

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(D) The total energy $E$ of an electron in the $n^{	ext{th}}$ orbit of a hydrogen-like atom with atomic number $Z$ is given by the formula: $E_n = -1

Vedclass · Accepted Answer

$\frac{1}{4}$

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(D) The total energy $E$ of an electron in the $n^{	ext{th}}$ orbit of a hydrogen-like atom with atomic number $Z$ is given by the formula: $E_n = -13.6 \frac{Z^2}{n^2} 	ext{ eV}$.For the hydrogen atom $(_1H^1)$,$Z_H = 1$ and $n = 2$. Thus,$E_H = -13.6 \frac{1^2}{2^2} = -13.6 	imes \frac{1}{4} 	ext{ eV}$.For the helium ion $(He^+)$,$Z_{He} = 2$ and $n = 2$. Thus,$E_{He} = -13.6 \frac{2^2}{2^2} = -13.6 	imes 1 	ext{ eV}$.The ratio of the total energy of the hydrogen atom to that of the helium ion is: $\frac{E_H}{E_{He}} = \frac{-13.6 	imes (1/4)}{-13.6 	imes 1} = \frac{1}{4}$.

The ratio of the total energy of the $2^{\text{nd}}$ orbit electron for the hydrogen atom $(_1H^1)$ to that of the helium ion $(He^+)$ $(_2^4He)$ is:

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