The binding energy of an e^- in the ground st | vedclass

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(D) The energy of an electron in the $n^{th}$ orbit of a hydrogen-like species is given by the formula: $E_n = -13.6 	imes \frac{Z^2}{n^2} \ eV$. For

Vedclass · Accepted Answer

$54.4, 13.6, 6.0$

Vedclass · Answer

(D) The energy of an electron in the $n^{th}$ orbit of a hydrogen-like species is given by the formula: $E_n = -13.6 	imes \frac{Z^2}{n^2} \ eV$. For a hydrogen atom $(Z=1)$ in the ground state $(n=1)$,$E_1 = -13.6 \ eV$. The binding energy is the energy required to remove the electron,which is $|E_1| = 13.6 \ eV$. For a $He^+$ ion,the atomic number $Z=2$. In the ground state $(n=1)$,the energy is $E_1 = -13.6 	imes \frac{2^2}{1^2} = -13.6 	imes 4 = -54.4 \ eV$. The energy required to remove the electron (binding energy) is $|-54.4 \ eV| = 54.4 \ eV$. Thus,the correct option is $D$.

The binding energy of an $e^-$ in the ground state of a hydrogen atom is $13.6 \ eV$. The energy required to remove an electron from the ground state of a $He^+$ ion is:

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