The energy and radius of an electron present | vedclass

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(D) For $He^{+}$ ion, the atomic number $Z = 2$ and the orbit number $n = 2$. The energy of an electron in the $n^{th}$ orbit is given by $E_n = -2.18

Vedclass · Accepted Answer

$-2.18 	imes 10^{-18} \ J, 105.8 \ pm$

Vedclass · Answer

(D) For $He^{+}$ ion, the atomic number $Z = 2$ and the orbit number $n = 2$. The energy of an electron in the $n^{th}$ orbit is given by $E_n = -2.18 	imes 10^{-18} 	imes \frac{Z^2}{n^2} \ J$. Substituting the values: $E_2 = -2.18 	imes 10^{-18} 	imes \frac{2^2}{2^2} = -2.18 	imes 10^{-18} \ J$. The radius of the $n^{th}$ orbit is given by $r_n = 52.9 	imes \frac{n^2}{Z} \ pm$. Substituting the values: $r_2 = 52.9 	imes \frac{2^2}{2} = 52.9 	imes 2 = 105.8 \ pm$. Thus, the energy is $-2.18 	imes 10^{-18} \ J$ and the radius is $105.8 \ pm$.

The energy and radius of an electron present in the second orbit of $He^{+}$ are respectively:

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