The energies of an electron in the first orbi | vedclass

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Question

(A) The energy of an electron in the $n$th orbit of a hydrogen-like species is given by the formula:$E_n = -2.18 	imes 10^{-18} 	imes \frac{Z^2}{n^2

Vedclass · Accepted Answer

$-8.72 	imes 10^{-18}, -2.18 	imes 10^{-18}$

Vedclass · Answer

(A) The energy of an electron in the $n$th orbit of a hydrogen-like species is given by the formula:$E_n = -2.18 	imes 10^{-18} 	imes \frac{Z^2}{n^2} \ J/	ext{ion}$For $He^{+}$ ion:Atomic number $Z = 2$,orbit number $n = 1$.$E_1 = -2.18 	imes 10^{-18} 	imes \frac{2^2}{1^2} = -2.18 	imes 10^{-18} 	imes 4 = -8.72 	imes 10^{-18} \ J$.For $Li^{2+}$ ion:Atomic number $Z = 3$,orbit number $n = 3$.$E_3 = -2.18 	imes 10^{-18} 	imes \frac{3^2}{3^2} = -2.18 	imes 10^{-18} 	imes 1 = -2.18 	imes 10^{-18} \ J$.Thus,the energies are $-8.72 	imes 10^{-18} \ J$ and $-2.18 	imes 10^{-18} \ J$ respectively.Therefore,option $A$ is correct.

Energy level	$K$ $(n=1)$	$L$ $(n=2)$	$M$ $(n=3)$	$N$ $(n=4...n=\infty)$
Energy	$-864 \ a.u.$	$-216 \ a.u.$	$-96 \ a.u.$	$0 \ a.u.$

The energies of an electron in the first orbit of $He^{+}$ and in the third orbit of $Li^{2+}$ in $J$ are respectively:

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