The electron present in an excited state of H | vedclass

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

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$5$

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(B) The energy of an electron in the $n$-th orbit of a hydrogen-like species is given by $E_n = -13.6 	imes \frac{Z^2}{n^2} \ eV$. For $He^{+}$,$Z = 2$,so $E_n = -13.6 	imes \frac{4}{n^2} = -\frac{54.4}{n^2} \ eV$. To ionize the electron,the energy supplied must be at least equal to the ionization energy,which is $|E_n| = \frac{54.4}{n^2} \ eV$. Given that the collision with an electron of $2.5 \ eV$ energy causes ionization,the energy of the electron in the excited state must be greater than or equal to $-2.5 \ eV$. Thus,$\frac{54.4}{n^2} \leq 2.5$. $n^2 \geq \frac{54.4}{2.5} = 21.76$. Since $n$ must be an integer,the minimum value of $n$ is $\sqrt{21.76} \approx 4.66$. Therefore,the minimum integer value for $n$ is $5$.

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 electron present in an excited state of $He^{+}$ ion shows separation (ionization) on collision with an electron having $2.5 \ eV$ energy. What will be the minimum value of $n$ for this electron of $He^{+}$ ion?

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