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(B) The energy of an electron in the $n$th orbit of a hydrogen-like atom is given by the formula:$E_{n} = -13.6 	ext{ eV} 	imes \frac{Z^{2}}{n^{2}}$

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$\frac{16}{9} E_{2}$

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(B) The energy of an electron in the $n$th orbit of a hydrogen-like atom is given by the formula:$E_{n} = -13.6 	ext{ eV} 	imes \frac{Z^{2}}{n^{2}}$This implies that $E_{n} \propto \frac{Z^{2}}{n^{2}}$.For a hydrogen atom,the atomic number $Z = 1$. Thus,for the second orbit $(n = 2)$:$E_{2} = k 	imes \frac{1^{2}}{2^{2}} = \frac{k}{4}$,where $k = -13.6 	ext{ eV}$.For a $He^{+}$ ion,the atomic number $Z = 2$. For the third orbit $(n = 3)$:$E_{3} = k 	imes \frac{2^{2}}{3^{2}} = \frac{4k}{9}$.Now,we find the ratio of $E_{3}$ to $E_{2}$:$\frac{E_{3}}{E_{2}} = \frac{4k/9}{k/4} = \frac{4}{9} 	imes 4 = \frac{16}{9}$.Therefore,$E_{3} = \frac{16}{9} E_{2}$.

Energy of an electron in the second orbit of a hydrogen atom is $E_{2}$. The energy of an electron in the third orbit of $He^{+}$ will be

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