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(B) The radius of an orbit in a hydrogen-like species is given by $r_n = a_0 	imes \frac{n^2}{Z}$, where $a_0 = 52.9 \ pm = 0.0529 \ nm$ and $Z = 2$

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$-5.45 	imes 10^{-19} \ J$

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(B) The radius of an orbit in a hydrogen-like species is given by $r_n = a_0 	imes \frac{n^2}{Z}$, where $a_0 = 52.9 \ pm = 0.0529 \ nm$ and $Z = 2$ for $He^{+}$.Given $r_n = 0.4232 \ nm$, we have $0.4232 = 0.0529 	imes \frac{n^2}{2}$.$n^2 = \frac{0.4232 	imes 2}{0.0529} = 8 	imes 2 = 16$, so $n = 4$.The energy of an electron in the $n^{th}$ orbit is $E_n = -2.18 	imes 10^{-18} 	imes \frac{Z^2}{n^2} \ J$.Substituting $Z = 2$ and $n = 4$, $E_4 = -2.18 	imes 10^{-18} 	imes \frac{2^2}{4^2} = -2.18 	imes 10^{-18} 	imes \frac{4}{16} = -2.18 	imes 10^{-18} 	imes 0.25 = -5.45 	imes 10^{-19} \ J$.

If the radius of an electron in the excited state of $He^{+}$ is $0.4232 \ nm$, the energy of the electron in that excited state in $J$ is: (The radius and energy of an electron in the first orbit of a hydrogen atom are $52.9 \ pm$ and $-2.18 \times 10^{-18} \ J$ respectively)

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