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(C) The energy of an electron in a hydrogen-like atom is given by the formula: $E_n = -13.6 \frac{Z^2}{n^2} 	ext{ eV}$.For the second orbit of a hydr

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$E_3 = \frac{16E}{9}$

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(C) The energy of an electron in a hydrogen-like atom is given by the formula: $E_n = -13.6 \frac{Z^2}{n^2} 	ext{ eV}$.For the second orbit of a hydrogen atom $(Z=1, n=2)$: $E = -13.6 \frac{1^2}{2^2} = -\frac{13.6}{4} 	ext{ eV}$.Therefore,$13.6 	ext{ eV} = 4E$.For the third orbit of a helium ion $(Z=2, n=3)$: $E_3 = -13.6 \frac{2^2}{3^2} = -13.6 \frac{4}{9} 	ext{ eV}$.Substituting $13.6 = -4E$ (considering magnitude or relative energy levels): $E_3 = -(-4E) \frac{4}{9} = \frac{16E}{9}$.

Energy of an electron in the second orbit of a hydrogen atom is $E$. The energy of an electron $E_3$ in the third orbit of a helium $(He^+)$ ion will be:

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