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(D) The energy of an electron in the $n^{	ext{th}}$ orbit of a hydrogen-like atom is given by $E_n = -13.6 \frac{Z^2}{n^2} 	ext{ eV}$.The energy rel

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$40.8$

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(D) The energy of an electron in the $n^{	ext{th}}$ orbit of a hydrogen-like atom is given by $E_n = -13.6 \frac{Z^2}{n^2} 	ext{ eV}$.The energy released when an electron jumps from an initial state $n_i$ to a final state $n_f$ is $\Delta E = E_{n_i} - E_{n_f} = 13.6 Z^2 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} ight) 	ext{ eV}$.Given $Z = 4$,$n_i = 4$,and $n_f = 2$:$\Delta E = 13.6 	imes (4)^2 \left( \frac{1}{2^2} - \frac{1}{4^2} ight) 	ext{ eV}$.$\Delta E = 13.6 	imes 16 \left( \frac{1}{4} - \frac{1}{16} ight) 	ext{ eV}$.$\Delta E = 13.6 	imes 16 \left( \frac{4-1}{16} ight) 	ext{ eV}$.$\Delta E = 13.6 	imes 3 	ext{ eV} = 40.8 	ext{ eV}$.

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