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(A) The energy of a photon emitted during a transition from $n_2$ to $n_1$ in a hydrogen-like atom is given by $\Delta E = 13.6 Z^2 \left[ \frac{1}{n_

Vedclass · Accepted Answer

$3$

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(A) The energy of a photon emitted during a transition from $n_2$ to $n_1$ in a hydrogen-like atom is given by $\Delta E = 13.6 Z^2 \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} ight] \ eV$.For the transition $n = 2$ to $n = 1$:$E_1 = 13.6 Z^2 \left[ \frac{1}{1^2} - \frac{1}{2^2} ight] = 13.6 Z^2 \left( 1 - \frac{1}{4} ight) = 13.6 Z^2 \left( \frac{3}{4} ight)$.For the transition $n = 3$ to $n = 2$:$E_2 = 13.6 Z^2 \left[ \frac{1}{2^2} - \frac{1}{3^2} ight] = 13.6 Z^2 \left( \frac{1}{4} - \frac{1}{9} ight) = 13.6 Z^2 \left( \frac{5}{36} ight)$.According to the problem,$E_1 - E_2 = 74.8 \ eV$:$13.6 Z^2 \left( \frac{3}{4} - \frac{5}{36} ight) = 74.8$.Calculating the term in the bracket:$\frac{3}{4} - \frac{5}{36} = \frac{27 - 5}{36} = \frac{22}{36} = \frac{11}{18}$.Substituting back:$13.6 Z^2 	imes \frac{11}{18} = 74.8$.$Z^2 = \frac{74.8 	imes 18}{13.6 	imes 11} = 5.5 	imes \frac{18}{11} = 0.5 	imes 18 = 9$.$Z = 3$.

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