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(B) The energy of a photon emitted during a transition is given by $E = \Delta E = 13.6 	ext{ eV} 	imes Z^2 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2}

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$x = 9/4$

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(B) The energy of a photon emitted during a transition is given by $E = \Delta E = 13.6 	ext{ eV} 	imes Z^2 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} ight)$.For the first transition (second excited state $n=3$ to first excited state $n=2$):$E_1 = 13.6 \left( \frac{1}{2^2} - \frac{1}{3^2} ight) = 13.6 \left( \frac{1}{4} - \frac{1}{9} ight) = 13.6 \left( \frac{5}{36} ight)$.For the second transition (first excited state $n=2$ to ground state $n=1$):$E_2 = 13.6 \left( \frac{1}{1^2} - \frac{1}{2^2} ight) = 13.6 \left( 1 - \frac{1}{4} ight) = 13.6 \left( \frac{3}{4} ight) = 13.6 \left( \frac{27}{36} ight)$.Ratio of energies $z = \frac{E_1}{E_2} = \frac{5/36}{27/36} = \frac{5}{27}$.Since $E = \frac{hc}{\lambda}$,we have $\frac{E_1}{E_2} = \frac{\lambda_2}{\lambda_1} = z$. Thus,$x = \frac{\lambda_1}{\lambda_2} = \frac{1}{z} = \frac{27}{5}$.Since $p = \frac{E}{c}$,the ratio of momenta $y = \frac{p_1}{p_2} = \frac{E_1/c}{E_2/c} = \frac{E_1}{E_2} = z = \frac{5}{27}$.Comparing with options:$(A)$ $z = 1/x$ is correct because $z = \frac{E_1}{E_2}$ and $x = \frac{\lambda_1}{\lambda_2} = \frac{E_2}{E_1} = 1/z$.$(B)$ $x = 9/4$ is incorrect (actual value is $27/5$).$(C)$ $y = 5/27$ is correct.$(D)$ $z = 5/27$ is correct.Therefore,the wrong answer is $(B)$.

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