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(B) The energy of a spectral line is given by $\Delta E = 13.6 	imes Z^2 (\frac{1}{n_1^2} - \frac{1}{n_2^2}) 	ext{ eV}$.For the Lyman series, the lo

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$54$

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(B) The energy of a spectral line is given by $\Delta E = 13.6 	imes Z^2 (\frac{1}{n_1^2} - \frac{1}{n_2^2}) 	ext{ eV}$.For the Lyman series, the lowest energy line $(L_1)$ corresponds to the transition from $n_2 = 2$ to $n_1 = 1$: $\Delta E(L_1) = 13.6 	imes 1^2 (\frac{1}{1^2} - \frac{1}{2^2}) = 13.6 	imes \frac{3}{4} 	ext{ eV}$.For the Balmer series, the lowest energy line $(B_1)$ corresponds to the transition from $n_2 = 3$ to $n_1 = 2$: $\Delta E(B_1) = 13.6 	imes 1^2 (\frac{1}{2^2} - \frac{1}{3^2}) = 13.6 	imes (\frac{1}{4} - \frac{1}{9}) = 13.6 	imes \frac{5}{36} 	ext{ eV}$.Given $\Delta E(L_1) = x 	imes \Delta E(B_1)$, we have $x = \frac{\Delta E(L_1)}{\Delta E(B_1)} = \frac{3/4}{5/36} = \frac{3}{4} 	imes \frac{36}{5} = \frac{27}{5} = 5.4$.To express $x$ as $x 	imes 10^{-1}$, we have $5.4 = 54 	imes 10^{-1}$.Therefore, the value is $54$.

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