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(B) The longest wavelength corresponds to the minimum energy transition in the Lyman series,which occurs from $n_2 = 2$ to $n_1 = 1$.Using the Rydberg

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$1215.8 \mathring{A}$

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(B) The longest wavelength corresponds to the minimum energy transition in the Lyman series,which occurs from $n_2 = 2$ to $n_1 = 1$.Using the Rydberg formula: $\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.For the $H$ atom,$Z = 1$,$n_1 = 1$,and $n_2 = 2$.$\frac{1}{\lambda} = R 	imes 1^2 \left( \frac{1}{1^2} - \frac{1}{2^2} ight) = R \left( 1 - \frac{1}{4} ight) = \frac{3}{4} R$.$\lambda = \frac{4}{3R}$.Given $\frac{1}{R} \approx 911.6 \mathring{A}$,we have $\lambda = \frac{4}{3} 	imes 911.6 \mathring{A} \approx 1215.5 \mathring{A}$.

The longest wavelength line in the Lyman series of the $H$ atom spectrum is $:-$

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