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(A) The Rydberg formula for the wavelength $\lambda$ of a spectral line is given by $\frac{1}{\lambda} = R Z^2 \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2

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$1215 \mathring A$

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(A) The Rydberg formula for the wavelength $\lambda$ of a spectral line is given by $\frac{1}{\lambda} = R Z^2 \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} ight]$.For the first spectral line of the Balmer series in a hydrogen atom $(Z=1, n_1=2, n_2=3)$:$\frac{1}{\lambda_1} = R (1)^2 \left[ \frac{1}{2^2} - \frac{1}{3^2} ight] = R \left[ \frac{1}{4} - \frac{1}{9} ight] = R \left( \frac{5}{36} ight) \implies \lambda_1 = \frac{36}{5R} = 6561 \mathring A$.For the second spectral line of the Balmer series in a singly-ionized helium atom $(Z=2, n_1=2, n_2=4)$:$\frac{1}{\lambda_2} = R (2)^2 \left[ \frac{1}{2^2} - \frac{1}{4^2} ight] = 4R \left[ \frac{1}{4} - \frac{1}{16} ight] = 4R \left( \frac{3}{16} ight) = R \left( \frac{3}{4} ight) \implies \lambda_2 = \frac{4}{3R}$.Dividing $\lambda_2$ by $\lambda_1$:$\frac{\lambda_2}{\lambda_1} = \frac{4/3R}{36/5R} = \frac{4}{3} 	imes \frac{5}{36} = \frac{5}{27}$.Therefore,$\lambda_2 = \frac{5}{27} 	imes 6561 \mathring A = 5 	imes 243 \mathring A = 1215 \mathring A$.

The wavelength of the first spectral line in the Balmer series of a hydrogen atom is $6561 \mathring A$. The wavelength of the second spectral line in the Balmer series of a singly-ionized helium atom is

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