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(D) The Rydberg formula for the wavelength of spectral lines is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.For

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$\frac{27}{5} \lambda$

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(D) The Rydberg formula for the wavelength of spectral lines is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.For the first line of the Lyman series,$n_1 = 1$ and $n_2 = 2$. Thus,$\frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{2^2} ight) = R \left( 1 - \frac{1}{4} ight) = \frac{3R}{4}$. Therefore,$\lambda = \frac{4}{3R}$.For the first line of the Balmer series,$n_1 = 2$ and $n_2 = 3$. Thus,$\frac{1}{\lambda_B} = R \left( \frac{1}{2^2} - \frac{1}{3^2} ight) = R \left( \frac{1}{4} - \frac{1}{9} ight) = R \left( \frac{9-4}{36} ight) = \frac{5R}{36}$. Therefore,$\lambda_B = \frac{36}{5R}$.Dividing $\lambda_B$ by $\lambda$,we get $\frac{\lambda_B}{\lambda} = \frac{36/5R}{4/3R} = \frac{36}{5R} 	imes \frac{3R}{4} = \frac{9 	imes 3}{5} = \frac{27}{5}$.Hence,$\lambda_B = \frac{27}{5} \lambda$.

The first line in the Lyman series has wavelength $\lambda$. The wavelength of the first line in the Balmer series is

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