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(D) According to Bohr's model,the wavelength $\lambda$ of the emitted radiation when an electron jumps from orbit $n_2$ to $n_1$ is given by the Rydbe

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$0.021$

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(D) According to Bohr's model,the wavelength $\lambda$ of the emitted radiation when an electron jumps from orbit $n_2$ to $n_1$ is given by the Rydberg formula:$\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$For the first line of the Lyman series,$n_1 = 1$ and $n_2 = 2$:$\frac{1}{\lambda_1} = R \left( \frac{1}{1^2} - \frac{1}{2^2} ight) = R \left( 1 - \frac{1}{4} ight) = \frac{3R}{4} \implies \lambda_1 = \frac{4}{3R}$For the first line of the Balmer series,$n_1 = 2$ and $n_2 = 3$:$\frac{1}{\lambda_2} = R \left( \frac{1}{2^2} - \frac{1}{3^2} ight) = R \left( \frac{1}{4} - \frac{1}{9} ight) = R \left( \frac{9-4}{36} ight) = \frac{5R}{36} \implies \lambda_2 = \frac{36}{5R}$The ratio of the wavelengths is given as $\frac{\lambda_1}{\lambda_2} = 9 \alpha$:$\frac{\lambda_1}{\lambda_2} = \left( \frac{4}{3R} ight) 	imes \left( \frac{5R}{36} ight) = \frac{4 	imes 5}{3 	imes 36} = \frac{20}{108} = \frac{5}{27}$Given $\frac{\lambda_1}{\lambda_2} = 9 \alpha$,we have:$9 \alpha = \frac{5}{27} \implies \alpha = \frac{5}{27 	imes 9} = \frac{5}{243} \approx 0.02057 \approx 0.021$

In hydrogen atom spectra,if the ratio of wavelengths corresponding to the first line of the Lyman series and the first line of the Balmer series is $9 \alpha$,the value of $\alpha$ is

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