In the Lyman series,the series limit of wavel | vedclass

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(D) According to Rydberg's formula,$\frac{1}{\lambda} = R \left( \frac{1}{n^2} - \frac{1}{m^2} \right)$.For the series limit of the Lyman series,$n=1,

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$\frac{1}{\lambda_1}-\frac{1}{\lambda_2}=\frac{1}{\lambda_3}$

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(D) According to Rydberg's formula,$\frac{1}{\lambda} = R \left( \frac{1}{n^2} - \frac{1}{m^2} \right)$.For the series limit of the Lyman series,$n=1, m=\infty$,so $\frac{1}{\lambda_1} = R(1 - 0) = R$.For the first line of the Lyman series,$n=1, m=2$,so $\frac{1}{\lambda_2} = R \left( 1 - \frac{1}{4} \right) = \frac{3R}{4}$.For the series limit of the Balmer series,$n=2, m=\infty$,so $\frac{1}{\lambda_3} = R \left( \frac{1}{2^2} - 0 \right) = \frac{R}{4}$.Comparing these values,we observe that $\frac{1}{\lambda_1} - \frac{1}{\lambda_2} = R - \frac{3R}{4} = \frac{R}{4}$.Since $\frac{R}{4} = \frac{1}{\lambda_3}$,we get the relation $\frac{1}{\lambda_1} - \frac{1}{\lambda_2} = \frac{1}{\lambda_3}$.

In the Lyman series,the series limit of wavelength is $\lambda_1$. The wavelength of the first line of the Lyman series is $\lambda_2$ and in the Balmer series,the series limit of wavelength is $\lambda_3$. Then the relation between $\lambda_1$,$\lambda_2$,and $\lambda_3$ is

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