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(B) For the Lyman series,the shortest wavelength corresponds to the transition from $n_2 = \infty$ to $n_1 = 1$. Using the Rydberg formula $\frac{1}{\

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$\frac{4(\lambda_2-\lambda_1)}{3 \lambda_1 \lambda_2}$

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(B) For the Lyman series,the shortest wavelength corresponds to the transition from $n_2 = \infty$ to $n_1 = 1$. Using the Rydberg formula $\frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$,we get:$\frac{1}{\lambda_1} = R_H \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) = R_H$ ...$(1)$For the Balmer series,the shortest wavelength corresponds to the transition from $n_2 = \infty$ to $n_1 = 2$. Using the Rydberg formula,we get:$\frac{1}{\lambda_2} = R_H \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) = \frac{R_H}{4}$ ...$(2)$From equation $(1)$,$R_H = \frac{1}{\lambda_1}$. Substituting this into equation $(2)$,we have $\frac{1}{\lambda_2} = \frac{1}{4\lambda_1}$. However,the question asks for the expression of $R_H$ in terms of $\lambda_1$ and $\lambda_2$. Subtracting the equations: $\frac{1}{\lambda_1} - \frac{1}{\lambda_2} = R_H - \frac{R_H}{4} = \frac{3R_H}{4}$.Therefore,$R_H = \frac{4}{3} \left( \frac{1}{\lambda_1} - \frac{1}{\lambda_2} \right) = \frac{4(\lambda_2 - \lambda_1)}{3 \lambda_1 \lambda_2}$.

In the hydrogen spectrum,the shortest wavelengths of the Lyman and Balmer series are $\lambda_1$ and $\lambda_2$ respectively. The Rydberg constant of hydrogen is

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