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(C) The shortest wavelength in the Balmer series occurs for the transition of an electron from $n = \infty$ to $n = 2$.Using the Rydberg formula: $\fr

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$4$

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(C) The shortest wavelength in the Balmer series occurs for the transition of an electron from $n = \infty$ to $n = 2$.Using the Rydberg formula: $\frac{1}{\lambda} = R \left[ \frac{1}{2^2} - \frac{1}{\infty^2} \right] = \frac{R}{4} \implies \lambda = \frac{4}{R}$.The shortest wavelength in the Brackett series occurs for the transition of an electron from $n = \infty$ to $n = 4$.Using the Rydberg formula: $\frac{1}{\lambda'} = R \left[ \frac{1}{4^2} - \frac{1}{\infty^2} \right] = \frac{R}{16} \implies \lambda' = \frac{16}{R}$.Dividing the two expressions: $\frac{\lambda'}{\lambda} = \frac{16/R}{4/R} = \frac{16}{4} = 4$.Therefore,$\lambda' = 4\lambda$.

In the hydrogen spectrum,the shortest wavelength in the Balmer series is $\lambda$. The shortest wavelength in the Brackett series is $........\,\lambda$.

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