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(C) The Rydberg formula for the wavelength $\lambda$ of spectral lines in the hydrogen atom is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2}

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$4 \lambda$

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(C) The Rydberg formula for the wavelength $\lambda$ of spectral lines in the hydrogen atom is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$,where $R$ is the Rydberg constant.For the Balmer series,$n_1 = 2$. The shortest wavelength occurs when $n_2 = \infty$.$\frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{\infty^2} ight) = \frac{R}{4} \implies \lambda = \frac{4}{R} \quad \dots (i)$For the Brackett series,$n_1 = 4$. The shortest wavelength occurs when $n_2 = \infty$.$\frac{1}{\lambda_{	ext{Brackett}}} = R \left( \frac{1}{4^2} - \frac{1}{\infty^2} ight) = \frac{R}{16}$$\lambda_{	ext{Brackett}} = \frac{16}{R}$Substituting $\frac{1}{R} = \frac{\lambda}{4}$ from equation $(i)$:$\lambda_{	ext{Brackett}} = 16 	imes \left( \frac{\lambda}{4} ight) = 4 \lambda$.

In the hydrogen spectrum,if the shortest wavelength in the Balmer series is $\lambda$,then the shortest wavelength in the Brackett series is:

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