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(C) For the shortest wavelength in the Brackett series,the electron transitions from $n_2 = \infty$ to $n_1 = 4$.Using the Rydberg formula: $\frac{1}{

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$1.46 \, \mu	ext{m}$

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(C) For the shortest wavelength in the Brackett series,the electron transitions from $n_2 = \infty$ to $n_1 = 4$.Using the Rydberg formula: $\frac{1}{\lambda} = R \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} ight]$.Substituting $n_1 = 4$ and $n_2 = \infty$:$\frac{1}{\lambda} = R \left[ \frac{1}{4^2} - \frac{1}{\infty^2} ight] = R \left[ \frac{1}{16} - 0 ight] = \frac{R}{16}$.Therefore,$\lambda = \frac{16}{R}$.Given Rydberg constant $R \approx 1.097 	imes 10^7 \, 	ext{m}^{-1}$.$\lambda = \frac{16}{1.097 	imes 10^7} \approx 14.58 	imes 10^{-7} \, 	ext{m} = 1.458 	imes 10^{-6} \, 	ext{m} \approx 1.46 \, \mu	ext{m}$.

The shortest wavelength of the spectral line in the Brackett series is .......

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