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(C) For the Brackett series,the transition is to $n_1 = 4$.For the third line,the electron jumps from $n_2 = 4 + 3 = 7$.Using the Rydberg formula: $\f

Vedclass · Accepted Answer

$\frac{784}{33R}$

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(C) For the Brackett series,the transition is to $n_1 = 4$.For the third line,the electron jumps from $n_2 = 4 + 3 = 7$.Using the Rydberg formula: $\frac{1}{\lambda} = R Z^2 \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} ight]$.For hydrogen,$Z = 1$,so $\frac{1}{\lambda} = R \left[ \frac{1}{4^2} - \frac{1}{7^2} ight]$.$\frac{1}{\lambda} = R \left[ \frac{1}{16} - \frac{1}{49} ight] = R \left[ \frac{49 - 16}{16 	imes 49} ight]$.$\frac{1}{\lambda} = R \left[ \frac{33}{784} ight]$.Therefore,$\lambda = \frac{784}{33R}$.

Find the wavelength of the third line in the Brackett series of the hydrogen spectrum.

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