If the wavelength of the first line in the Ba | vedclass

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(B) According to the Rydberg formula for the hydrogen spectrum: $\frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.For the Bal

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$485.9 \ nm$ and $364.4 \ nm$

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(B) According to the Rydberg formula for the hydrogen spectrum: $\frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.For the Balmer series,$n_1 = 2$.For the first line,$n_2 = 3$. Given $\lambda_1 = 656 \ nm$,we have $\frac{1}{656} = R_H \left( \frac{1}{4} - \frac{1}{9} ight) = R_H \left( \frac{5}{36} ight) \dots (i)$.For the second line,$n_2 = 4$. So,$\frac{1}{\lambda_2} = R_H \left( \frac{1}{4} - \frac{1}{16} ight) = R_H \left( \frac{3}{16} ight) \dots (ii)$.Dividing $(i)$ by $(ii)$,we get $\frac{\lambda_2}{656} = \frac{5/36}{3/16} = \frac{5}{36} 	imes \frac{16}{3} = \frac{20}{27}$.$\lambda_2 = 656 	imes \frac{20}{27} \approx 485.9 \ nm$.For the limiting line,$n_2 = \infty$. So,$\frac{1}{\lambda_{\infty}} = R_H \left( \frac{1}{4} - 0 ight) = \frac{R_H}{4}$.From $(i)$,$R_H = \frac{36}{5 	imes 656}$.$\frac{1}{\lambda_{\infty}} = \frac{36}{5 	imes 656 	imes 4} = \frac{9}{5 	imes 656} = \frac{9}{3280}$.$\lambda_{\infty} = \frac{3280}{9} \approx 364.4 \ nm$.

If the wavelength of the first line in the Balmer series is $656 \ nm$,then the wavelength of its second line and limiting line respectively are

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