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

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(C) For the Balmer series,the Rydberg formula is $\frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$,where $n_1 = 2$. For the f

Vedclass · Accepted Answer

$486$

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(C) For the Balmer series,the Rydberg formula is $\frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$,where $n_1 = 2$. For the first line,$n_2 = 3$: $\frac{1}{\lambda_1} = R_H \left( \frac{1}{2^2} - \frac{1}{3^2} ight) = R_H \left( \frac{1}{4} - \frac{1}{9} ight) = R_H \left( \frac{5}{36} ight)$. For the second line,$n_2 = 4$: $\frac{1}{\lambda_2} = R_H \left( \frac{1}{2^2} - \frac{1}{4^2} ight) = R_H \left( \frac{1}{4} - \frac{1}{16} ight) = R_H \left( \frac{3}{16} ight)$. Taking the ratio: $\frac{\lambda_2}{\lambda_1} = \frac{5/36}{3/16} = \frac{5}{36} 	imes \frac{16}{3} = \frac{80}{108} = \frac{20}{27}$. Therefore,$\lambda_2 = \lambda_1 	imes \frac{20}{27} = 656.1 	imes \frac{20}{27} = 486 \ nm$.

If the wavelength of the first line of the Balmer series is $656.1 \ nm$,then the wavelength of the second line of this series will be ............. $nm$.

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