The wavelength of the second Balmer line in t | vedclass

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(D) The Rydberg formula is given by $\frac{1}{\lambda} = R \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$.For the Balmer series,$n_1 = 2$. The firs

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None of the above

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(D) The Rydberg formula is given by $\frac{1}{\lambda} = R \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$.For the Balmer series,$n_1 = 2$. The first line corresponds to $n_2 = 3$,and the second line corresponds to $n_2 = 4$.$\frac{1}{\lambda_B} = R \left[ \frac{1}{2^2} - \frac{1}{4^2} \right] = R \left[ \frac{1}{4} - \frac{1}{16} \right] = R \left( \frac{3}{16} \right)$.Given $\lambda_B = 600 \ nm$,so $\frac{1}{600} = R \left( \frac{3}{16} \right) \implies R = \frac{16}{1800} \ nm^{-1}$.For the Lyman series,$n_1 = 1$. The third line corresponds to $n_2 = 4$ (since $n_2 = 2, 3, 4, \dots$).$\frac{1}{\lambda_L} = R \left[ \frac{1}{1^2} - \frac{1}{4^2} \right] = R \left[ 1 - \frac{1}{16} \right] = R \left( \frac{15}{16} \right)$.Substituting the value of $R$:$\frac{1}{\lambda_L} = \left( \frac{16}{1800} \right) \left( \frac{15}{16} \right) = \frac{15}{1800} = \frac{1}{120}$.Therefore,$\lambda_L = 120 \ nm$.Since $120 \ nm$ is not among the options,the correct choice is $D$.

The wavelength of the second Balmer line in the hydrogen spectrum is $600 \ nm$. The wavelength for its third line in the Lyman series is:

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