The difference in wavelength of first and sec | vedclass

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(D) For a hydrogen-like species,the Rydberg formula is given by: $\frac{1}{\lambda} = R Z^2 (\frac{1}{n_1^2} - \frac{1}{n_2^2})$.For the Lyman series,

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$21.1$

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(D) For a hydrogen-like species,the Rydberg formula is given by: $\frac{1}{\lambda} = R Z^2 (\frac{1}{n_1^2} - \frac{1}{n_2^2})$.For the Lyman series,$n_1 = 1$. For $Li^{2+}$,$Z = 3$.First line of Lyman series ($n_1 = 1$ to $n_2 = 2$): $\frac{1}{\lambda_1} = R (3)^2 (\frac{1}{1^2} - \frac{1}{2^2}) = 9R (1 - \frac{1}{4}) = 9R (\frac{3}{4}) = \frac{27R}{4}$. So,$\lambda_1 = \frac{4}{27R}$.Second line of Lyman series ($n_1 = 1$ to $n_2 = 3$): $\frac{1}{\lambda_2} = R (3)^2 (\frac{1}{1^2} - \frac{1}{3^2}) = 9R (1 - \frac{1}{9}) = 9R (\frac{8}{9}) = 8R$. So,$\lambda_2 = \frac{1}{8R}$.Given $R \approx 1.097 	imes 10^7 \ m^{-1} = 1.097 	imes 10^{-2} \ \mathring{A}^{-1}$.$\lambda_1 = \frac{4}{27 	imes 1.097 	imes 10^{-2}} \approx 135.05 \ \mathring{A}$.$\lambda_2 = \frac{1}{8 	imes 1.097 	imes 10^{-2}} \approx 113.95 \ \mathring{A}$.Difference $\Delta \lambda = \lambda_1 - \lambda_2 = 135.05 - 113.95 = 21.1 \ \mathring{A}$.

The difference in wavelength of first and second spectral lines of Lyman series in $Li^{2+}$ ion spectrum is ................ $\mathring{A}$

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