The ratio of the largest to shortest waveleng | vedclass

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(D) For the Lyman series,the wavelength $\lambda$ is given by the Rydberg formula: $\frac{1}{\lambda} = R \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \ri

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$4/3$

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(D) For the Lyman series,the wavelength $\lambda$ is given by the Rydberg formula: $\frac{1}{\lambda} = R \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$,where $n_1 = 1$ and $n_2 = 2, 3, 4, \dots$ For the largest wavelength $(\lambda_{\max})$,we take the transition from $n_2 = 2$ to $n_1 = 1$: $\frac{1}{\lambda_{\max}} = R \left[ \frac{1}{1^2} - \frac{1}{2^2} \right] = R \left[ 1 - \frac{1}{4} \right] = \frac{3R}{4} \implies \lambda_{\max} = \frac{4}{3R}$. For the shortest wavelength $(\lambda_{\min})$,we take the transition from $n_2 = \infty$ to $n_1 = 1$: $\frac{1}{\lambda_{\min}} = R \left[ \frac{1}{1^2} - \frac{1}{\infty^2} \right] = R [1 - 0] = R \implies \lambda_{\min} = \frac{1}{R}$. The ratio of the largest to the shortest wavelength is: $\frac{\lambda_{\max}}{\lambda_{\min}} = \frac{4/3R}{1/R} = \frac{4}{3}$.

The ratio of the largest to shortest wavelengths in the Lyman series of hydrogen spectra is

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