The ratio of the highest to the lowest wavele | vedclass

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(A) For the Lyman series,the transition is from $n_2$ to $n_1 = 1$. The Rydberg formula is given by $\frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} -

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$4 : 3$

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(A) For the Lyman series,the transition is from $n_2$ to $n_1 = 1$. The Rydberg formula is given by $\frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.For the longest wavelength (lowest energy),the transition is from $n_2 = 2$ to $n_1 = 1$: $\frac{1}{\lambda_{max}} = R_H \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R_H \left( 1 - \frac{1}{4} \right) = \frac{3}{4} R_H$,so $\lambda_{max} = \frac{4}{3 R_H}$.For the shortest wavelength (highest energy),the transition is from $n_2 = \infty$ to $n_1 = 1$: $\frac{1}{\lambda_{min}} = R_H \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) = R_H (1 - 0) = R_H$,so $\lambda_{min} = \frac{1}{R_H}$.The ratio of the highest wavelength to the lowest wavelength is $\frac{\lambda_{max}}{\lambda_{min}} = \frac{4 / (3 R_H)}{1 / R_H} = \frac{4}{3}$.

The ratio of the highest to the lowest wavelength of Lyman series is

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