The ratio of the minimum frequencies of the L | vedclass

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Question

(C) The frequency $
u$ is given by $
u = \frac{c}{\lambda}$.For the minimum frequency,the wavelength $\lambda$ must be maximum.For the Lyman series

Vedclass · Accepted Answer

$5.4$

Vedclass · Answer

(C) The frequency $
u$ is given by $
u = \frac{c}{\lambda}$.For the minimum frequency,the wavelength $\lambda$ must be maximum.For the Lyman series $(n_1 = 1)$,the maximum wavelength corresponds to the transition from $n_2 = 2$ to $n_1 = 1$: $\frac{1}{\lambda_{L, 	ext{max}}} = R \left( \frac{1}{1^2} - \frac{1}{2^2} ight) = R \left( \frac{3}{4} ight) \implies \lambda_{L, 	ext{max}} = \frac{4}{3R}$.For the Balmer series $(n_1 = 2)$,the maximum wavelength corresponds to the transition from $n_2 = 3$ to $n_1 = 2$: $\frac{1}{\lambda_{B, 	ext{max}}} = R \left( \frac{1}{2^2} - \frac{1}{3^2} ight) = R \left( \frac{1}{4} - \frac{1}{9} ight) = R \left( \frac{5}{36} ight) \implies \lambda_{B, 	ext{max}} = \frac{36}{5R}$.The ratio of minimum frequencies is $\frac{
u_{L, 	ext{min}}}{
u_{B, 	ext{min}}} = \frac{c / \lambda_{L, 	ext{max}}}{c / \lambda_{B, 	ext{max}}} = \frac{\lambda_{B, 	ext{max}}}{\lambda_{L, 	ext{max}}} = \frac{36 / 5R}{4 / 3R} = \frac{36}{5} 	imes \frac{3}{4} = \frac{27}{5} = 5.4$.

The ratio of the minimum frequencies of the Lyman and Balmer series will be $......$.

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