The ratio of the frequencies of the long wave | vedclass

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Question

(A) The frequency $
u$ is given by $
u = \frac{c}{\lambda}$. The long wavelength limit corresponds to the minimum energy transition for a series.For

Vedclass · Accepted Answer

$27 : 5$

Vedclass · Answer

(A) The frequency $
u$ is given by $
u = \frac{c}{\lambda}$. The long wavelength limit corresponds to the minimum energy transition for a series.For the Lyman series,the transition is from $n_2 = 2$ to $n_1 = 1$:$
u_L = Rc \left( \frac{1}{1^2} - \frac{1}{2^2} ight) = Rc \left( 1 - \frac{1}{4} ight) = \frac{3Rc}{4}$For the Balmer series,the transition is from $n_2 = 3$ to $n_1 = 2$:$
u_B = Rc \left( \frac{1}{2^2} - \frac{1}{3^2} ight) = Rc \left( \frac{1}{4} - \frac{1}{9} ight) = Rc \left( \frac{9-4}{36} ight) = \frac{5Rc}{36}$The ratio of the frequencies is:$\frac{
u_L}{
u_B} = \frac{3Rc/4}{5Rc/36} = \frac{3}{4} 	imes \frac{36}{5} = \frac{3 	imes 9}{5} = \frac{27}{5}$Thus,the ratio is $27:5$.

The ratio of the frequencies of the long wavelength limits of Lyman and Balmer series of hydrogen spectrum is

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