The ratio of the frequencies of the long wave | vedclass

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(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

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$27 : 5$

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(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

Similar Questions

Specify the quantum number for the limit wavelength of the spectral series of a hydrogen atom.

If an electron in a hydrogen atom jumps from an orbit of level $n=3$ to an orbit of level $n=2$,the emitted radiation has a frequency ($R=$ Rydberg constant,$C=$ velocity of light).

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Reason : Lyman series constitute spectral lines corresponding to transition from higher energy to ground state of hydrogen atom.

Assertion : Balmer series lies in the visible region of the electromagnetic spectrum.
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