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

Similar Questions

Which series is found in the visible light region?

The difference between the frequencies of the second and first Paschen lines of the hydrogen atom is (where $R$ is the Rydberg constant and $c$ is the speed of light in vacuum).

Hydrogen atoms are excited from the ground state to the principal quantum number $n = 5$. The number of spectral lines observed will be:

If the wavelength of the first member of the Lyman series of hydrogen is $\lambda$,what is the wavelength of the second member?

Assertion: In the Lyman series,the ratio of minimum and maximum wavelength is $\frac{3}{4}$.
Reason: The Lyman series constitutes spectral lines corresponding to transitions from higher energy levels to the ground state of the hydrogen atom.

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