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Question

(D) The frequency $
u$ of a spectral line in the hydrogen atom is given by $
u = c \cdot \bar{
u} = Rc \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} i

Vedclass · Accepted Answer

$\frac{5 Rc}{36}$

Vedclass · Answer

(D) The frequency $
u$ of a spectral line in the hydrogen atom is given by $
u = c \cdot \bar{
u} = Rc \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.For the Lyman series,$n_1 = 1$.The first Lyman line corresponds to the transition from $n_2 = 2$ to $n_1 = 1$:$
u_1 = Rc \left( \frac{1}{1^2} - \frac{1}{2^2} ight) = Rc \left( 1 - \frac{1}{4} ight) = \frac{3Rc}{4}$.The second Lyman line corresponds to the transition from $n_2 = 3$ to $n_1 = 1$:$
u_2 = Rc \left( \frac{1}{1^2} - \frac{1}{3^2} ight) = Rc \left( 1 - \frac{1}{9} ight) = \frac{8Rc}{9}$.The difference between the frequencies is $\Delta 
u = 
u_2 - 
u_1 = \frac{8Rc}{9} - \frac{3Rc}{4}$.Taking the common denominator as $36$:$\Delta 
u = \frac{32Rc - 27Rc}{36} = \frac{5Rc}{36}$.

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

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