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(D) The frequency of radiation emitted during an electron transition is given by $
u = R_H c \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.For th

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$\frac{5 X}{36}$

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(D) The frequency of radiation emitted during an electron transition is given by $
u = R_H c \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.For the transition from an excited state $n$ to the ground state $n_1 = 1$,we have $
u_1 = R_H c \left( 1 - \frac{1}{n^2} ight) = \frac{3X}{4}$.Solving for $n$,$1 - \frac{1}{n^2} = \frac{3}{4} \implies \frac{1}{n^2} = \frac{1}{4} \implies n = 2$.The excited state is $n = 2$.The next immediate excited state is $n = 3$.The frequency of radiation absorbed for the transition from $n = 2$ to $n = 3$ is $
u_2 = R_H c \left( \frac{1}{2^2} - \frac{1}{3^2} ight) = R_H c \left( \frac{1}{4} - \frac{1}{9} ight) = R_H c \left( \frac{5}{36} ight)$.Since $R_H c = X$,we have $
u_2 = \frac{5X}{36} \ Hz$.

If the frequency of radiation emitted for the electron transition from an excited state to ground state of hydrogen atom is $\frac{3 X}{4} \ Hz$,the frequency of radiation absorbed for the electron transition from the above excited state to the next immediate excited state in $Hz$ is:

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