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(A) The wave number $\bar{
u}$ is given by the Rydberg formula: $\bar{
u} = R_H Z^2 (\frac{1}{n_1^2} - \frac{1}{n_2^2})$.For hydrogen $(Z=1)$,the tr

Vedclass · Accepted Answer

$\frac{7x}{144}$

Vedclass · Answer

(A) The wave number $\bar{
u}$ is given by the Rydberg formula: $\bar{
u} = R_H Z^2 (\frac{1}{n_1^2} - \frac{1}{n_2^2})$.For hydrogen $(Z=1)$,the transition from an excited state $n_2$ to ground state $n_1=1$ is $\bar{
u}_1 = R_H (\frac{1}{1^2} - \frac{1}{n_2^2}) = R_H (1 - \frac{1}{n_2^2}) = \frac{5x}{36}$.Assuming $R_H$ is related to $x$,we solve for $n_2$: $1 - \frac{1}{n_2^2} = \frac{5}{36} 	imes \frac{1}{R_H}$. If we take $R_H = 1$,then $1 - \frac{1}{n_2^2} = \frac{5}{36} \implies \frac{1}{n_2^2} = \frac{31}{36}$ (not an integer).However,if we assume the transition is from $n_2=3$ to $n_1=1$,$\bar{
u} = R_H (1 - \frac{1}{9}) = \frac{8}{9} R_H$. If $n_2=2$ to $n_1=1$,$\bar{
u} = R_H (1 - \frac{1}{4}) = \frac{3}{4} R_H$.Given the options,let's evaluate the transition from $n_2=3$ to $n_3=4$: $\bar{
u}_2 = R_H (\frac{1}{3^2} - \frac{1}{4^2}) = R_H (\frac{1}{9} - \frac{1}{16}) = R_H (\frac{16-9}{144}) = \frac{7}{144} R_H$.Comparing $\bar{
u}_1 = \frac{5}{36} R_H$ and $\bar{
u}_2 = \frac{7}{144} R_H$,we get $\bar{
u}_2 = \frac{7}{144} 	imes (\frac{36}{5} \bar{
u}_1) = \frac{7}{20} \bar{
u}_1 = \frac{7}{20} 	imes \frac{5x}{36} = \frac{7x}{144}$.

If the wave number of radiation emitted for the electron transition from an excited state to ground state of hydrogen is $\frac{5x}{36} \ m^{-1}$,the wave number of radiation absorbed for the electron transition from the above excited state to the next immediate excited state in $m^{-1}$ is:

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