Find the wave number of the shortest waveleng | vedclass

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Question

(A) For the Balmer series,the transition occurs to the $n_1 = 2$ energy level.For the shortest wavelength,the transition must occur from $n_2 = \infty

Vedclass · Accepted Answer

$2.7419 	imes 10^{4} \ cm^{-1}$

Vedclass · Answer

(A) For the Balmer series,the transition occurs to the $n_1 = 2$ energy level.For the shortest wavelength,the transition must occur from $n_2 = \infty$ to $n_1 = 2$.The Rydberg formula for wave number $(\bar{
u})$ is given by: $\bar{
u} = R_H 	imes Z^2 	imes (\frac{1}{n_1^2} - \frac{1}{n_2^2})$.For hydrogen,$Z = 1$ and $R_H = 109677 \ cm^{-1}$.Substituting the values: $\bar{
u} = 109677 	imes (\frac{1}{2^2} - \frac{1}{\infty^2}) = 109677 	imes \frac{1}{4} = 27419.25 \ cm^{-1}$.Thus,$\bar{
u} \approx 2.7419 	imes 10^{4} \ cm^{-1}$.

Find the wave number of the shortest wavelength in the Balmer series.

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