The wave number of the first emission line in | vedclass

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Question

(A) The Rydberg formula for the wave number is given by $\bar{
u} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.For the Balmer series,$n_1

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$\frac{5R}{36} \ cm^{-1}$

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(A) The Rydberg formula for the wave number is given by $\bar{
u} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.For the Balmer series,$n_1 = 2$.The first emission line corresponds to the transition from $n_2 = 3$ to $n_1 = 2$.Substituting the values for the $H$ atom $(Z = 1)$:$\bar{
u} = R(1)^2 \left( \frac{1}{2^2} - \frac{1}{3^2} ight) = R \left( \frac{1}{4} - \frac{1}{9} ight)$.$\bar{
u} = R \left( \frac{9 - 4}{36} ight) = \frac{5R}{36} \ cm^{-1}$.

The wave number of the first emission line in the Balmer series of the $H$ atom spectrum will be .......

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