In terms of Rydberg's constant R, the wave nu | vedclass

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Question

(C) The wave number $\bar{
u}$ is given by the Rydberg formula: $\bar{
u} = \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.

Vedclass · Accepted Answer

$\frac{5R}{36}$

Vedclass · Answer

(C) The wave number $\bar{
u}$ is given by the Rydberg formula: $\bar{
u} = \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.For the Balmer series,the transition occurs to the $n_1 = 2$ energy level.The first line of the Balmer series corresponds to the transition from $n_2 = 3$ to $n_1 = 2$.Substituting these values into the formula:$\bar{
u} = R \left( \frac{1}{2^2} - \frac{1}{3^2} ight)$$\bar{
u} = R \left( \frac{1}{4} - \frac{1}{9} ight)$$\bar{
u} = R \left( \frac{9 - 4}{36} ight) = \frac{5R}{36}$.

In terms of Rydberg's constant $R,$ the wave number of the first Balmer line is

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