The wave number of the first line in the Lyma | vedclass

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Question

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

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$\frac{3 R}{4}$

Vedclass · Answer

(B) The wave number $\bar{
u}$ for a hydrogen atom is given by the Rydberg formula: $\bar{
u} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.For the Lyman series,the transition occurs to the ground state,so $n_1 = 1$.The first line of the Lyman series corresponds to the transition from the first excited state to the ground state,so $n_2 = 2$.Substituting these values into the formula:$\bar{
u} = R \left( \frac{1}{1^2} - \frac{1}{2^2} ight)$$\bar{
u} = R \left( 1 - \frac{1}{4} ight)$$\bar{
u} = R \left( \frac{3}{4} ight) = \frac{3 R}{4}$.

The wave number of the first line in the Lyman series of a hydrogen atom is ($R$ is Rydberg's constant).

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