What is the wave number of the lowest energy | vedclass

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(A) For the Lyman series,the electron transitions to the ground state,so $n_1 = 1$.The lowest energy transition in the Lyman series corresponds to the

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$\bar{v}=R_{H}\left(\frac{3}{4}\right)$

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(A) For the Lyman series,the electron transitions to the ground state,so $n_1 = 1$.The lowest energy transition in the Lyman series corresponds to the transition from the nearest higher energy level,which is $n_2 = 2$.The Rydberg formula for the wave number $\bar{v}$ is given by $\bar{v} = R_{H} \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.Substituting the values $n_1 = 1$ and $n_2 = 2$:$\bar{v} = R_{H} \left( \frac{1}{1^2} - \frac{1}{2^2} \right) \ cm^{-1}$$\bar{v} = R_{H} \left( 1 - \frac{1}{4} \right) \ cm^{-1}$$\bar{v} = R_{H} \left( \frac{3}{4} \right) \ cm^{-1}$

What is the wave number of the lowest energy transition associated with the Lyman series?

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