A spectral line of Lyman series of hydrogen spectrum has a frequency of 2.466*10¹⁵s^-1.What transition is responsible for this spherical line.

Dear Student,

Please find below the solution to the asked query:


In Lyman series the lower stationary state n_i = 1
We have to find out the higher stationary state n_f = ?

The frequency can be related to the absorption or emission of photon as:

$$\begin{gathered} \nu =\frac{R_H}{h}\left(\frac{1}{n_i^2}-\frac{1}{n_f^2}\right) \\ or \nu = 3.29\times {10}^{15}\left(\frac{1}{n_i^2}-\frac{1}{n_f^2}\right) Hz \\ 2.466\times {10}^{15} =3.29\times {10}^{15}\left(\frac{1}{1^2}-\frac{1}{n_f^2}\right) \\ 2.466\times {10}^{15} -3.29\times {10}^{15}=-\frac{3.29\times {10}^{15}}{n_f^2} \\ 8.24\times {10}^{14}=\frac{3.29\times {10}^{15}}{n_f^2} \\ {n}_f^2 =\frac{3.29\times {10}^{15}}{8.24\times {10}^{14}}=4 \\ {n}_f=\sqrt{4}=2 \end{gathered}$$

Hence, the transition which is responsible for the spectral line is between n_i  = 1 and n_f = 2.

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