In a hydrogen spectrum, $\lambda$ is the wavelength of the first transition line of the Lyman series. The wavelength difference will be $a\lambda$ between the wavelength of the 3rd transition line of the Paschen series and that of the 2nd transition line of the Balmer series, where $a$ = ______.
Numerical value type. Enter your answer.
Answer: 5
Lyman first line ($2\to1$): $\dfrac1\lambda = \dfrac{3R}{4}\Rightarrow\lambda = \dfrac{4}{3R}$.
Paschen 3rd line ($6\to3$): $\dfrac1{\lambda_P} = R\left(\dfrac19 - \dfrac1{36}\right) = \dfrac{R}{12}\Rightarrow\lambda_P = \dfrac{12}{R}$.
Balmer 2nd line ($4\to2$): $\dfrac1{\lambda_B} = R\left(\dfrac14 - \dfrac1{16}\right) = \dfrac{3R}{16}\Rightarrow\lambda_B = \dfrac{16}{3R}$.
$$\lambda_P - \lambda_B = \frac{36 - 16}{3R} = \frac{20}{3R} = 5\times\frac{4}{3R} = 5\lambda\ \Rightarrow\ a = 5$$
Solution by Sreeraj P, M.Sc Physics