Q 12-12-059JEE MainJEE Main 2024 (31 Jan, Shift 1)Medium
If the wavelength of the first member of the Lyman series of hydrogen is $\lambda$, the wavelength of the second member will be
Answer: (A) $\dfrac{27}{32}\lambda$
$\dfrac1\lambda = R\left(1 - \dfrac14\right) = \dfrac{3R}{4}$ and $\dfrac1{\lambda'} = R\left(1 - \dfrac19\right) = \dfrac{8R}{9}$.
$$\frac{\lambda'}{\lambda} = \frac{3/4}{8/9} = \frac{27}{32}$$
Solution by Sreeraj P, M.Sc Physics