Q 12-12-045JEE MainJEE Main 2024 (6 Apr, Shift 1)Easy
The ratio of the shortest wavelength of the Balmer series to the shortest wavelength of the Lyman series for the hydrogen atom is
Answer: (A) $4:1$
Shortest wavelengths are the series limits ($n_2 \to \infty$):
Balmer: $\dfrac1\lambda = \dfrac R4 \Rightarrow \lambda_B = \dfrac4R$. Lyman: $\dfrac1\lambda = R \Rightarrow \lambda_L = \dfrac1R$.
Ratio $= 4:1$.
Solution by Sreeraj P, M.Sc Physics