Q 12-12-134JEE MainJEE Main 2025 (7 Apr, Shift 1)Easy
For a hydrogen atom, the ratio of the largest wavelength of the Lyman series to that of the Balmer series is:
Answer: (B) $5 : 27$
The largest wavelength in each series comes from the smallest energy jump.
Lyman ($2 \to 1$): $\dfrac1{\lambda_L} = R\left(1 - \dfrac14\right) = \dfrac{3R}{4}$.
Balmer ($3 \to 2$): $\dfrac1{\lambda_B} = R\left(\dfrac14 - \dfrac19\right) = \dfrac{5R}{36}$.
$$\frac{\lambda_L}{\lambda_B} = \frac{4/3R}{36/5R} = \frac{20}{108} = \frac{5}{27}$$
Solution by Sreeraj P, M.Sc Physics