Q 12-12-027JEE MainJEE Main 2026 (6 Apr, Shift 1)Medium
The ratio of momentum of the photons of the $1^{\text{st}}$ and $2^{\text{nd}}$ line of Balmer series of Hydrogen atoms is $\alpha/\beta$. The possible values of $\alpha$ and $\beta$ are:-
Answer: (D) $20$ and $27$
Concept. Photon momentum $p = \dfrac{h}{\lambda} \propto \dfrac{1}{\lambda} \propto \left(\dfrac{1}{2^2} - \dfrac{1}{n^2}\right)$ for Balmer lines.
Step 1. First line ($n = 3$). $\dfrac14 - \dfrac19 = \dfrac{5}{36}$.
Step 2. Second line ($n = 4$). $\dfrac14 - \dfrac{1}{16} = \dfrac{3}{16}$.
Step 3. Ratio.
$$\frac{p_1}{p_2} = \frac{5/36}{3/16} = \frac{80}{108} = \frac{20}{27}$$
Answer. $\alpha = 20$, $\beta = 27$ (Option D).
Solution by Sreeraj P, M.Sc Physics
