Physics Q44 – JEE-MAIN 2026

The ratio of momentum of the photons of the

1^{s t}

and

2^{n d}

line of Balmer series of Hydrogen atoms is

α / β

. The possible values of

α

and

β

are:-

Clue 1 of 3: Photon Momentum and Energy

Recall the relationship between photon momentum and its energy.

🥷

Ninja Strategy•Energy Comparison and Ratio Magnitude

Compare the energy differences for the two transitions to determine if the ratio $p_{1} / p_{2}$ should be greater or less than 1, then eliminate options that don't fit this criterion or are not actual ratios.

🧠 Full Solution Path

Step 1: Relate Momentum to Energy and Identify Transitions✦ Active

The momentum of a photon $p$ is related to its energy $E$ by $p = E / c$ , where $c$ is the speed of light. The energy of a photon emitted from a hydrogen atom (Z=1) during a transition from $n_{i}$ to $n_{f}$ is $E = R_{H} h c (\frac{1}{n_{f}^{2}} - \frac{1}{n_{i}^{2}})$ . Thus, $p = R_{H} h (\frac{1}{n_{f}^{2}} - \frac{1}{n_{i}^{2}})$ , implying $p \propto (\frac{1}{n_{f}^{2}} - \frac{1}{n_{i}^{2}})$ . For the Balmer series, $n_{f} = 2$ .

The $1^{s t}$ line of the Balmer series corresponds to the transition $n_{i} = 3 \to n_{f} = 2$ . The $2^{n d}$ line corresponds to $n_{i} = 4 \to n_{f} = 2$ .

Step 2: Calculate Proportionality Factors for Each Line○ Expand

For the $1^{s t}$ line ( $p_{1}$ ):

p_{1} \propto (\frac{1}{2^{2}} - \frac{1}{3^{2}}) = (\frac{1}{4} - \frac{1}{9}) = \frac{9 - 4}{36} = \frac{5}{36}

For the $2^{n d}$ line ( $p_{2}$ ):

p_{2} \propto (\frac{1}{2^{2}} - \frac{1}{4^{2}}) = (\frac{1}{4} - \frac{1}{16}) = \frac{4 - 1}{16} = \frac{3}{16}

💡 Teacher's Secret Hint

Ensure correct identification of initial and final energy levels for each line of the Balmer series.

Step 3: Determine the Ratio of Momenta○ Expand

The ratio of the momentum of the photons of the $1^{s t}$ and $2^{n d}$ line is $\frac{p_{1}}{p_{2}}$ :

\frac{p_{1}}{p_{2}} = \frac{5 / 36}{3 / 16} = \frac{5}{36} \times \frac{16}{3} = \frac{5 \times 4}{9 \times 3} = \frac{20}{27}

Thus, $α / β = 20 / 27$ . The possible values for $α$ and $β$ are $20$ and $27$ respectively.

💡 Teacher's Secret Hint

Simplify the fraction carefully to avoid calculation errors.

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Physics Question 44 – JEE-MAIN 2026

Clue 1 of 3: Photon Momentum and Energy

🧠 Full Solution Path