JEE Main 2026 — Physics Modern Physics

2026mcqmedium

The ratio of momentum of the photons of the $1^{st}$ and $2^{nd}$ line of Balmer series of Hydrogen atoms is $\alpha/\beta$ . The possible values of $\alpha$ and $\beta$ are:-

Official previous-year question

Held on 6 Apr 2026 · Verified 6 Jul 2026.

Options

A
$27$ and $20$
B
$3$ and $16$
C
$5$ and $36$
D
$20$ and $27$

Did you get this right?

Solution

The momentum of a photon is given by $p = \dfrac{h}{\lambda}$ .

From Rydberg's formula, the wave number is $\dfrac{1}{\lambda} = R \left( \dfrac{1}{n_1^2} - \dfrac{1}{n_2^2} \right)$ .

For the first line of the Balmer series, the transition is from $n_2 = 3$ to $n_1 = 2$ .

$p_1 = \dfrac{h}{\lambda_1} = hR \left( \dfrac{1}{2^2} - \dfrac{1}{3^2} \right) = hR \left( \dfrac{1}{4} - \dfrac{1}{9} \right) = \dfrac{5hR}{36}$

For the second line of the Balmer series, the transition is from $n_2 = 4$ to $n_1 = 2$ .

$p_2 = \dfrac{h}{\lambda_2} = hR \left( \dfrac{1}{2^2} - \dfrac{1}{4^2} \right) = hR \left( \dfrac{1}{4} - \dfrac{1}{16} \right) = \dfrac{3hR}{16}$

The ratio of the momenta is:

$\dfrac{p_1}{p_2} = \dfrac{\dfrac{5hR}{36}}{\dfrac{3hR}{16}} = \dfrac{5}{36} \times \dfrac{16}{3} = \dfrac{20}{27}$

Comparing with $\dfrac{\alpha}{\beta}$ , the possible values are $\alpha = 20$ and $\beta = 27$ .

Answer: $20$ and $27$

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JEE Main 2026 — Physics Modern Physics

2026mcqmedium

The ratio of momentum of the photons of the $1^{st}$ and $2^{nd}$ line of Balmer series of Hydrogen atoms is $\alpha/\beta$ . The possible values of $\alpha$ and $\beta$ are:-

Official previous-year question

Held on 6 Apr 2026 · Verified 6 Jul 2026.

Options

A
$27$ and $20$
B
$3$ and $16$
C
$5$ and $36$
D
$20$ and $27$

Did you get this right?

Solution

The momentum of a photon is given by $p = \dfrac{h}{\lambda}$ .

From Rydberg's formula, the wave number is $\dfrac{1}{\lambda} = R \left( \dfrac{1}{n_1^2} - \dfrac{1}{n_2^2} \right)$ .

For the first line of the Balmer series, the transition is from $n_2 = 3$ to $n_1 = 2$ .

$p_1 = \dfrac{h}{\lambda_1} = hR \left( \dfrac{1}{2^2} - \dfrac{1}{3^2} \right) = hR \left( \dfrac{1}{4} - \dfrac{1}{9} \right) = \dfrac{5hR}{36}$

For the second line of the Balmer series, the transition is from $n_2 = 4$ to $n_1 = 2$ .

$p_2 = \dfrac{h}{\lambda_2} = hR \left( \dfrac{1}{2^2} - \dfrac{1}{4^2} \right) = hR \left( \dfrac{1}{4} - \dfrac{1}{16} \right) = \dfrac{3hR}{16}$

The ratio of the momenta is:

$\dfrac{p_1}{p_2} = \dfrac{\dfrac{5hR}{36}}{\dfrac{3hR}{16}} = \dfrac{5}{36} \times \dfrac{16}{3} = \dfrac{20}{27}$

Comparing with $\dfrac{\alpha}{\beta}$ , the possible values are $\alpha = 20$ and $\beta = 27$ .

Answer: $20$ and $27$

Your note

JEE Main Modern Physics in other years

Keep practicing Modern Physics

Work through every JEE Main Modern Physics PYQ, year by year.

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JEE Main 2026 — Physics Modern Physics

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Solution

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JEE Main Modern Physics in other years

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JEE Main 2026 — Physics Modern Physics

Options

Did you get this right?

Solution

Your note

JEE Main Modern Physics in other years

More JEE Main Modern Physics questions

Other Physics topics for JEE Main

Mechanics

Electromagnetism

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Modern Physics