Momentum of Photon NEET 2026: Electron-to-Photon Ratio

What is the correct answer to the momentum of photon NEET 2026 question?

The electron’s momentum is 225 times the photon’s momentum, so option C is correct in the momentum of photon NEET 2026 question. The electron’s stated energy is kinetic energy; the photon follows a different energy–momentum relation.

An electron with kinetic energy 20 eV and a photon with energy 20 eV travel freely. Calculate electron momentum divided by photon momentum, not the reverse.

The supplied constants are the speed of light, electron charge and electron mass:

c = 3 \times 10^{8} m / s, q_{e} = - 1.6 \times 10^{- 19} C, m = 9 \times 10^{- 31} k g .

The options are:

Source: NEET UG 2026, held on 21 June 2026, Code-50, Physics, Dual Nature of Matter and Radiation. “Medium” is the question bank’s difficulty classification.

Which momentum formulas should you use for the photon and electron?

Use energy divided by light speed for the photon, and the non-relativistic kinetic-energy relation for the electron. The first decision is the energy type: the electron’s 20 eV is kinetic energy, not total relativistic energy. Equal energies do not imply equal momenta for these different particles.

For the photon:

E = p_{P h} c \Rightarrow p_{P h} = \frac{E}{c} .

For the electron:

K = \frac{p_{e}^{2}}{2 m} \Rightarrow p_{e} = \sqrt{2 m K} .

These relations appear in the photon momentum and de Broglie treatment in NCERT Class 12, Dual Nature of Matter and Radiation. Applying the photon formula to the electron would incorrectly give equal momenta.

The second decision is to convert both energies to joules. Use the magnitude of the electron’s charge; its negative charge does not make the energy negative.

1 e V = 1.6 \times 10^{- 19} J .

K = E = 20 \times 1.6 \times 10^{- 19} = 3.2 \times 10^{- 18} J .

Check the non-relativistic approximation by comparing kinetic energy with rest energy. Using the supplied constants:

m c^{2} = (9 \times 10^{- 31}) (3 \times 10^{8})^{2} = 8.1 \times 10^{- 14} J .

\frac{K}{m c^{2}} = \frac{3.2 \times 10^{- 18}}{8.1 \times 10^{- 14}} \approx 3.95 \times 10^{- 5} .

The kinetic energy is much smaller than the rest energy, so the non-relativistic formula is appropriate. A relativistic calculation is unnecessary here.

How do you calculate electron momentum divided by photon momentum?

The ratio is 225 using the supplied constants. The third decision is ratio direction: keep electron momentum in the numerator and write “electron ÷ photon” beside the ratio before substituting. Use the common energy in joules, then rewrite the square root with an even power of ten.

Start with the requested order:

\frac{p_{e}}{p_{P h}} = \frac{\sqrt{2 m K}}{E / c} .

Since photon energy equals electron kinetic energy: $E = K .$

Therefore:

\frac{p_{e}}{p_{P h}} = \frac{c \sqrt{2 m K}}{K} = c \sqrt{\frac{2 m}{K}} .

Evaluate the quantity inside the square root first:

\frac{2 m}{K} = \frac{2 \times 9 \times 10^{- 31}}{3.2 \times 10^{- 18}} = \frac{18}{3.2} \times 10^{- 13} = 5.625 \times 10^{- 13} .

Rewrite the quantity so the power of ten has an even exponent. This lets you take the square root without guessing the decimal:

\sqrt{5.625 \times 10^{- 13}} = \sqrt{0.5625 \times 10^{- 12}} = 0.75 \times 10^{- 6} = 7.5 \times 10^{- 7} .

Now multiply by the speed of light:

\frac{p_{e}}{p_{P h}} = (3 \times 10^{8}) (7.5 \times 10^{- 7}) = 22.5 \times 10 = 225 .

The ratio is dimensionless. As a units check, the square-root factor has units of seconds per metre:

[\sqrt{\frac{m}{K}}] = \sqrt{\frac{k g}{k g m^{2} s^{- 2}}} = s / m .

These cancel the metres per second supplied by light speed. The result matches option C exactly with the given constants, without rounding to a nearby option.

Why does reversing the ratio give option A?

Option A answers the reversed question: photon momentum divided by electron momentum. This error can survive correct formulas, energy conversion and arithmetic. The numerator and denominator simply answer the wrong question.

Taking the reciprocal of the solved result gives:

\frac{p_{P h}}{p_{e}} = \frac{1}{225} = \frac{2}{450} .

That is option A, not an alternative form of option C. The solved result says the electron has greater momentum here, so the requested electron-to-photon ratio must exceed 1.

Prevent this by writing “electron ÷ photon” beside your starting ratio, then checking your final expression against those words. A fraction below 1 is a warning that you have reversed the order.

Can you solve three related momentum and wavelength questions?

The answers are 112.5, 225 and a doubling of momentum, respectively. These are original practice variations, not verified past-paper questions. They test energy scaling, ratio direction and wavelength’s inverse dependence on momentum.

What is the momentum ratio if both energies are 80 eV?

The ratio falls to 112.5 because it varies inversely with the square root of the common energy. Here, the electron’s 80 eV again means kinetic energy.

R = \frac{p_{e}}{p_{P h}} = c \sqrt{\frac{2 m}{K}} \Rightarrow R \propto \frac{1}{\sqrt{K}} .

\frac{R_{n e w}}{R_{o l d}} = \sqrt{\frac{20}{80}} = \frac{1}{2}, R_{n e w} = 225 \times \frac{1}{2} = 112.5 .

Answer: 112.5. Quadrupling both energies halves their momentum ratio.

What is photon wavelength divided by electron wavelength for the 20 eV pair?

The wavelength ratio is 225 because wavelength is inversely proportional to momentum. Planck’s constant cancels, so you do not need its numerical value. $λ = \frac{h}{p} .$

\frac{λ_{P h}}{λ_{e}} = \frac{h / p_{P h}}{h / p_{e}} = \frac{p_{e}}{p_{P h}} = 225 .

Answer: 225. The electron has the shorter wavelength because it has the greater momentum.

What happens when a photon’s energy rises from 20 eV to 40 eV?

Its momentum doubles because photon momentum is directly proportional to energy. Do not use the electron’s square-root scaling here.

p = \frac{E}{c}, \frac{p_{n e w}}{p_{o l d}} = \frac{40}{20} = 2 .

Answer: the photon’s momentum doubles. Keep the scaling rules separate:

Photon: $p_{P h} \propto E .$
Non-relativistic electron: $p_{e} \propto \sqrt{K} .$

Cover the working and redo these three variations. Write the requested ratio in words before substituting.

Next step: the past-paper archive on NEET JEEnius AI and search past NEET papers by year, subject or chapter, each with a worked solution (100 free searches a month).

Keep going with How to Study Properties of Matter NEET: A 6-Step Plan.

Frequently asked questions

What is the answer to the photon momentum question in NEET 2026?

For an electron with kinetic energy 20 eV and a photon with energy 20 eV, electron momentum divided by photon momentum is 225 using the supplied constants. Option C is correct; the electron has the greater momentum.

Which momentum formulas should I use for an electron and a photon?

For a photon, use p = E/c; for a non-relativistic electron, use p = √(2mK), where K is kinetic energy. Convert energies from eV to joules before substituting SI constants. At 20 eV, the electron's kinetic energy is much smaller than its rest energy, so the non-relativistic formula is appropriate.

Why is option A wrong in the 20 eV electron and photon question?

Option A, 2/450, equals 1/225 and gives photon momentum divided by electron momentum. The question asks for electron momentum divided by photon momentum, which is 225. Write the requested ratio in words before substituting to avoid this reversal.

What is the electron-to-photon momentum ratio if both energies are 80 eV?

The ratio is 112.5 using the same supplied constants, with 80 eV representing the electron's kinetic energy and the photon's energy. For equal energies, the electron-to-photon momentum ratio varies inversely with the square root of the common energy. Increasing both energies from 20 eV to 80 eV therefore halves the ratio from 225 to 112.5.