Dual Nature NEET 2026: Electron-to-Photon Momentum Ratio

What is the answer to the Dual Nature NEET 2026 photon–electron question?

The electron-to-photon momentum ratio is 225, option C. This Dual Nature NEET 2026 question tests one distinction: equal energies do not mean equal momenta.

An electron with kinetic energy 20 eV and a photon with energy 20 eV travel through free space. Calculate the electron’s momentum divided by the photon’s momentum, keeping the electron in the numerator.

The supplied constants are:

c = 3 \times 10^{8} m s^{- 1}, m = 9 \times 10^{- 31} k g

q_{e} = - 1.6 \times 10^{- 19} C

The four options are:

A) & \frac{2}{450} B) & \frac{1}{250} C) & 225 D) & 275

The chapter is Dual Nature of Matter and Radiation. The question bank labels this question medium, with an expected solving time of 90 seconds, not an observed student completion time.

Which energy–momentum formula applies to the photon and the electron?

Use the photon’s energy divided by light speed for its momentum, but use the non-relativistic kinetic-energy relation for the electron. In the official method, the symbol below represents the common numerical energy: photon energy for the photon and kinetic energy for the electron.

E = E_{p h o t o n} = K_{e l e c t r o n}

For the photon, energy is momentum multiplied by light speed. Rearranging gives its momentum: $E = p_{P h} c$

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

For the electron, start with kinetic energy and momentum. Eliminate speed to express momentum in terms of kinetic energy:

K = \frac{m v^{2}}{2}, p_{e} = m v

E = \frac{p_{e}^{2}}{2 m}

$p_{e} = \sqrt{2 m E}$ The non-relativistic approximation is justified because 20 eV is far below the electron’s rest energy. Calculate that rest energy from the supplied constants:

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

m c^{2} = \frac{8.1 \times 10^{- 14}}{1.6 \times 10^{- 19}} e V \approx 5.06 \times 10^{5} e V

The photon relation below is not the appropriate kinetic-energy relation for this electron. Using it for both particles would incorrectly predict equal momenta: $E = p c$

How do you calculate the momentum ratio and get 225?

Form the electron-to-photon ratio first, then convert the common energy into joules. This gives 225 without calculating the two momenta separately. Keep the particle labels through substitution so the electron stays in the numerator:

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

Convert the energy before using the SI mass and speed. The electron’s negative charge does not make this conversion negative: an electronvolt uses the magnitude of the elementary charge.

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

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

Substitute the supplied mass, light speed and converted energy into the ratio. Keep the numerator and denominator inside the square root together:

\frac{p_{e}}{p_{P h}} = 3 \times 10^{8} \sqrt{\frac{2 \times 9 \times 10^{- 31}}{3.2 \times 10^{- 18}}}

Inside the square root, divide the coefficients and subtract the exponents. The resulting power of ten still sits inside the root:

\frac{18 \times 10^{- 31}}{3.2 \times 10^{- 18}} = \frac{18}{3.2} \times 10^{- 13} = 5.625 \times 10^{- 13}

Rewrite the power as an even exponent before taking the square root. Then take the square root of both the coefficient and the power of ten:

\sqrt{5.625 \times 10^{- 13}} = \sqrt{56.25 \times 10^{- 14}} = 7.5 \times 10^{- 7}

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

Option C is correct. The ratio is dimensionless because momentum units cancel: the electron has 225 times the photon’s momentum at the stated energies.

Why does reversing the ratio give option A?

Option A is exactly the reciprocal of the correct answer. Calculating photon momentum divided by electron momentum produces it, even if both energy–momentum formulas and every arithmetic step are correct. This is a numerator–denominator error, not evidence that the physical relations were wrong.

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

The same reversal matters when applying de Broglie’s relation. Wavelength varies inversely with momentum: $λ = \frac{h}{p}$

\frac{λ_{e}}{λ_{P h}} = \frac{p_{P h}}{p_{e}} \neq \frac{p_{e}}{p_{P h}}

Write the requested ratio with particle labels before substitution. Preserve those labels through the calculation, and check the final line against the question, not just against the available options.

What is the electron-to-photon wavelength ratio at 20 eV?

The electron-to-photon wavelength ratio is one divided by 225, the reciprocal of the momentum ratio. Original related practice question, not another verified NEET PYQ: for the same electron with kinetic energy 20 eV and photon with energy 20 eV, calculate the electron wavelength divided by the photon wavelength.

Apply de Broglie’s relation separately, then divide in the requested order. Keep the electron wavelength in the numerator:

λ_{e} = \frac{h}{p_{e}}, λ_{P h} = \frac{h}{p_{P h}}

\frac{λ_{e}}{λ_{P h}} = \frac{h / p_{e}}{h / p_{P h}} = \frac{p_{P h}}{p_{e}} = \frac{1}{225}

Planck’s constant cancels, so its numerical value is unnecessary. The electron has the smaller wavelength, consistent with its larger momentum.

What happens to the momentum ratio when both energies become 80 eV?

The electron-to-photon momentum ratio halves to 112.5. Original related practice question, not a sourced past-paper question: find the ratio when the electron’s kinetic energy and photon’s energy both increase from 20 eV to 80 eV.

Start from the expression already derived for a non-relativistic electron. Compare the new ratio with the old one:

R = c \sqrt{\frac{2 m}{E}}, R \propto E^{- 1 / 2}

\frac{R_{n e w}}{R_{o l d}} = \sqrt{\frac{20}{80}} = \frac{1}{2}

R_{n e w} = \frac{225}{2} = 112.5

Quadrupling energy doubles electron momentum but quadruples photon momentum, so their ratio halves. At 80 eV, the electron’s kinetic energy remains far below its rest energy, keeping the approximation valid.

For a changed-energy question, use the scaling ratio before repeating the full SI calculation. Keep the electron label in the numerator throughout.

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).

If that step was the hard part, work through How to Study Electromagnetic Waves NEET: A Study Plan.

Frequently asked questions

What is the electron-to-photon momentum ratio at 20 eV?

For an electron with kinetic energy 20 eV and a photon with energy 20 eV, the electron-to-photon momentum ratio is 225 using the supplied constants, so option C is correct. Use p_e = sqrt(2mE) and p_photon = E/c to obtain p_e/p_photon = c sqrt(2m/E). Convert 20 eV to 3.2 × 10^-18 J before substituting SI values.

Why can't I use E = pc for the electron in this question?

E = pc is the photon energy–momentum relation, not the kinetic-energy relation for this electron. The electron's kinetic energy of 20 eV is far below its rest energy of approximately 5.06 × 10^5 eV, so K = p²/(2m) is appropriate. Equal photon energy and electron kinetic energy therefore do not imply equal momenta.

What is the electron-to-photon wavelength ratio at 20 eV?

The electron-to-photon wavelength ratio is 1/225, the reciprocal of the momentum ratio. Since wavelength equals h/p, Planck's constant cancels when the two wavelengths are divided. The electron has the smaller wavelength because it has the larger momentum.

What happens to the momentum ratio if both energies increase to 80 eV?

The electron-to-photon momentum ratio falls from 225 to 112.5. For a non-relativistic electron, this ratio varies as E^(-1/2), so quadrupling both energies halves it. The non-relativistic approximation remains valid at 80 eV.

Dual Nature NEET 2026: Electron-to-Photon Momentum Ratio

What is the answer to the Dual Nature NEET 2026 photon–electron question?

Which energy–momentum formula applies to the photon and the electron?

How do you calculate the momentum ratio and get 225?

Why does reversing the ratio give option A?

What is the electron-to-photon wavelength ratio at 20 eV?

What happens to the momentum ratio when both energies become 80 eV?

Frequently asked questions

What is the electron-to-photon momentum ratio at 20 eV?

Why can't I use E = pc for the electron in this question?

What is the electron-to-photon wavelength ratio at 20 eV?

What happens to the momentum ratio if both energies increase to 80 eV?

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