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Comparing de Broglie Wavelengths of Different Particles

Physics · Dual Nature Of Radiation And Matter · NEET

The de Broglie wavelength is lambda = h/p, so a heavier or faster particle always has a SMALLER wavelength. The key is to know what is kept the same. Same momentum p means same lambda for all particles. Same kinetic energy E means lambda = h/sqrt(2mE), so lambda is proportional to 1/sqrt(m) (heavier = smaller lambda). Same speed v means lambda = h/(mv), so lambda is proportional to 1/m. Memory hook: "Small p, big wave; big mass, small wave."
de Broglie wavelength: what is kept the same?Same momentum plambda = h / pNo mass in formulaAll particles: SAME lambdaSame energy Elambda = h / sqrt(2mE)lambda ~ 1 / sqrt(m)Lighter = longer lambdaSame speed vlambda = h / (mv)lambda ~ 1 / mLighter = longer lambdaRead the fixed quantity first, then write p, then lambda = h/p
Three cases for comparing de Broglie wavelengths. Same momentum gives equal wavelengths for all particles; same energy gives lambda proportional to 1/sqrt(m); same speed gives lambda proportional to 1/m. Always find momentum first.

Your doubts, answered

Two particles have the SAME momentum. Which one has the larger de Broglie wavelength?

Neither. lambda = h/p depends only on momentum p, not on mass. If p is the same, lambda is the same for an electron, a proton, or any particle. Mass does not appear in lambda = h/p. This is the most common trap: mass only matters when energy or speed is fixed, not when momentum is fixed.

At the SAME kinetic energy, does a proton or an electron have the bigger wavelength?

The electron. When energy E is fixed, use lambda = h/sqrt(2mE), so lambda is proportional to 1/sqrt(m). The lighter particle (electron) has the larger wavelength. A proton is about 1836 times heavier than an electron, so the electron wavelength is sqrt(1836) which is about 43 times longer than the proton wavelength at the same energy.

When is lambda proportional to 1/sqrt(m) and when is it 1/m?

It depends on what is held constant. Same kinetic energy: p = sqrt(2mE), so lambda = h/sqrt(2mE), which is 1/sqrt(m). Same speed v: p = mv, so lambda = h/(mv), which is 1/m. Same accelerating voltage V (for a charge q): p = sqrt(2mqV), so lambda is 1/sqrt(mq). Always write p first, then divide h by it.

A proton and an alpha particle get the same kinetic energy. What is the ratio of their wavelengths?

lambda = h/sqrt(2mE). Same E means lambda is proportional to 1/sqrt(m). The alpha mass is 4 times the proton mass, so lambda_proton / lambda_alpha = sqrt(m_alpha/m_proton) = sqrt(4) = 2. So lambda_p : lambda_alpha = 2 : 1. The lighter proton has the longer wave.

For a thermal neutron, what energy do I use in the formula?

A thermal neutron is in thermal equilibrium, so its average kinetic energy is (3/2)kT. Then p = sqrt(2m x (3/2)kT) = sqrt(3mkT), giving lambda = h/sqrt(3mkT). Do NOT use (1/2)kT here; the 3/2 factor comes from three directions of motion in a gas.

⚠️ The NEET trap
✗ Same momentum, so the lighter electron must have the longer de Broglie wavelength.
✓ Same momentum means SAME wavelength for every particle, because lambda = h/p has no mass in it. Mass only changes lambda when energy, speed, or voltage is fixed.
🧠 NTA loves swapping 'same momentum' with 'same energy'. Read which quantity is fixed FIRST, then pick the right form: p fixed to equal lambda, E fixed to 1/sqrt(m), v fixed to 1/m.

Real NEET questions

NEET 2019 (Odisha)

A proton and an alpha-particle are accelerated from rest to the same energy. The de Broglie wavelengths lambda_p and lambda_alpha are in the ratio:

A · sqrt(2) : 1
B · 1 : 1
C · 2 : 1 ✓
D · 4 : 1
Solution: Use lambda = h/sqrt(2mE). Same energy E, so lambda is proportional to 1/sqrt(m). Alpha mass = 4 x proton mass. lambda_p / lambda_alpha = sqrt(m_alpha/m_p) = sqrt(4) = 2. So lambda_p : lambda_alpha = 2 : 1.
NEET 2017

The de Broglie wavelength of a neutron in thermal equilibrium with heavy water at a temperature T (kelvin) and mass m is:

A · h/sqrt(mkT)
B · h/sqrt(3mkT) ✓
C · 2h/sqrt(3mkT)
D · 2h/sqrt(mkT)
Solution: Thermal kinetic energy KE = (3/2)kT. Momentum p = sqrt(2m x KE) = sqrt(2m x (3/2)kT) = sqrt(3mkT). So lambda = h/p = h/sqrt(3mkT).

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Frequently asked

What is the single rule to compare de Broglie wavelengths?

Always write momentum p first for the given condition, then use lambda = h/p. Smaller p gives a larger wavelength.

Do heavier particles always have a smaller de Broglie wavelength?

Only when energy, speed, or voltage is fixed. At the SAME momentum, all particles have the same wavelength regardless of mass.

How much longer is the electron wavelength than the proton at the same energy?

About 43 times longer, because lambda is proportional to 1/sqrt(m) and the proton is about 1836 times heavier, and sqrt(1836) is about 43.

What formula do I use for the same accelerating voltage?

For charge q accelerated through voltage V, p = sqrt(2mqV), so lambda = h/sqrt(2mqV), which is proportional to 1/sqrt(mq).

Why does a thermal neutron use (3/2)kT and not (1/2)kT?

A thermal particle moves in three directions, so its average kinetic energy is (3/2)kT. This gives lambda = h/sqrt(3mkT).

Related concepts:
de Broglie Wavelength: Proton vs Alpha Particlede Broglie Wavelength: Photon vs Electronde Broglie Wavelength of a Thermal Neutronde Broglie Wavelength in Terms of Kinetic EnergyWhat Is the de Broglie Wavelength
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