JEE Mains Physics · Dual Nature of Radiation and Matter

Comparing de Broglie Wavelengths

To compare two wavelengths, write λ = h/p and ask what is held fixed: at the same kinetic energy λ ∝ 1/√m, at the same voltage λ ∝ 1/√(mq), at the same speed λ ∝ 1/m, and the same wavelength means the same momentum.

All Dual Nature of Radiation and Matter notes JEE Mains Physics strategy

Why this matters

Thirty-four PYQs, thirty-three of them multiple choice, and two from 2026, the largest page in the chapter. Sixteen compare particles at given kinetic energies or accelerating voltages, eight of each. Eleven compare through momentum: seven at equal or given wavelengths, two for the pieces of a nucleus that splits at rest, and two at given speeds. Seven compare a particle with a photon. Each one is a single ratio; the marks go to knowing what the question holds fixed.

Concept 1 of 3: Comparing particles at given kinetic energies or voltages

Write λ = h/√(2mK) for both particles and divide; h and 2 cancel. At the same kinetic energy, only the masses are left, so the heavier particle has the shorter wavelength. Through the same voltage, the energy is qV, so the charge enters too: an alpha particle gains twice a proton's energy.

Definition

$\dfrac{\lambda_1}{\lambda_2} = \sqrt{\dfrac{m_2K_2}{m_1K_1}}$ .
Same kinetic energy: $\lambda \propto 1/\sqrt{m}$ .
Accelerated from rest through V: $K = qV$ , so $\dfrac{\lambda_1}{\lambda_2} = \sqrt{\dfrac{m_2q_2V_2}{m_1q_1V_1}}$ . Same V: $\lambda \propto 1/\sqrt{mq}$ .
Momentum after acceleration: $p = \sqrt{2mqV}$ .
Masses and charges: proton $m_p$ , e; neutron $\approx m_p$ , 0; deuteron $2m_p$ , e; alpha $4m_p$ , 2e; electron $\approx m_p/1836$ , −e.

Ratio of wavelengths

\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{m_2K_2}{m_1K_1}} = \sqrt{\frac{m_2q_2V_2}{m_1q_1V_1}}

Worked example

A deuteron is accelerated from rest through 50 V and an alpha particle through 25 V. Find the ratio of their de Broglie wavelengths, λ_d : λ_α.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 12 Apr 2023 · Q6Moderate

Example 1 · Dual Nature of Radiation and Matter · Comparing de Broglie Wavelengths

A proton and an

\alpha

-particle are accelerated from rest by

2\text{ }V

and

4\text{ }V

potentials, respectively. The ratio of their de-Broglie wavelength is:

The same voltage is not the same energy

Through the same V, a particle of charge q gains qV. An alpha particle, charge 2e, gains twice what a proton gains, so its mass AND its charge both enter the ratio.

Read the ratio the right way round

λ₁/λ₂ has the second particle's mass and energy on top. Inverting it gives the reciprocal, which is always one of the options.

At the same kinetic energy, heavier means shorter

λ ∝ 1/√m when K is fixed, so an alpha particle has the shortest wavelength and an electron the longest. A proton and a neutron come out nearly equal.

Drill 15 more on comparing particles at given kinetic energies or voltages

Concept 2 of 3: Same de Broglie wavelength means same momentum

λ = h/p, so two particles with the same wavelength have the same momentum, whatever their masses. Their speeds and energies then differ: with p fixed, v = p/m and K = p²/2m both go as 1/m. The same idea settles a nucleus that splits from rest: the two pieces fly apart with equal and opposite momenta.

Definition

Equal λ means equal p.
At equal p: $v \propto 1/m$ and $K = \dfrac{p^{2}}{2m} \propto 1/m$ .
In general $\dfrac{\lambda_1}{\lambda_2} = \dfrac{m_2v_2}{m_1v_1}$ ; at the same speed, $\lambda \propto 1/m$ .
A body at rest that splits in two: momentum is conserved, so the pieces have equal momenta and equal wavelengths, whatever their masses.

Momentum from wavelength

p = \frac{h}{\lambda}, \qquad K = \frac{p^{2}}{2m} = \frac{h^{2}}{2m\lambda^{2}}

Worked example

A deuteron and an alpha particle have the same de Broglie wavelength. Find the ratio, deuteron to alpha, of their (a) momenta, (b) speeds and (c) kinetic energies.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 1 February 2024 · Q11Moderate

Example 2 · Dual Nature of Radiation and Matter · Comparing de Broglie Wavelengths

The de Broglie wavelengths of a proton and an

\alpha

particle are

\lambda

and

2\lambda

respectively. The ratio of the velocities of proton and

\alpha

particle will be:

Equal wavelengths do not mean equal energies

The same λ means the same momentum. Kinetic energy is p²/2m, so the lighter particle carries more energy, in the inverse ratio of the masses.

Pieces from rest share one wavelength

A body at rest that splits sends its pieces off with equal and opposite momenta, so their wavelengths are equal. The masses do not enter.

At the same speed, λ goes as 1/m, not 1/√m

The square root belongs to the same-energy comparison. At equal speed, p = mv, so the wavelength ratio is the inverse mass ratio itself.

Drill 10 more on same de broglie wavelength means same momentum

Concept 3 of 3: Comparing a particle with a photon

Both obey λ = h/p, but their energies depend on momentum differently. A photon's energy is pc. A slow particle's kinetic energy is p²/2m. So write each energy from its own formula and never use p²/2m for light. At the same wavelength they share a momentum, and the particle carries far less energy; at the same energy, the photon has the far longer wavelength.

Definition

Photon: $E = pc$ , so $\lambda = \dfrac{h}{p} = \dfrac{hc}{E}$ .
Particle at non-relativistic speed: $K = \dfrac{p^{2}}{2m}$ , so $\lambda = \dfrac{h}{\sqrt{2mK}}$ .
Same wavelength: the momenta are equal. Write $K = p^{2}/2m$ for the particle and $E = pc$ for the photon, then divide.
Same energy: write each wavelength from its own formula, then divide.

Photon against particle

\lambda_{\text{photon}} = \frac{hc}{E}, \qquad \lambda_{\text{particle}} = \frac{h}{\sqrt{2mK}}

Worked example

An electron moves at

6 \times 10^{6}

m/s. A photon has the same de Broglie wavelength. Find the ratio of the electron's kinetic energy to the photon's energy. (m = 9.1 × 10⁻³¹ kg, c = 3 × 10⁸ m/s)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 28 January 2025 · Q18Moderate

Example 3 · Dual Nature of Radiation and Matter · Comparing de Broglie Wavelengths

A proton of mass '

m_{p}^{'}

' has same energy

E

as that of a photon of wavelength '

\lambda

'. If the proton is moving at non-relativistic speed, then ratio of its de Broglie wavelength to the wavelength of photon is.

A photon's energy is pc, not p²/2m

Light has no rest mass, so p²/2m means nothing for it. Use E = pc, and λ = hc/E, for the photon; keep p²/2m for the particle.

Same wavelength, same momentum, different energies

When a particle and a photon share a wavelength, they share a momentum, but the slow particle's kinetic energy is much smaller than the photon's energy.

Drill 6 more on comparing a particle with a photon

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Comparing particles at given kinetic energies or voltages
Ratio of wavelengths
$\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{m_2K_2}{m_1K_1}} = \sqrt{\frac{m_2q_2V_2}{m_1q_1V_1}}$
Same de Broglie wavelength means same momentum
Momentum from wavelength
$p = \frac{h}{\lambda}, \qquad K = \frac{p^{2}}{2m} = \frac{h^{2}}{2m\lambda^{2}}$
Comparing a particle with a photon
Photon against particle
$\lambda_{\text{photon}} = \frac{hc}{E}, \qquad \lambda_{\text{particle}} = \frac{h}{\sqrt{2mK}}$

Watch out for (8)

The same voltage is not the same energy
→ Comparing particles at given kinetic energies or voltages
Read the ratio the right way round
→ Comparing particles at given kinetic energies or voltages
At the same kinetic energy, heavier means shorter
→ Comparing particles at given kinetic energies or voltages
Equal wavelengths do not mean equal energies
→ Same de Broglie wavelength means same momentum
Pieces from rest share one wavelength
→ Same de Broglie wavelength means same momentum
At the same speed, λ goes as 1/m, not 1/√m
→ Same de Broglie wavelength means same momentum
A photon's energy is pc, not p²/2m
→ Comparing a particle with a photon
Same wavelength, same momentum, different energies
→ Comparing a particle with a photon

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