de Broglie Wavelength — Worked Problems

The Core Idea

In 1924, Louis de Broglie proposed that every moving particle behaves like a wave with wavelength $\lambda = h/p$ , where $p$ is the particle’s momentum and $h$ is Planck’s constant. This wave-particle duality unified the dual nature of light (photons) with matter (electrons, neutrons, atoms). It directly led to Schrödinger’s wave equation and the entire framework of quantum mechanics.

For Class 12 boards, JEE Main, and NEET, the de Broglie wavelength shows up in problems about electrons accelerated through potential differences, neutrons of given kinetic energy, and protons in particle accelerators. The math is short — a single formula plus careful unit handling. The conceptual subtlety lies in connecting kinetic energy, momentum, and accelerating voltage.

We’ll walk through the derivations once, drill three standard problem types, and flag the unit traps that catch students out.

Key Terms & Definitions

de Broglie wavelength: The wavelength associated with the matter wave of a particle of momentum $p$ :

$\lambda = \frac{h}{p}$

where $h = 6.626 \times 10^{-34}$ J s.

Momentum-Energy Relation (non-relativistic): For kinetic energy $K$ and mass $m$ :

$p = \sqrt{2 m K}$

So the de Broglie wavelength becomes:

$\lambda = \frac{h}{\sqrt{2 m K}}$

Accelerating Potential: When a charged particle of charge $q$ is accelerated through a potential difference $V$ , it gains kinetic energy $K = qV$ . So $p = \sqrt{2 m q V}$ and:

$\lambda = \frac{h}{\sqrt{2 m q V}}$

For an electron specifically, this simplifies to a memorable formula:

$\lambda \text{ (in Å)} = \sqrt{\frac{150}{V}}$

when $V$ is in volts. So $V = 150$ V gives $\lambda = 1$ Å.

Methods & Concepts

Three Standard Problem Setups

Type 1: Given speed. Use $p = m v$ directly, then $\lambda = h/(mv)$ . Straightforward.

Type 2: Given kinetic energy. Compute $p = \sqrt{2mK}$ , then $\lambda = h/p$ . Watch units — convert $K$ to joules if you’re using $h$ in SI.

Type 3: Given accelerating potential. Use $K = qV$ , then proceed as Type 2. For electrons, the $\sqrt{150/V}$ shortcut saves time.

$\lambda = \frac{h}{p} = \frac{h}{m v} = \frac{h}{\sqrt{2 m K}} = \frac{h}{\sqrt{2 m q V}}$

For electrons accelerated through $V$ volts: $\lambda \approx \sqrt{150/V}$ Å.

Why the Wavelength is So Small for Macroscopic Objects

Plug in a $1$ kg ball moving at $1$ m/s: $\lambda = h/p = 6.6 \times 10^{-34}$ m. That’s $10^{19}$ times smaller than the radius of an atomic nucleus. The wave nature is utterly undetectable — that’s why classical mechanics works for everyday objects.

For an electron at $10^6$ m/s: $\lambda = 7.3 \times 10^{-10}$ m, comparable to atomic spacing. That’s why electron diffraction works and why electrons in atoms must be treated as waves.

Solved Examples

Easy (CBSE)

Find the de Broglie wavelength of an electron moving at $v = 10^6$ m/s. ( $m_e = 9.1 \times 10^{-31}$ kg.)

$\lambda = \frac{6.63 \times 10^{-34}}{9.1 \times 10^{-31} \times 10^6} = \frac{6.63 \times 10^{-34}}{9.1 \times 10^{-25}} \approx 7.28 \times 10^{-10} \text{ m}$

That’s $7.28$ Å, comparable to atomic dimensions.

Medium (JEE Main)

An electron is accelerated from rest through a potential difference of $100$ V. Find its de Broglie wavelength.

Use the shortcut: $\lambda = \sqrt{150/V}$ Å $= \sqrt{1.5} \approx 1.225$ Å.

Or the long way: $K = qV = 1.6 \times 10^{-19} \times 100 = 1.6 \times 10^{-17}$ J. Then $p = \sqrt{2 m K} = \sqrt{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-17}} \approx 5.4 \times 10^{-24}$ kg m/s. So $\lambda = h/p \approx 1.23 \times 10^{-10}$ m $= 1.23$ Å. Same answer.

Hard (JEE Advanced)

A proton and an alpha particle are accelerated through the same potential difference $V$ . Find the ratio of their de Broglie wavelengths.

For both, $\lambda = h/\sqrt{2 m q V}$ . So $\lambda \propto 1/\sqrt{m q}$ . For proton: $m_p$ , $q = e$ . For alpha: $m_\alpha = 4 m_p$ , $q = 2e$ . So $m_\alpha q_\alpha = 8 m_p e$ .

$\frac{\lambda_p}{\lambda_\alpha} = \sqrt{\frac{m_\alpha q_\alpha}{m_p q_p}} = \sqrt{8} = 2\sqrt{2}$

The proton’s wavelength is $2\sqrt{2} \approx 2.83$ times larger.

Exam-Specific Tips

For JEE Main: The shortcut $\lambda = \sqrt{150/V}$ Å for electrons saves you 30 seconds per problem. Memorise it and recognise when to apply it.

For NEET: Often combined with the photoelectric effect. A typical problem: “Light of wavelength $\lambda_p$ ejects electrons with KE $K$ . Find their de Broglie wavelength $\lambda_e$ .” You’ll need to compute $K = h c/\lambda_p - \phi$ , then $\lambda_e = h/\sqrt{2 m K}$ .

For CBSE Boards: Derivation of the relation $\lambda = h/\sqrt{2 m q V}$ is a 3-mark question. Practice the algebra cleanly.

Common Mistakes to Avoid

Mistake 1: Using $p = mv$ even when speeds approach $c$ . For relativistic particles (high-energy electrons in accelerators), use $p = \gamma m v$ or $p = \sqrt{E^2/c^2 - m^2 c^2}$ instead. JEE rarely asks this, but flag it.

Mistake 2: Confusing wavelength of the photon ( $\lambda = hc/E$ ) with the de Broglie wavelength of a particle ( $\lambda = h/p$ ). Photons are massless — different formula.

Mistake 3: Forgetting to convert eV to joules when using $h$ in SI. $1$ eV $= 1.6 \times 10^{-19}$ J.

Mistake 4: Plugging $V$ into $\sqrt{150/V}$ in non-volts (e.g., kV without conversion). The formula needs $V$ in volts and gives $\lambda$ in angstroms.

Mistake 5: Treating $\sqrt{m q}$ wrong for an alpha particle. Alpha is doubly charged ( $q = 2e$ ) and four times heavier ( $m = 4m_p$ ). Both factors enter.

Practice Questions

Q1. An electron has kinetic energy $50$ eV. Find its de Broglie wavelength.

$\lambda = \sqrt{150/50}$ Å $= \sqrt{3} \approx 1.73$ Å.

Q2. A neutron has kinetic energy $0.025$ eV (thermal energy at room temperature). Find $\lambda$ . ( $m_n \approx 1.67 \times 10^{-27}$ kg.)

$K = 4 \times 10^{-21}$ J, $p = \sqrt{2 m K} \approx 1.16 \times 10^{-24}$ kg m/s, $\lambda \approx 1.79$ Å. Comparable to atomic spacing — that’s why thermal neutrons are used in crystallography.

Q3. Two electrons have de Broglie wavelengths in ratio $2:1$ . What’s the ratio of their kinetic energies?

$\lambda \propto 1/\sqrt{K}$ , so $K \propto 1/\lambda^2$ . Ratio is $1:4$ .

Q4. A proton accelerated through $V = 1000$ V. Find its de Broglie wavelength.

$\lambda = h/\sqrt{2 m q V}$ . For proton, $m = 1.67 \times 10^{-27}$ kg, $q = e$ . $\lambda \approx 9.06 \times 10^{-13}$ m $= 9.06 \times 10^{-3}$ Å. Much smaller than electron’s at same voltage — because proton is much heavier.

Q5. A photon and an electron have the same wavelength. Compare their momenta.

Both satisfy $p = h/\lambda$ . So momenta are equal — even though their energies are vastly different.

FAQs

Why don’t we see matter waves for cricket balls? The wavelength is so small ( $\sim 10^{-34}$ m) that no detector can resolve it. Wave behaviour is completely masked by classical motion.

Is the de Broglie wavelength a real wave? It’s a probability amplitude wave — the modulus squared gives the probability density of finding the particle. It’s not a vibration in a medium.

How was de Broglie’s hypothesis verified? By Davisson-Germer in 1927, who observed electron diffraction off a nickel crystal. The diffraction pattern matched the predicted wavelength exactly.

Does the de Broglie wavelength change in different reference frames? Yes — momentum is frame-dependent, so $\lambda$ is too. But the formula $\lambda = h/p$ holds in any frame.

Why is $h$ so small? It’s a fundamental constant of nature, $\sim 10^{-34}$ J s. If $h$ were larger, quantum effects would be visible at human scales — and the world would look very strange.