de Broglie WavelengthJEE Main

Concept

de Broglie proposed that every moving particle has a wavelength $\lambda = h/p$. For everyday objects it is far too small to notice, but for electrons it is comparable to atomic spacings — which is why electrons diffract, and why electron microscopes work.

Key formula

Derivation

Combining the photon relations $E = h\nu$ and $p = E/c$ with $c = \nu\lambda$ gives $\lambda = h/p$. de Broglie extended this to matter.

A charge $q$ accelerated through $V$ gains $K = qV$, so its momentum is $p = \sqrt{2mK} = \sqrt{2mqV}$. For an electron this gives the handy result $\lambda = 12.27/\sqrt{V}\;\text{Å}$ with $V$ in volts.

Scenarios to explore

• de Broglie Wavelength — Matter waves: λ = h/√(2mqV).

Real-world applications

• Electron microscopes (far finer resolution than light).
• Electron & neutron diffraction for crystal structure.
• Confirming wave–particle duality (Davisson–Germer experiment).

JEE exam tips

• Electron shortcut: $\lambda = \dfrac{12.27}{\sqrt{V}}\,\text{Å}$ (V in volts) — e.g. 100 V → 1.227 Å.
• At the same KE, $\lambda \propto 1/\sqrt{m}$, so a proton's wave is ~43× shorter than an electron's.

Common mistakes

• Using $K$ in eV instead of joules when computing $p = \sqrt{2mK}$.
• Forgetting an alpha particle carries charge $2e$, so $K = 2eV$.
• Thinking only charged particles have a wavelength — all matter does.

Exam traps to avoid

• $\lambda$ depends on momentum, not energy directly — relate them carefully.
• Doubling the voltage shortens the wavelength by only $\sqrt2$, not 2.