λ = h/p = h/(mv). Find wavelength for electron at 100 eV.

de Broglie Wavelength of an Electron

Question

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

Given: mass of electron $m = 9.1 \times 10^{-31}$ kg, charge $e = 1.6 \times 10^{-19}$ C, Planck’s constant $h = 6.63 \times 10^{-34}$ J·s.

Solution — Step by Step

When a charge $e$ moves through potential difference $V$ , it gains kinetic energy:

KE = eV = 1.6 \times 10^{-19} \times 100 = 1.6 \times 10^{-17} \text{ J}

This is the work done by the electric field — all of it converts to kinetic energy.

We need momentum $p$ , not velocity. Why use $p$ directly? Because de Broglie’s formula is $\lambda = h/p$ , and going through $p$ avoids one calculation step.

KE = \frac{p^2}{2m} \implies p = \sqrt{2mKE}

p = \sqrt{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-17}}

p = \sqrt{2.912 \times 10^{-47}} = 5.396 \times 10^{-24} \text{ kg·m/s}

\lambda = \frac{h}{p} = \frac{6.63 \times 10^{-34}}{5.396 \times 10^{-24}}

\boxed{\lambda \approx 1.23 \times 10^{-10} \text{ m} = 1.23 \text{ Å}}

Why This Works

de Broglie proposed that every particle with momentum $p$ has an associated wavelength $\lambda = h/p$ . This isn’t just theory — electron diffraction experiments (Davisson-Germer, 1927) confirmed it directly, which is why this formula carries marks in every JEE and CBSE paper.

The key physical insight: we accelerated the electron, so the electric field did work on it. All that work became kinetic energy ( $eV = \frac{1}{2}mv^2$ ). From kinetic energy we get momentum, and from momentum we get wavelength. Every step follows from energy conservation.

At 100 eV, the wavelength comes out around 1.2 Å — comparable to atomic spacing in crystals. That’s exactly why electrons can diffract off crystal lattices, making them useful in electron microscopy.

Alternative Method

There’s a shortcut formula worth memorising for competitive exams — it appears in JEE Main 2024 and saves 40 seconds:

\lambda = \frac{12.27}{\sqrt{V}} \text{ Å}

where $V$ is the accelerating voltage in volts.

For $V = 100$ V:

\lambda = \frac{12.27}{\sqrt{100}} = \frac{12.27}{10} = 1.227 \text{ Å}

Memorise $\lambda = \frac{12.27}{\sqrt{V}}$ Å for electrons. It’s derived by combining $eV = p^2/2m$ and $\lambda = h/p$ with numerical substitution — same as what we did above, but pre-computed. Saves time in MCQ rounds.

This shortcut only works for electrons accelerated from rest. For protons or other particles, go back to first principles.

Common Mistake

Using $\lambda = h/mv$ and calculating $v$ separately — then students compute $v = \sqrt{2eV/m}$ and plug in. This is valid but introduces a rounding error midway. If you round $v$ to 3 significant figures before computing $\lambda$ , your answer shifts. Always go through $p = \sqrt{2mKE}$ directly, keeping full precision until the final step.

A second trap: using $KE$ in eV directly inside $p = \sqrt{2mKE}$ without converting to joules. The formula needs SI units — joules, not electron-volts.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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