How is the de Broglie wavelength of an electron accelerated through a potential difference V derived?

In one line

An electron accelerated from rest through a potential difference \(V\) gains kinetic energy \(eV\), so its de Broglie wavelength is \(\lambda = h/\sqrt{2meV} \approx 12.27/\sqrt{V}\) Å.

Explained simply

In an electron gun (like in an old TV or an electron microscope), electrons start nearly at rest and are pulled by a voltage. The bigger the voltage, the faster they move and the larger their momentum. Since wavelength is \(h/p\), higher voltage means shorter wavelength.

How it works

1. Work done on the electron: \(eV\). This becomes kinetic energy: \(\dfrac{1}{2}mv^2 = eV\).
2. Momentum: \(p = mv = \sqrt{2m\cdot\tfrac{1}{2}mv^2} = \sqrt{2meV}\).
3. de Broglie wavelength: \(\lambda = \dfrac{h}{p} = \dfrac{h}{\sqrt{2meV}}\).
4. Substitute \(h = 6.63\times10^{-34}\) J s, \(m = 9.11\times10^{-31}\) kg, \(e = 1.6\times10^{-19}\) C: \(\lambda = \dfrac{1.227\times10^{-9}}{\sqrt{V}}\) m \(= \dfrac{12.27}{\sqrt{V}}\) Å.
5. In terms of kinetic energy \(E\): \(\lambda = h/\sqrt{2mE}\). For any charged particle of charge \(q\): \(\lambda = h/\sqrt{2mqV}\).
6. This is valid when \(V\) is not too large (non-relativistic, roughly below a few kV for good accuracy).

Key formula

\(\lambda = \dfrac{h}{\sqrt{2meV}} = \dfrac{12.27}{\sqrt{V}}\) Å. \(m\): electron mass (kg), \(e\): electron charge (C), \(V\): accelerating voltage (V), 1 Å \(= 10^{-10}\) m.

Quick example

For \(V = 100\) V: \(\lambda = 12.27/\sqrt{100} = 1.227\) Å. For \(V = 10\) kV: \(\lambda = 12.27/100 = 0.1227\) Å, far smaller than visible light (about 5000 Å), which is why electron microscopes resolve much finer detail.

Common mistakes

• Using \(12.27/V\) instead of \(12.27/\sqrt{V}\).
• Forgetting that the answer from 12.27/\(\sqrt{V}\) is in Å, not m.
• Using the electron formula for protons or alpha particles without changing \(m\) and \(q\).

How MU asks it

Asked as "Show that the de Broglie wavelength of an electron accelerated through V volts is \(12.27/\sqrt{V}\) Å" (3 to 5 marks), and as numericals. Show each step from \(eV = \frac{1}{2}mv^2\), substitute constants, and state units clearly.