Calculate de Broglie wavelength of electron accelerated through 100V potential

Question

Calculate the de Broglie wavelength of an electron accelerated through a potential difference of $100\,\text{V}$ .

(JEE Main 2023, similar pattern)

Solution — Step by Step

When an electron (charge $e$ ) is accelerated through potential $V$ :

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

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

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

= \sqrt{29.12 \times 10^{-48}} = 5.396 \times 10^{-24}\,\text{kg m/s}

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

= 1.228 \times 10^{-10}\,\text{m} \approx \mathbf{1.23\,\text{\AA}}

Why This Works

De Broglie’s hypothesis says every moving particle has an associated wavelength $\lambda = h/p$ . For an electron accelerated through voltage $V$ , the electric field gives it kinetic energy $eV$ , which determines its momentum, which determines its wavelength.

The result $\lambda \approx 1.23\,\text{\AA}$ for $V = 100\,\text{V}$ is comparable to atomic spacings in crystals — which is why electron diffraction works and why the electron microscope is useful.

Alternative Method — Use the shortcut formula

For electrons accelerated through $V$ volts:

\lambda = \frac{12.27}{\sqrt{V}}\,\text{\AA}

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

This shortcut is derived from combining $\lambda = h/\sqrt{2meV}$ with the known constants. It’s a massive time-saver in MCQs.

Memorise $\lambda = 12.27/\sqrt{V}$ (in angstroms) for electrons. For JEE, this formula converts a 5-step calculation into a 1-step answer. Also remember: $1\,\text{\AA} = 10^{-10}\,\text{m} = 0.1\,\text{nm}$ .

Common Mistake

Students sometimes use the relativistic momentum formula for this problem. At $V = 100\,\text{V}$ , the electron’s kinetic energy is $100\,\text{eV}$ , which is tiny compared to its rest energy of $0.511\,\text{MeV}$ . The non-relativistic approximation is perfectly valid here. Relativistic corrections become necessary only above ~ $10\,\text{kV}$ or so. Using the relativistic formula unnecessarily complicates the calculation for no gain in accuracy.

Question

Solution — Step by Step

Why This Works

Alternative Method — Use the shortcut formula

Common Mistake

Try These Next