De Broglie Wavelength Of Accelerated Electron

When a charge $e$ is accelerated through a potential difference $V$, it gains kinetic energy $eV=\frac{p^2}{2m}$, so its momentum becomes $p=\sqrt{2meV}$ and its de Broglie wavelength is $\lambda =\frac{h}{\sqrt{2meV}}$. For electrons this reduces to the handy formula $\lambda =\frac{12.27}{\sqrt{V}}\text{A˚}$ with $V$ in volts. Substituting $V=100\text{V}$ gives $\lambda =\frac{12.27}{\sqrt{100}}=\frac{12.27}{10}=1.23\text{A˚}$. The option $0.123\text{A˚}$ misplaces a power of ten. The option $12.3\text{A˚}$ forgets the square root and divides by $V$ instead of $\sqrt{V}$. The option $0.39\text{A˚}$ wrongly assumes a much higher accelerating voltage. The fact that the wavelength is around one angstrom, matching crystal-lattice spacings, is precisely why such electrons diffract in the Davisson-Germer arrangement, as emphasised in NCERT. A magnitude check confirms the result is comparable to interatomic distances, validating the use of crystals as natural diffraction gratings for electrons. It should be noted that this simple formula assumes the electron remains non-relativistic, which holds well for accelerating voltages up to several kilovolts; at much higher energies a relativistic correction to the momentum becomes necessary. The remarkable practical consequence is that the electron microscope, by accelerating electrons to short wavelengths, resolves structures far finer than any light microscope can ever reach.