Electrons in a diffraction experiment are accelerated through a potential difference of $200\ \mathrm{V}$.
What is the de Broglie wavelength of these electrons?
$\lambda =0.087\ \mathrm{nm}$
By definition, the kinetic energy of these electrons is 200 eV. Since the relationship between linear momentum and kinetic energy is
$$p=\sqrt{2mK}$$
$$\lambda =\frac{h}{p}$$
$$\lambda =\frac{h}{\sqrt{2mK}}$$
$$\lambda =\frac{6.63\times 10^{-34}\ \mathrm{J\cdot s}}{\sqrt{2\left(9.11\times 10^{-31}\ \mathrm{kg}\right)\left[200\ \mathrm{eV}\cdot \displaystyle \frac{1.6\times 10^{-19}\ \mathrm{J}}{1\ \mathrm{eV}}\right]}}$$
$$\lambda =8.7\times 10^{-11}\ \mathrm{m}$$
$$\lambda =0.087\ \mathrm{nm}$$
This wavelength is characteristic of X‑rays.