An electron having de-Broglie wavelength $$\lambda$$ is incident on a target in a X-ray tube. Cut-off wavelength of emitted X-ray is:

An electron with de-Broglie wavelength $$\lambda$$ has momentum $$p = \frac{h}{\lambda}$$.

The kinetic energy of the electron is $$K = \frac{p^2}{2m} = \frac{h^2}{2m\lambda^2}$$.

When this electron strikes the target and comes to a complete stop, all of its kinetic energy can be converted into a single X-ray photon. This gives the maximum energy photon, i.e., the cut-off (minimum wavelength) X-ray.

Setting the photon energy equal to the kinetic energy of the electron:

$$\frac{hc}{\lambda_0} = \frac{h^2}{2m\lambda^2}$$

Solving for the cut-off wavelength $$\lambda_0$$:

$$\lambda_0 = \frac{hc \cdot 2m\lambda^2}{h^2} = \frac{2mc\lambda^2}{h}$$