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

Since, we know that de-Broglie wavelength of an electron can be given as

$\mathrm{\lambda }=\frac{\mathrm{h}}{\mathrm{mv}}$           ...(i)

Also, we know that,

Kinetic energy, $\mathrm{K}=\frac{{\mathrm{p}}^2}{2\mathrm{m}}$          ...(ii)

where, p is the linear momentum.

$\because \mathrm{Linear} \mathrm{momentum}, \mathrm{p}=\frac{\mathrm{h}}{\mathrm{\lambda }}$       ...(iii)

From Eq. (ii) and Eq. (iii), we get

$\begin{array}{l}K=\frac{h^2}{2{\lambda }^{2}m}\\ \Rightarrow hv=\frac{h^2}{2{\lambda }^{2}m} \left\{\because K=hv=\frac{hc}{\lambda }\right\}\\ \Rightarrow \frac{hc}{{\lambda }_c}=\frac{h^2}{2{\lambda }^{2}m}\end{array}$

where, ${\mathrm{\lambda }}_{\mathrm{C}}$ = cut off wavelength.

$\Rightarrow {\mathrm{\lambda }}_{\mathrm{C}}=\frac{2{\mathrm{m\lambda }}^2\mathrm{C}}{\mathrm{h}}$