The de-Broglie wavelength of electron falling on the target in an...

The de-Broglie wavelength is given by $\lambda=\frac{h}{p}=\frac{h}{\sqrt{2 m E}}$ ...(i) where, $E$ is the energy of the electron. The cut-off wavelength $\lambda_0$ is given by $\lambda_0=\frac{h C}{E}$ ...(ii) From Eq. (i), $\lambda^2=\frac{h^2}{2 m E}$ $\Rightarrow \quad E=\frac{h^2}{2 m X^2}$ ... (iii) Substituting the value of $E$ from Eq. (ii), we get $\lambda_0=\frac{h c}{\frac{h^2}{2 m \lambda^2}}=\frac{2 m c \lambda^2}{h}$