Cut-off Wavelength of X-rays from Incident Electrons

To find the cut-off wavelength ($\lambda_0$) of the emitted X-rays, we relate the kinetic energy of the incident electrons to the maximum energy of the emitted photons.

1. Kinetic Energy of the Electron:
   The de-Broglie wavelength $\lambda$ of an electron of mass $m$ and momentum $p$ is given by:
   $\lambda = \frac{h}{p} \implies p = \frac{h}{\lambda}$

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

1. Cut-off Wavelength of the Emitted X-rays:
   In an X-ray tube, the minimum (cut-off) wavelength $\lambda_0$ corresponds to the maximum photon energy, which occurs when the entire kinetic energy of the incident electron is converted into a single X-ray photon:
   $E_{\text{max}} = K$
   $\frac{hc}{\lambda_0} = \frac{h^2}{2m\lambda^2}$

2. Solving for $\lambda_0$:
   $\frac{c}{\lambda_0} = \frac{h}{2m\lambda^2}$
   $\lambda_0 = \frac{2mc\lambda^2}{h}$

Therefore, the correct option is (A).