Q.Explain the physical origin of the continuous X-ray spectrum produced in an X-ray tube, and derive the expression for its short-wavelength limit $\lambda_{\min}$ in terms of the accelerating voltage $V$.

As a fast electron passes close to a target atom's nucleus, the strong Coulomb attraction deflects and decelerates it, and this sudden loss of kinetic energy is radiated as a photon (bremsstrahlung, "braking radiation"). Because an electron can lose ANY fraction of its kinetic energy in a given encounter -- a little in a distant near-miss, up to virtually all of it in a very close encounter -- the resulting photons span a continuous range of energies, producing the continuous X-ray spectrum.

This spectrum has a sharp lower cut-off, $\lambda_{\min}$, reached in the limiting case where an electron loses its ENTIRE kinetic energy $eV$ (gained by accelerating through the tube voltage $V$) in a single encounter, converting it all into ONE photon:

$eV = \frac{hc}{\lambda_{\min}} \quad\Longrightarrow\quad \boxed{\lambda_{\min} = \frac{hc}{eV}}$ …

Explain the deceleration (bremsstrahlung) mechanism qualitatively, identify the maximum-energy special case (all KE into one photon) …

• Believing $\lambda_{\min}$ depends on the target material -- it is set entirely by the accelerating voltage. …