Suppose we produce X-rays not by smashing electrons into targets but by smashing protons, which are far more massive. If the same accelerating potential difference were used for both, how would the cut off wavelengths of the two X-ray spectra compare? Explain.

X-rays are produced by smashing protons instead of electrons using the same potential difference.

The x-rays are energetic electromagnetic radiation having shorter wavelengths, which are produced X-rays are emitted from the metal surface when the energy of the bombarding particles is more than the work function of the given metal.

The following equation represents the relation between kinetic energy, work function, and wavelength of spectra.

$\begin{array}{rcl}\mathrm{hv}& =& {\mathrm{E}}_{\mathrm{k}}+\mathrm{\varphi }\\ \mathrm{v}& =& \frac{\mathrm{c}}{\mathrm{\lambda }}\\ \mathrm{h}\frac{\mathrm{c}}{\mathrm{\lambda }}& =& {\mathrm{E}}_{\mathrm{k}}+\mathrm{\varphi }\\ \mathrm{h}\frac{\mathrm{c}}{\mathrm{\lambda }}& =& \mathrm{qV}+\mathrm{\varphi }\\ \lambda & =& \frac{hc}{\left(qV+\varphi \right)}\end{array}$

Where,

data-custom-editor="chemistry" $\mathrm{\lambda }$ is the wavelength of the spectra, c is the speed of light in the vacuum, E_k=eV is the kinetic energy of a particle, q is the charge of the particle, V is the potential difference in Volts, data-custom-editor="chemistry" $\mathrm{\varphi }$is the work function of the metal, and h is the Planck's constant.

The cut-off wavelength depends upon the charge of the particle, potential difference, and work function of the metal. As the charge of a proton is equal to that of an electron, the cut-off wavelength would not be affected by using protons instead of electrons under the same potential difference. Also, the wavelength of spectra does not depend upon the mass of the particle. So, it will not change with the mass of the electron or proton.