### Select your language

Suggested languages for you:

Americas

Europe

4CQ

Expert-verified
Found in: Page 92

### Modern Physics

Book edition 2nd Edition
Author(s) Randy Harris
Pages 633 pages
ISBN 9780805303087

# 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.

The cut-off wavelengths will be the same in both cases.

See the step by step solution

## Step 1: Given Data

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

## Step 2: Emission of X-rays

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.

## Step 3: Relation between kinetic energy, work function, and wavelength

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, Ek=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.

## Step 4: Wavelength of spectra

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.