explain that X-rays are produced by electron bombardment of a metal target and calculate the minimum wavelength of X-rays produced from the accelerating p.d.

Published by Patrick Mutisya · 14 days ago

Production and Use of X‑rays – Cambridge A‑Level Physics 9702

Production and Use of X‑rays

Learning Objective

Explain how X‑rays are produced by electron bombardment of a metal target and calculate the minimum wavelength of X‑rays emitted from a given accelerating potential.

1. How X‑rays are Generated

In an X‑ray tube a high‑voltage source creates an accelerating potential \$V\$ between a cathode and an anode (metal target). Electrons are emitted from the heated cathode, accelerated through the potential difference, and strike the target. Two processes occur:

  • Bremsstrahlung (braking radiation): Deceleration of electrons in the electric field of the nuclei produces a continuous spectrum of X‑rays.

  • Characteristic radiation: Electrons knock inner‑shell electrons from the target atoms; outer‑shell electrons fill the vacancies, emitting X‑rays of discrete energies.

2. The Accelerating Potential and Minimum Wavelength

The maximum energy that an electron can transfer to a photon is equal to its kinetic energy after acceleration:

\$\$

E_{\text{max}} = eV

\$\$

where \$e = 1.602\times10^{-19}\,\text{C}\$ is the elementary charge.

If all this energy is converted into a single photon, the photon’s energy is \$E = hc/\lambda_{\min}\$, giving the minimum wavelength:

\$\$

\lambda_{\min} = \frac{hc}{eV}

\$\$

Using \$h = 6.626\times10^{-34}\,\text{J·s}\$ and \$c = 3.00\times10^{8}\,\text{m·s}^{-1}\$, the expression can be written in convenient units:

\$\$

\lambda_{\min}(\text{nm}) = \frac{1240}{V(\text{kV})}

\$\$

3. Example Calculation

Find the minimum wavelength of X‑rays produced when the accelerating potential is \$30\ \text{kV}\$.

  1. Insert \$V = 30\ \text{kV}\$ into the convenient formula:

    2. \$\lambda_{\min} = \frac{1240}{30} \text{ nm}\$

  2. Calculate:

    3. \$\lambda_{\min} \approx 41.3\ \text{pm} = 0.041\ \text{nm}\$

This wavelength corresponds to the highest‑energy (shortest‑wavelength) photon that can be produced in the tube.

4. Typical \cdot alues

Accelerating Potential \$V\$ (kV)Minimum Wavelength \$\lambda_{\min}\$ (nm)
100.124
200.062
300.041
500.025
1000.012

5. Uses of X‑rays

  • Medical imaging (radiography, CT scans)

  • Crystallography – determination of crystal structures

  • Industrial inspection – weld and material testing

  • Security – baggage scanning

6. Suggested Diagram

Suggested diagram: Schematic of an X‑ray tube showing the cathode, accelerating potential, electron beam, metal target (anode), and emitted X‑ray beam.

7. Summary

• X‑rays are produced when high‑energy electrons strike a metal target.

• The accelerating potential \$V\$ determines the maximum photon energy and thus the minimum wavelength \$\lambda_{\min} = hc/eV\$.

• Using the practical formula \$\lambda{\min}(\text{nm}) = 1240/V(\text{kV})\$ allows quick calculation of \$\lambda{\min}\$ for any tube voltage.

• Understanding this relationship is essential for selecting tube voltages appropriate to the required X‑ray penetration and resolution in various applications.