Describe the deflection of α-particles, β-particles and γ-radiation in electric fields and magnetic fields

5.2 Radioactivity – Emissions, Deflection and Safety

5.2.1 Detection of Radioactivity

• Geiger‑Müller (GM) counter – gas‑filled tube that gives a click for each ionising event; useful for measuring count‑rate (counts s⁻¹).
• Scintillation detector – phosphor converts radiation to light, which is amplified by a photomultiplier.
• Background radiation – natural ionising radiation that must be measured and subtracted from the experimental count‑rate.
• Typical procedure: record the background count for a fixed time, then record the sample count under identical conditions; net count = sample – background.

5.2.2 The Three Types of Nuclear Emission

5.2.3 Deflection in a Uniform Electric Field

The electric force on a charged particle is

\(F = qE\)

With mass \(m\) the resulting acceleration is

\(a = \dfrac{qE}{m}\)

If the particle enters the field perpendicular to the field lines, the trajectory is a circular arc of radius

\(r_{\!E}= \dfrac{mv^{2}}{|q|E}\)

where \(v\) is the speed of the particle as it enters the field.

5.2.4 Deflection in a Uniform Magnetic Field

A moving charge experiences the Lorentz force

\(\mathbf{F}=q\mathbf{v}\times\mathbf{B}\)

The force is perpendicular to both \(\mathbf{v}\) and \(\mathbf{B}\), forcing the particle into a circular (or helical) path of radius

\(r_{\!B}= \dfrac{mv}{|q|B}\)

5.2.5 Radioactive Decay Processes

• α‑decay: a nucleus emits an α‑particle
  

  \(\displaystyle {Z}^{A}\text{X} \;\rightarrow\; {Z-2}^{A-4}\text{Y} + \alpha\)

• β⁻‑decay: a neutron converts to a proton + electron + antineutrino
  

  \(\displaystyle {Z}^{A}\text{X} \;\rightarrow\; {Z+1}^{A}\text{Y} + \beta^{-} + \bar{\nu}_{e}\)

• γ‑decay: an excited nucleus releases excess energy as a photon
  

  \(\displaystyle {Z}^{A}\text{X}^{*} \;\rightarrow\; {Z}^{A}\text{X} + \gamma\)

Each decay changes the atomic number (Z) and/or mass number (A) as shown, conserving charge and nucleon number.

5.2.6 Half‑Life and Decay Calculations

The number of undecayed nuclei after time \(t\) is

\(N = N{0}\left(\dfrac{1}{2}\right)^{t/t{1/2}}\)

or, using the decay constant \(\lambda = \dfrac{\ln 2}{t_{1/2}}\),

\(N = N_{0}e^{-\lambda t}\)

Worked example

A sample contains \(1.0\times10^{6}\) atoms of a radionuclide with a half‑life of 30 min. How many atoms remain after 90 min?

Solution:

\(t/t_{1/2}=90/30=3\) half‑lives.

\(N = 1.0\times10^{6}\,(1/2)^{3}=1.0\times10^{6}\times\frac{1}{8}=1.25\times10^{5}\) atoms.

5.2.7 Safety, Shielding and Applications

• ALARA principle – keep exposure As Low As Reasonably Achievable by controlling time, distance and shielding.
• Shielding materials:

  • α – thin paper, a few cm of air.
  • β – aluminium sheet (≈ 1 mm) or plastic.
  • γ – dense materials such as lead (several cm) or concrete (metres).

• Common applications:

  • α – smoke detectors (americium‑241).
  • β – medical imaging (PET), thickness gauging, sterilisation of medical equipment.
  • γ – radiography, cancer radiotherapy, food irradiation, sterilisation of pharmaceuticals.

5.2.8 Summary Comparison of the Three Emissions

5.2.9 Key Points to Remember (AO1–AO3)

1. Deflection in an electric field depends on the sign of the charge and on the ratio \(\dfrac{m}{|q|}\) (through \(r_{\!E}=mv^{2}/|q|E\)).
2. In a magnetic field the radius of curvature is given by \(r_{\!B}=mv/|q|B\); the sense of rotation follows the right‑hand rule for positive charges and the opposite for negative charges.
3. α‑particles are heavy and doubly charged → small curvature, low penetration.
   β‑particles are light, single‑charged → large curvature, moderate penetration.
   γ‑rays are uncharged photons → no curvature, high penetration.
4. Radioactive decay changes the nucleus according to the three decay equations; the number of undecayed nuclei follows the half‑life law.
5. Safety relies on limiting time, increasing distance and using appropriate shielding; each radiation type requires a different shielding material.

5.2.10 Sample Exam Question (IGCSE style)

A beam of α‑particles and a beam of β⁻‑particles, each travelling at \(2.0\times10^{7}\ \text{m s}^{-1}\), pass through a uniform magnetic field of strength \(0.5\ \text{T}\) directed into the page. Which beam has the smaller radius of curvature and why?

Answer: The β⁻‑beam has the smaller radius. Using \(r_{\!B}=mv/|q|B\):

\(r_{\alpha}= \dfrac{(6.6\times10^{-27}\,\text{kg})(2.0\times10^{7}\,\text{m s}^{-1})}{(2e)(0.5\,\text{T})}\approx 1.3\times10^{-2}\,\text{m}\).

\(r_{\beta}= \dfrac{(9.1\times10^{-31}\,\text{kg})(2.0\times10^{7}\,\text{m s}^{-1})}{(e)(0.5\,\text{T})}\approx 2.3\times10^{-5}\,\text{m}\).

Because the electron’s mass is many orders of magnitude smaller, its radius is far smaller despite the same speed and a comparable charge magnitude.