IGCSE Physics 0625 – 5.2.2 The Three Types of Nuclear Emission
5.2.2 The Three Types of Nuclear Emission
In nuclear decay three common types of radiation are emitted: alpha (α) particles, beta (β) particles and gamma (γ) radiation. Their behaviour in electric and magnetic fields is a key diagnostic tool in the laboratory and helps us understand their physical properties.
β‑particles: high‑speed electrons (β⁻) or positrons (β⁺), charge \$-e\$ (electron) or \$+e\$ (positron), mass \$≈9.1\times10^{-31}\,\text{kg}\$, speed close to \$c\$.
γ‑radiation: high‑energy photons, no charge, no rest mass, travels at \$c\$.
Suggested diagram: Schematic showing an α‑particle, a β⁻‑particle and a γ‑photon emitted from a nucleus.
2. Deflection in an electric field
The force on a charged particle in a uniform electric field \$\mathbf{E}\$ is \$\mathbf{F}=q\mathbf{E}\$. The direction of the force is along the field for a positive charge and opposite for a negative charge.
α‑particles: \$q=+2e\$, so they are deflected toward the negative plate. The large mass means the curvature of the path is small for a given field strength.
β⁻‑particles: \$q=-e\$, deflected toward the positive plate. Their very small mass produces a large curvature, often a pronounced arc.
β⁺‑particles: \$q=+e\$, deflected in the same direction as α‑particles but with a larger curvature because of the smaller mass.
γ‑radiation: \$q=0\$, experiences no electric force; the beam passes straight through the field unchanged.
Suggested diagram: Parallel-plate capacitor with arrows showing the direction of deflection for α, β⁻, β⁺ and γ radiation.
3. Deflection in a magnetic field
In a uniform magnetic field \$\mathbf{B}\$, a moving charge experiences the Lorentz force \$\mathbf{F}=q\mathbf{v}\times\mathbf{B}\$, which is perpendicular to both the velocity \$\mathbf{v}\$ and the field direction. The particle follows a circular (or helical) path of radius
\$r=\frac{mv}{|q|B}\$
where \$m\$ is the particle’s mass and \$v\$ its speed.
α‑particles: \$q=+2e\$, relatively large \$m\$, so \$r\$ is relatively large; the path is a gentle curve.
β⁻‑particles: \$q=-e\$, very small \$m\$, high \$v\$, giving a small radius; the trajectory is a tight circle (or helix).
β⁺‑particles: \$q=+e\$, same magnitude of curvature as β⁻ but opposite sense of rotation (right‑hand rule).
γ‑radiation: \$q=0\$, no magnetic force; the photon continues in a straight line.
Suggested diagram: Uniform magnetic field into the page with circular trajectories for α, β⁻ and β⁺ particles; γ‑ray shown as a straight line.
4. Comparison of deflection behaviour
Particle
Charge (\$q\$)
Mass (\$m\$)
Deflection in Electric Field
Deflection in Magnetic Field (radius \$r\$)
α‑particle
\$+2e\$
\$\approx4\,\text{u}\$
Deflected toward negative plate; small curvature due to large mass.
Large radius \$r=\dfrac{4u\,v}{2eB}\$ (gentle curve).
Small radius \$r=\dfrac{m_ev}{eB}\$ (tight circle).
β⁺‑particle (positron)
\$+e\$
Same as electron
Deflected toward negative plate; curvature similar to β⁻ but opposite rotation.
Same magnitude of \$r\$ as β⁻, opposite sense of rotation.
γ‑ray
0
0 (photon)
No deflection; travels straight.
No deflection; travels straight.
5. Key points to remember
The direction of deflection in an electric field depends only on the sign of the charge.
In a magnetic field the sense of rotation follows the right‑hand rule for positive charges; negative charges rotate oppositely.
Mass and speed determine the curvature: lighter, faster particles curve more strongly.
γ‑radiation, being uncharged, is unaffected by either field and therefore provides a useful reference beam.
6. Sample exam question
A beam of α‑particles and a beam of β⁻‑particles are passed through a uniform magnetic field of strength \$0.5\ \text{T}\$, both travelling at \$2.0\times10^{7}\ \text{m s}^{-1}\$. Which beam will have the smaller radius of curvature and why?
Answer: The β⁻‑beam will have the smaller radius because its mass \$m_e\$ is many orders of magnitude smaller than the α‑particle mass, and the radius \$r=mv/|q|B\$ is directly proportional to \$m\$.