42-4 Alpha Decay

42-4 Alpha Decay#

Prompts

  • What is an alpha particle? Write the general alpha-decay equation and identify parent, daughter, and the alpha.

  • Define the decay energy \(Q\). How do you calculate \(Q\) from atomic masses? Why must \(Q > 0\) for decay to occur?

  • Classically, an alpha particle inside the nucleus has energy less than the Coulomb barrier height. Why doesn’t it stay trapped? Explain quantum tunneling in this context.

  • The alpha and the daughter nucleus share the kinetic energy \(Q\). Which gets more — and why?

  • Why do alpha emitters with higher \(Q\) typically have shorter half-lives?

Lecture Notes#

Overview#

  • Alpha decay is the emission of an alpha particle — a \(^4\text{He}\) nucleus (2 protons, 2 neutrons) — from a heavy nucleus. The parent becomes a lighter daughter nucleus.

  • The decay energy \(Q\) is the total kinetic energy released; it equals the mass defect times \(c^2\). Conservation of momentum determines how \(Q\) is shared between the alpha and the daughter.

  • Quantum tunneling through the Coulomb barrier explains why alpha particles can escape even when their kinetic energy is less than the barrier height — a purely classical picture would forbid the decay.

Alpha Decay and the Alpha Particle#

An alpha particle (\(\alpha\)) is a \(^4\text{He}\) nucleus: \(Z = 2\), \(N = 2\), \(A = 4\). It is a tightly bound cluster (high binding energy per nucleon) and is emitted as a unit.

The general alpha-decay reaction is

(448)#\[ _Z^A\text{X} \to _{Z-2}^{A-4}\text{Y} + _2^4\text{He} \]

Example: \(^{238}\text{U} \to ^{234}\text{Th} + \alpha\). Alpha decay occurs mainly for heavy nuclei (\(A \gtrsim 150\)) where the Coulomb repulsion among protons makes emission energetically favorable.

Decay Energy (Q-Value)#

The decay energy \(Q\) is the total kinetic energy released in the decay. By conservation of energy:

(449)#\[ Q = (M_{\text{parent}} - M_{\text{daughter}} - M_\alpha)c^2 \]

where the masses are nuclear masses (or atomic masses; the electron masses cancel in the difference). For decay to be possible, we need \(Q > 0\) — the parent must be heavier than the sum of the products.

Using atomic masses

When using atomic masses, the \(Z\) electrons on the parent cancel with \((Z-2)\) on the daughter plus 2 on the alpha (as \(^4\text{He}\) atom). The same \(Q\) results.

Energy sharing: The alpha and daughter are emitted back-to-back (conservation of momentum). The lighter alpha carries most of the kinetic energy. In the nonrelativistic limit:

(450)#\[ K_\alpha \approx \frac{A-4}{A}\,Q \]

For \(^{238}\text{U}\), \(K_\alpha \approx 0.98\,Q\) — the alpha gets nearly all of \(Q\).

Quantum Tunneling Through the Coulomb Barrier#

Inside the nucleus, the alpha feels the strong force (attractive, short range) and the Coulomb repulsion (repulsive, long range) from the daughter. Outside, only Coulomb remains. The potential has a barrier: the alpha must overcome Coulomb repulsion to escape.

Classically, if the alpha’s kinetic energy \(E\) is less than the barrier height, it cannot escape. Yet alpha decay occurs with \(E < E_{\text{barrier}}\). The explanation is quantum tunneling (section 38-9): the alpha has a nonzero probability to “tunnel” through the classically forbidden region.

  • Higher \(Q\) → alpha has more energy → tunneling probability increases → shorter half-life.

  • Lower \(Q\) → alpha has less energy → tunneling is less likely → longer half-life.

This Geiger–Nuttall relationship (half-life vs. alpha energy) was observed empirically before quantum mechanics; tunneling provides the theoretical basis.

Why alpha and not single nucleons?

An alpha is very tightly bound; ejecting a single proton or neutron would require more energy (the nucleus would have to break a strong bond). Alpha emission is favored when the mass of (daughter + alpha) is less than the parent — a condition met for many heavy nuclei.

Summary#

  • Alpha particle: \(^4\text{He}\) nucleus; alpha decay: parent \(\to\) daughter \(+ \alpha\).

  • Decay energy: \(Q = (M_{\text{parent}} - M_{\text{daughter}} - M_\alpha)c^2\); \(Q > 0\) required.

  • Alpha carries most of \(K\) (lighter particle); \(K_\alpha \approx \frac{A-4}{A}Q\).

  • Quantum tunneling through the Coulomb barrier explains escape when \(E < E_{\text{barrier}}\).

  • Higher \(Q\) → shorter half-life (Geiger–Nuttall).