10.4 Nuclear reactions  (Page 2/19)

Beta decay

In most $\beta$ particle decays (or beta decay    ), either an electron ( ${\beta }^{-}$ ) or positron ( ${\beta }^{+}$ ) is emitted by a nucleus. A positron has the same mass as the electron, but its charge is $\text{+}e$ . For this reason, a positron is sometimes called an antielectron. How does $\beta$ decay occur? A possible explanation is the electron (positron) is confined to the nucleus prior to the decay and somehow escapes. To obtain a rough estimate of the escape energy, consider a simplified model of an electron trapped in a box (or in the terminology of quantum mechanics, a one-dimensional square well) that has the width of a typical nucleus ( ${10}^{\mathrm{-14}}\text{m}$ ). According to the Heisenberg uncertainty principle    in Quantum Mechanics , the uncertainty of the momentum of the electron is:

Taking this momentum value (an underestimate) to be the “true value,” the kinetic energy of the electron on escape is approximately

Experimentally, the electrons emitted in ${\beta }^{-}$ decay are found to have kinetic energies of the order of only a few MeV. We therefore conclude that the electron is somehow produced in the decay rather than escaping the nucleus. Particle production (annihilation) is described by theories that combine quantum mechanics and relativity, a subject of a more advanced course in physics.

Nuclear beta decay involves the conversion of one nucleon into another. For example, a neutron can decay to a proton by the emission of an electron ( ${\beta }^{-}$ ) and a nearly massless particle called an antineutrino    ( $\bar{\nu }$ ):

The notation ${}_{\mathrm{-1}}^0\text{e}$ is used to designate the electron. Its mass number is 0 because it is not a nucleon, and its atomic number is $\mathrm{-1}$ to signify that it has a charge of $\text{−}e$ . The proton is represented by ${}_1^1\text{p}$ because its mass number and atomic number are 1. When this occurs within an atomic nucleus, we have the following equation for beta decay: