Alpha and beta decay schemes

Problem

Show how the atomic number $Z$ and the mass number $A$ change in an $\alpha$ decay of $^{238}$U and in a $\beta^-$ decay of $^{14}$C. State the products and the particles emitted.

On the left, the $\alpha$ decay of $^{238}$U (emission of a helium nucleus); on the right, the $\beta^-$ decay of $^{14}$C (emission of an electron and an antineutrino).

Solution

$\alpha$ decay of $^{238}$U. The nucleus emits an $\alpha$ particle, i.e. a helium nucleus $^{4}_{2}\mathrm{He}$. The atomic number decreases by 2 and the mass number decreases by 4:

$^{238}_{92}\mathrm{U} \rightarrow\ ^{234}_{90}\mathrm{Th} + {}^{4}_{2}\mathrm{He}$

$\beta^-$ decay of $^{14}$C. Inside the nucleus a neutron transforms into a proton, emitting an electron and an electron antineutrino. The atomic number increases by 1, while the mass number stays unchanged:

$^{14}_{6}\mathrm{C} \rightarrow\ ^{14}_{7}\mathrm{N} + e^- + \bar\nu_e$

For comparison, in a $\gamma$ decay neither $Z$ nor $A$ changes: it is simply the de-excitation of the nucleus, which emits a photon.

$\ev{\alpha:\ Z{-}2,A{-}4;\quad \beta^-:\ Z{+}1,A\ \text{unchanged}}$

Collegamenti

Argomenti: Fisica nucleare Concetti: Decadimento radioattivo