Radioactive Decays

Radioactivity

• Many nuclei are stable and remain unchanged for a very long time; these have a particular ratio of mass number to atomic number. Other naturally occurring nuclei are not stable and change their structure by emitting particles, decaying to other nuclei with different $A$ and $Z$.
• The decaying nucleus is the parent nucleus, the nucleus produced after the decay is the daughter nucleus, and the process is radioactive decay or radioactivity, discovered by Becquerel. The term radioactivity was coined by Marie Curie, who with Pierre Curie discovered radium; the 1903 Nobel Prize went jointly to Becquerel and the Curies.
• Decays occur because the parent nucleus is unstable and gets converted to a more stable daughter by emitting particles. There are three types.

Alpha decay

• The parent nucleus emits an alpha particle, which is the nucleus of a helium atom, so it loses two protons and two neutrons.

$${}^{A}_{Z}\text{X} \longrightarrow {}^{A-4}_{Z-2}\text{Y} + \alpha$$

• All nuclei with $A > 210$ undergo alpha decay. The reason is that such nuclei have a large number of protons; the electrostatic repulsion between them is very large and the attractive nuclear forces cannot cope with it, so the nucleus reduces its proton number by ejecting protons two at a time as alpha particles.
• Example: ${}^{212}_{83}\text{Bi} \longrightarrow {}^{208}_{81}\text{Tl} + \alpha$.
• The total mass of the products is always less than the mass of the parent atom; the excess appears as the kinetic energy of the products. The $Q$-value is the difference in the energy equivalents:

$$Q = \left[m_X - m_Y - m_{He}\right]c^2$$

• where $m_X$, $m_Y$ and $m_{He}$ are the atomic masses of the parent, the daughter and helium. Note that atomic masses are used here.

Beta decay

• Beta-minus decay: the nucleus emits an electron, produced by converting one of its neutrons into a proton. The basic process inside the nucleus is

$$n \longrightarrow p + e^- + \bar\nu$$

• Neutrino and antineutrino are particles with very little mass and no charge. The mass number is unchanged; the daughter has one less neutron and one more proton, so $Z$ increases by one and $N$ decreases by one:

$${}^{A}_{Z}\text{X} \longrightarrow {}^{A}_{Z+1}\text{Y} + e^- + \bar\nu$$

• Example: ${}^{60}_{27}\text{Co} \longrightarrow {}^{60}_{28}\text{Ni} + e^- + \bar\nu$.
• Beta-plus decay: a proton is converted into a neutron by emitting a positron and a neutrino. A positron has all the properties of an electron except that its charge is positive; it is the antiparticle of the electron.

$$p \longrightarrow n + e^+ + \nu,\qquad {}^{A}_{Z}\text{X} \longrightarrow {}^{A}_{Z-1}\text{Y} + e^+ + \nu$$

• Example: ${}^{22}_{11}\text{Na} \longrightarrow {}^{22}_{10}\text{Ne} + e^+ + \nu$. Here $A$ is unchanged, $Z$ decreases by one and $N$ increases by one.
• An important point: the neutron is heavier than the proton, so beta-plus decay cannot take place for a free proton. Inside a nucleus it can, because the extra energy needed to produce a neutron is supplied by the rest of the nucleus.
• In beta decay too the total mass of the products is less than the mass of the parent atom, the excess going into kinetic energy:

$$Q = \left[m_X - m_Y - m_e\right]c^2$$

• with the neutrino mass neglected as negligible. This $Q$ is the maximum kinetic energy the beta particle can have — the case in which the neutrino carries away zero energy.
• Worked check: for ${}^{22}_{11}\text{Na} \to {}^{22}_{10}\text{Ne} + e^+ + \nu$ with masses $21.994437\ \text{u}$, $21.991385\ \text{u}$ and $m_e = 0.00055\ \text{u}$, $Q = 0.002502\times931.5 = 2.3306\ \text{MeV}$.

Gamma decay

• Gamma rays, which are high-energy photons, are emitted by the parent nucleus. No other particle is emitted, so the daughter nucleus is the same nuclide as the parent, only with less energy.
• The mechanism parallels atomic transitions: nucleons occupy nuclear energy levels, and a nucleon making a transition from a higher to a lower level emits a photon.
• The difference between atomic and nuclear levels is one of scale. Energies and level separations in an atom are of the order of a few eV, in a nucleus a few keV to a few MeV. That is why atoms radiate in the ultraviolet-to-radio range while nuclei radiate gamma rays.
• Nucleons are normally in the lowest possible state and cannot easily be excited, since keV to MeV would be needed. A nucleus usually ends up excited as a result of an alpha or beta decay, so gamma decay usually follows one of those.
• Example: ${}^{57}\text{Co}$ undergoes beta-plus decay to ${}^{57}\text{Fe}$ in an excited state $136\ \text{keV}$ above the ground state. It can reach the ground state either by emitting one gamma ray of $136\ \text{keV}$, or by emitting a gamma ray of $122\ \text{keV}$ to an intermediate state and then a photon of $14\ \text{keV}$. Both routes have been observed.

Decay series

• Which decay a nucleus undergoes depends on which daughter is more stable. Often the daughter is itself unstable and decays further, so a chain of decays takes place until a stable nucleus is reached.
• Example: ${}^{238}\text{U}$ undergoes a series of alpha and beta decays, fourteen in all, to finish as stable ${}^{206}\text{Pb}$.

Counting alpha and beta particles in a series

• An alpha emission reduces $A$ by $4$ and $Z$ by $2$; a beta-minus emission leaves $A$ unchanged and increases $Z$ by $1$.
• So fix the number of alphas from the change in mass number, then fix the number of betas from what is left over in atomic number:

$$n_\alpha = \dfrac{A_{\text{parent}}-A_{\text{daughter}}}{4},\qquad n_\beta = 2n_\alpha - \left(Z_{\text{parent}}-Z_{\text{daughter}}\right)$$

• Worked check, the board's own example: ${}^{232}_{90}\text{Th}$ disintegrating to ${}^{200}_{82}\text{Pb}$. Number of alpha particles $= \dfrac{232-200}{4} = 8$. Emission of $8$ alphas would take $Z$ from $90$ down to $90-16 = 74$, but the product has $Z = 82$; the difference is made up by beta emission, so the number of beta particles is $82-74 = 8$.

Completing decay equations

• Balance both the mass number and the atomic number on the two sides. Standard completions:
• ${}^{226}_{88}\text{Ra} \longrightarrow {}^{222}_{86}\text{Rn} + \alpha$
• ${}^{19}_{8}\text{O} \longrightarrow {}^{19}_{9}\text{F} + e^- + \bar\nu$
• ${}^{228}_{90}\text{Th} \longrightarrow {}^{224}_{88}\text{Ra} + \alpha$
• ${}^{12}_{7}\text{N} \longrightarrow {}^{12}_{6}\text{C} + e^+ + \nu$

Standard results collected