Find the Q-value and the kinetic energy of the emitted α-particle in the α-decay of (a) ²²⁶₈₈Ra and (b) ²²⁰₈₆R

Exercises · 13.12

\alpha

(a)

^{226}_{88}\text{Ra}

and

(b)

^{220}_{86}\text{Rn}

.
Given

m(^{226}_{88}\text{Ra}) = 226.02540

m(^{222}_{86}\text{Rn}) = 222.01750

m(^{220}_{86}\text{Rn}) = 220.01137

m(^{216}_{84}\text{Po}) = 216.00189

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Concept understanding — Alpha Beta Gamma Decay

Alpha Beta Gamma Decay

Why Nuclei Decay at All

A nucleus is a tight ball of protons and neutrons held together by the short-range strong nuclear force, which must overcome the electric repulsion between protons. For very large nuclei, or ones with the wrong proton-to-neutron ratio, this balance is unstable, and the nucleus sheds energy or particles to reach a more stable state. This is radioactive decay, and it happens in three main forms: alpha, beta, and gamma.

Alpha Decay — The Nucleus Ejects a Helium Core

An alpha particle is a helium-4 nucleus (2 protons + 2 neutrons), a very stable package. A heavy nucleus (atomic number greater than about 82), where the strong force can no longer hold together against the mutual repulsion of so many protons, lowers its energy by ejecting one.

The nucleus $X$ loses 4 units of mass number and 2 units of atomic number, becoming a different element two places to the left on the periodic table:

_{Z}^{A}X \;\rightarrow\; _{Z-2}^{A-4}Y \;+\; _{2}^{4}\alpha

Note

The alpha particle is written as $_{2}^{4}\text{He}^{2+}$ — a helium ion. It is positively charged and relatively heavy, so it travels only a few centimetres in air and is stopped by a sheet of paper.

Example: Uranium-238 decays to thorium-234:

_{92}^{238}\text{U} \;\rightarrow\; _{90}^{234}\text{Th} \;+\; _{2}^{4}\alpha

Beta Decay — A Neutron Turns Into a Proton (or Vice Versa)

Beta decay is more subtle: the nucleus doesn't eject a pre-existing particle. A neutron inside it, in a nucleus with too many neutrons for stability, converts into a proton, an electron, and an antineutrino, lowering the nucleus's energy. The electron is ejected as a beta particle ( $\beta^-$ ); the new proton stays behind.

_{0}^{1}n \;\rightarrow\; _{1}^{1}p \;+\; _{-1}^{0}e \;+\; \bar{\nu}_e

Mass number stays the same (a neutron and proton weigh almost the same), but atomic number increases by 1 — the element moves one place to the right:

_{Z}^{A}X \;\rightarrow\; _{Z+1}^{A}Y \;+\; _{-1}^{0}\beta \;+\; \bar{\nu}_e

Watch out

The beta particle is not an orbital electron — it is created inside the nucleus at the moment of decay. The antineutrino carries away some energy and momentum; it is nearly massless and barely interacts with matter.

Example: Carbon-14 decays to nitrogen-14:

_{6}^{14}\text{C} \;\rightarrow\; _{7}^{14}\text{N} \;+\; _{-1}^{0}\beta \;+\; \bar{\nu}_e

A proton-rich nucleus does the reverse via beta-plus decay ( $\beta^+$ ): a proton converts into a neutron and emits a positron (the electron's antimatter twin) plus a neutrino.

_{1}^{1}p \;\rightarrow\; _{0}^{1}n \;+\; _{+1}^{0}e \;+\; \nu_e

Gamma Decay — The Nucleus Sheds Excess Energy

After alpha or beta decay, the daughter nucleus is often left in an excited state — extra energy, same number of protons and neutrons. It sheds this energy by emitting a high-energy photon, a gamma ray ( $\gamma$ ), dropping to a lower energy level much as an electron drops to a lower orbit — except here the energy is millions of times larger. Neither the mass number nor the atomic number changes:

_{Z}^{A}X^* \;\rightarrow\; _{Z}^{A}X \;+\; \gamma

Important

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$Q = [m_{\text{parent}} - m_{\text{daughter}} - m_{\alpha}]c^2$ , and the alpha particle's kinetic energy is $Q$ scaled by $\frac{A_{\text{daughter}}}{A_{\text{daughter}}+4}$ from momentum conservation. Both decays are exothermic with $Q\approx4.9$ and $6.4$ MeV respectively.

(a) $^{226}_{88}\text{Ra} \rightarrow {}^{222}_{86}\text{Rn} + {}^{4}_{2}\text{He}$

Q = \left[m(^{226}\text{Ra}) - m(^{222}\text{Rn}) - m(^{4}\text{He})\right] \times 931.5\ \text{MeV}

= [226.02540 - 222.01750 - 4.002603] \times 931.5

= 0.005297\ \text{u} \times 931.5 = 4.934\ \text{MeV}

Since the recoiling Rn-222 nucleus must carry away momentum equal and opposite to the alpha's, the kinetic energy splits in inverse proportion to mass. Treating mass numbers as proportional to mass:

KE_\alpha = Q \times \frac{A_{\text{daughter}}}{A_{\text{daughter}} + 4} = 4.934 \times \frac{222}{226} = 4.847\ \text{MeV}

(b) $^{220}_{86}\text{Rn} \rightarrow {}^{216}_{84}\text{Po} + {}^{4}_{2}\text{He}$

Q = \left[m(^{220}\text{Rn}) - m(^{216}\text{Po}) - m(^{4}\text{He})\right] \times 931.5

= [220.01137 - 216.00189 - 4.002603] \times 931.5

$= 0.006877\ \text{u} \times 931.5 = 6.406\ \text{MeV}$ …

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