IGCSE Physics 0625: Balancing Alpha and Beta Decay Equations

A sealed source manufactured for industrial radiography contains a freshly made, highly unstable nuclide. For this question, this nuclide is represented as ${}^{224}_{88}\text{X}$ (X is used here only as a generic label, not a real chemical symbol).

${}^{224}_{88}\text{X}$ decays by emitting an alpha particle to form a new nuclide, Y. The nuclide Y is itself unstable and decays further by emitting a beta particle to form a new nuclide, W.

(a) Write a balanced nuclide equation for the decay of X to Y, showing the nucleon number and proton number of Y and including the symbol for the alpha particle emitted. [2]

(b) Write a balanced nuclide equation for the decay of Y to W, showing the nucleon number and proton number of W and including the symbol for the beta particle emitted. [2]

(c) During the decay of Y to W, a neutron inside the nucleus changes into other particles. Describe this change, and explain how it helps to make the nucleus more stable. [2]

Part (a): Balancing the alpha-decay equation

An alpha particle is a helium nucleus, ${}^{4}_{2}\text{He}$ , carrying away 4 from the nucleon number and 2 from the proton number.

Starting nucleon and proton numbers of X: $A = 224, \quad Z = 88$

Subtract the alpha particle’s numbers to find Y: $A_Y = 224 - 4 = 220 \qquad Z_Y = 88 - 2 = 86$

So the balanced equation is: ${}^{224}_{88}\text{X} \rightarrow {}^{220}_{86}\text{Y} + {}^{4}_{2}\text{He}$

Check: nucleon numbers: $220 + 4 = 224$ ✓. Proton numbers: $86 + 2 = 88$ ✓.

Part (b): Balancing the beta-decay equation

A beta particle is a fast electron, ${}^{0}_{-1}\text{e}$ , which carries away a nucleon number of $0$ and a proton number of $-1$ .

Starting nucleon and proton numbers of Y (from part (a)): $A = 220, \quad Z = 86$

The nucleon number is unchanged by beta decay (the emitted electron has negligible mass), while the proton number increases by 1 (a neutron converts into a proton): $A_W = 220 - 0 = 220 \qquad Z_W = 86 - (-1) = 87$

So the balanced equation is: ${}^{220}_{86}\text{Y} \rightarrow {}^{220}_{87}\text{W} + {}^{0}_{-1}\text{e}$

Check: nucleon numbers: $220 + 0 = 220$ ✓. Proton numbers: $87 + (-1) = 86$ ✓.

Part (c): The change inside the nucleus and why it increases stability

During beta decay, a neutron inside the nucleus changes into a proton and an electron: $\text{neutron} \rightarrow \text{proton} + \text{electron}$

The electron is emitted from the nucleus as the beta particle, while the newly formed proton remains inside the nucleus (raising the proton number from 86 to 87, as found in part (b)).

This process converts one of the nucleus’s neutrons into a proton, so it reduces the number of excess neutrons compared with protons. Nuclei with a proton-to-neutron ratio far from the stable range tend to be unstable; by shifting a neutron into a proton, beta decay moves nucleus Y towards a more stable proton-to-neutron ratio, so nucleus W is more stable than nucleus Y.

Final answers

(a) ${}^{224}_{88}\text{X} \rightarrow {}^{220}_{86}\text{Y} + {}^{4}_{2}\text{He}$
(b) ${}^{220}_{86}\text{Y} \rightarrow {}^{220}_{87}\text{W} + {}^{0}_{-1}\text{e}$
(c) A neutron changes into a proton (which stays in the nucleus) and an electron (emitted as the beta particle), reducing the number of excess neutrons and increasing nuclear stability.