3.2.1.2
Alpha decay ejects a helium nucleus
Atomic Structure & Nuclear Stability — AQA A-Level Physics
- Alpha decay occurs in large, unstable nuclei with too many nucleons.
- The nucleus emits an alpha particle: 2 protons + 2 neutrons (a helium-4 nucleus).
- The parent nucleus becomes a daughter nucleus of a different element.
$${}_{Z}^{A}X \longrightarrow {}_{Z-2}^{A-4}Y + {}_{2}^{4}\alpha$$
- Proton number decreases by 2.
- Nucleon number decreases by 4.
- Example: ${}_{84}^{212}\text{Po} \longrightarrow {}_{82}^{208}\text{Pb} + {}_{2}^{4}\alpha$
Worked Example
${}_{86}^{222}\text{Rn}$ undergoes alpha decay to form polonium (Po). Find the nucleon number and proton number of Po.
Show Solution
1
Write the general alpha decay equation
$${}_{Z}^{A}X \longrightarrow {}_{Z-2}^{A-4}Y + {}_{2}^{4}\alpha$$
2
Apply to radon-222
$${}_{86}^{222}\text{Rn} \longrightarrow {}_{84}^{218}\text{Po} + {}_{2}^{4}\alpha$$
3
Read off the values
Nucleon number of Po $= 222 - 4 = 218$
Proton number of Po $= 86 - 2 = 84$
Answer
Po has nucleon number 218 and proton number 84.
- Alpha particles have discrete energies — the energyThe capacity to do work. Measured in joules (J). spectrum shows sharp spikes.
- This is evidence that nuclear energyThe capacity to do work. Measured in joules (J). levels are quantisedExisting only in discrete values. ChargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C). is quantised because it can only exist as whole-number multiples of the elementary chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C). e..