Q.Under certain circumstances, a nucleus can decay by emitting a particle more massive than an $\alpha$-particle. Consider the following decay processes:

[!FORMULA] $^{223}_{88}\text{Ra} \rightarrow {}^{209}_{82}\text{Pb} + {}^{14}_{6}\text{C}$

[!FORMULA] $^{223}_{88}\text{Ra} \rightarrow {}^{219}_{86}\text{Rn} + {}^{4}_{2}\text{He}$

Calculate the Q-values for these decays and determine that both are energetically allowed.

Why Nuclear Reaction Balancing Works

Nuclear reaction balancing rests on a single, non-negotiable principle: conservation laws are absolute. In every nuclear reaction — whether natural decay, artificial transmutation, or fission/fusion — two quantities never change:

1. Total mass number ($A$) — the sum of protons + neutrons
2. Total atomic number ($Z$) — the sum of protons

These aren't arbitrary rules. They follow from deeper physics: baryon number conservation (protons and neutrons are baryons, and their total count is fixed) and charge conservation (electric charge cannot be created or destroyed). A nuclear reaction is just a rearrangement of nucleons; the number of nucleons stays constant, and the total charge stays constant.

The Reasoning Behind Each Conservation Law

Mass number conservation ($A$ conserved):

A nucleon (proton or neutron) can change identity — a neutron can beta-decay into a proton, or a proton can capture an electron and become a neutron — but it cannot vanish or appear from nothing. The total count of nucleons before the reaction equals the total count after. This is why, for example, in alpha decay:

The left side has $A = 238$; the right side has $234 + 4 = 238$. The alpha particle carries away exactly 4 nucleons.

Atomic number conservation ($Z$ conserved):

Charge is strictly conserved. The total positive charge (proton count) before equals the total after. In the same alpha decay, $Z$ goes from 92 to $90 + 2 = 92$. If charge weren't conserved, atoms would spontaneously change their chemical identity — which never happens in a closed system.

How to Apply It: A Worked Example

Suppose you see: $_{92}^{235}\text{U} + _{0}^{1}n \rightarrow _{56}^{141}\text{Ba} + _{?}^{?}\text{Kr} + 3\, _{0}^{1}n$

You know the total $A$ on the left: $235 + 1 = 236$.

On the right, you have $141 + A_{\text{Kr}} + 3(1) = 144 + A_{\text{Kr}}$. …