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.

Concept and Intuition

In any nuclear reaction the total mass number $A$ (superscript) and the total atomic number $Z$ (subscript) are each conserved. Balance them on both sides to identify the unknown particle.

Step-by-step solution

1. Conserve mass number $A$:

2. Conserve atomic number $Z$: $0+80=79+Z_X\ \Rightarrow\ Z_X=1$ …

Concept & Intuition

Nuclear reactions obey two fundamental conservation laws:

• Conservation of mass number (total number of protons + neutrons)
• Conservation of atomic number (total number of protons)

When a neutron strikes a boron-10 nucleus, the compound nucleus rearranges and breaks apart. The unknown emitted particle must balance the equation so that the sums on both sides match. By writing the reaction with symbols and solving for the missing particle’s mass and atomic numbers, we can identify it.

Step-by-step reasoning

1. Write the reaction in standard nuclear notation The target is boron-10: $^{10}_{\;5}\mathrm{B}$. The projectile is a neutron: $^{1}_{0}n$. One product is lithium-7: $^{7}_{3}\mathrm{Li}$. Let the unknown emitted particle be $^{A}_{Z}X$. The reaction is:

2. Apply conservation of mass number

   Left side: $10 + 1 = 11$.

   Right side: $7 + A$.

   So $7 + A = 11$ → $A = 4$.

3. Apply conservation of atomic number

   Left side: $5 + 0 = 5$.

   Right side: $3 + Z$. …

Gamma decay occurs when a nucleus in an excited state releases its excess energy as a high-energy photon ($\gamma$-ray) and settles to a lower energy state. A photon has no charge and no rest mass, and no protons or neutrons leave the nucleus. Therefore:

• Mass number $A$ — unchanged
• Atomic number $Z$ — unchanged …

1. The concept — mass defect

A bound nucleus is lighter than the sum of its free constituents. The missing mass, the mass defect $\Delta m$, has been released as the binding energy that holds the nucleons together (Einstein: $E = \Delta m\,c^{2}$). Equivalently, $E_b$ is the energy you must supply to pull the nucleus apart into free nucleons.

Why $m_H$ and not $m_p$: the quoted mass $14.00307$ u is the atomic mass of $^{14}N$, which includes its 7 electrons. Using the hydrogen atom mass ($1.00783$ u = proton + 1 electron) on the other side means the 7 electron masses cancel automatically — the standard bookkeeping trick.

2. Identify the nucleons

For $^{14}_{7}N$: $Z = 7$ protons, $A - Z = 14 - 7 = 7$ neutrons.

Standard masses:

3. Compute the mass defect

4. Convert to energy …

Concept: α-decay emits a helium nucleus ⁴₂He.

ᴬ_Z X → ᴬ⁻⁴_(Z−2) Y + ⁴₂He …

The reaction is

Mass number: $10 + A = 7 + 4 = 11 \Rightarrow A = 1$. …

Step 1 — The fundamental process.

and at the level of the whole nucleus,

The mass number $A$ is unchanged (a nucleon is merely converted), while the atomic number increases by 1 — the fingerprint of $\beta^-$ decay. It occurs in neutron-rich nuclei, because converting a neutron to a proton moves them towards the stability line.

Step 2 — Why (B) is wrong (the key conceptual point).

Electrons do not exist inside the nucleus. Two arguments: (i) confining an electron to a nuclear radius ($\sim 10^{-15}\ \mathrm m$) would, by the uncertainty principle, give it an energy of tens of MeV — far above the few-MeV energies actually observed;

(ii) nuclear spin statistics forbid it. The $\beta$ particle is created at the instant of decay, exactly as a photon is created when an atom de-excites.

Step 3 — Why (A) and (D) are wrong.

• (A) An atomic (orbital) electron is not involved: ejecting one would ionise the atom, not change $Z$. (The related process where a nucleus captures an orbital electron is electron capture, and it lowers $Z$.) …