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

Binding energy per nucleon measures how tightly bound, on average, each nucleon is. To get the total binding energy of a nucleus you must multiply by the mass number $A$ (the total nucleon count), because binding energy is an extensive (whole-nucleus) quantity while BE/nucleon is an intensive (per-particle) one. When lighter nuclei fuse into a more tightly-bound heavier nucleus, the difference in total binding energy is released as energy (mass converts to energy per $E=mc^2$, already baked into the BE values).

Step-by-Step Solution

1. Deuteron ($^2_1\mathrm{H}$) has mass number $A=2$ and BE/nucleon $=1.1$ MeV. Total BE of one deuteron $=2\times1.1=2.2$ MeV.
2. Two deuterons together carry total BE $=2\times2.2=4.4$ MeV.
3. Helium ($^4_2\mathrm{He}$) has mass number $A=4$ and BE/nucleon $=7$ MeV. …

Concept and Intuition

An $\alpha$ particle ($_2^4\text{He}$) removes 2 protons and 2 neutrons, so it always changes $A$ by $-4$, $Z$ by $-2$, $N$ by $-2$. A $\beta^-$ (electron) emission converts a neutron to a proton: $Z$ by $+1$, $N$ by $-1$, $A$ unchanged. A $\beta^+$ (positron) emission does the reverse: $Z$ by $-1$, $N$ by $+1$, $A$ unchanged. We're told the overall changes are $N$ down by 5 and $A$ down by 12, so $\Delta Z_{decrease}=\Delta A-\Delta N=12-5=7$.

Step-by-Step Solution

1. Only $\alpha$ particles change $A$, each by $-4$. To get a total $A$ decrease of 12 we need exactly 3 $\alpha$ particles.
2. Three $\alpha$ particles alone give a $Z$ decrease of 6 and an $N$ decrease of 6.
3. We still need one more unit of $Z$-decrease (to reach 7) while $N$'s decrease must come back up from 6 to 5 (i.e., $N$ needs +1 relative to the 3-alpha result).
4. A positron emission does exactly that: $Z$ decreases by 1 more (reaching 7 total) and $N$ increases by 1 (bringing the $N$-decrease from 6 down to 5). …

Concept and Intuition

In $\beta^-$ (negatron) decay, one neutron inside the nucleus transforms into a proton, emitting an electron ($e^-$) and an electron-antineutrino ($\bar\nu$):

Since a neutron is replaced by a proton, the mass number $A$ stays the same but the atomic number $Z$ increases by 1. The daughter nucleus therefore sits one place to the right in the periodic table.

Step-by-Step Solution

1. Parent nucleus: ${}_{10}\mathrm{Ne}^{23}$ (Z = 10, A = 23).
2. $\beta^-$-decay converts one neutron to a proton: $Z \to Z+1 = 11$, $A$ unchanged $= 23$.
3. Element with Z = 11 is sodium (Na). So the daughter is ${}_{11}\mathrm{Na}^{23}$.
4. Conservation of charge and lepton number requires the emitted particles to be $e^-$ and $\bar\nu$ (not a positron, which would occur in $\beta^+$-decay and would decrease Z).
5. Full equation: ${}_{10}\mathrm{Ne}^{23} \to {}_{11}\mathrm{Na}^{23} + e^{-} + \bar{\nu}$.

Common Mistakes …

Concept and Intuition

In any nuclear reaction, the energy released equals the increase in total binding energy: reactants with less total binding energy fuse into a product with more total binding energy, and the difference is released as $Q$. Binding energy per nucleon must first be converted to total binding energy (multiply by mass number) before adding/subtracting, since binding energy is an extensive (additive) quantity, not per-nucleon directly comparable across different nuclei.

Step-by-Step Solution

1. Each deuteron $_1H^2$ has mass number 2 and BE/nucleon $=1.15\,\mathrm{MeV}$, so total BE per deuteron $= 2\times1.15 = 2.3\,\mathrm{MeV}$.
2. Two deuterons together: total initial BE $= 2\times2.3 = 4.6\,\mathrm{MeV}$.
3. The product $_2He^4$ has mass number 4 and BE/nucleon $=7.1\,\mathrm{MeV}$, so total BE $= 4\times7.1=28.4\,\mathrm{MeV}$. …

Concept and Intuition

Every nuclear reaction must independently conserve total mass number $A$ (nucleon count) and total charge/atomic number $Z$ on both sides of the arrow (along with energy/momentum, but those aren't testable from the symbolic equation alone). This is a quick, purely bookkeeping check that instantly rules out fabricated-looking reactions.

Step-by-Step Solution

1. Option (A): $A$: $10+4=14$, $13+1=14$ — matches. $Z$: $5+2=7$, but $7+1=8$ — mismatch. Not possible as written (correct version would emit a neutron, not a proton).
2. Option (B): $A$: $24+1=25$ vs $20+4=24$ — mismatch. Not possible.
3. Option (C): $A$: $239$ vs $239+0+0=239$ — matches. $Z$: $93$ vs $94+(-1)+0=93$ — matches. This is the well-known genuine $\beta^-$ decay $^{239}_{93}Np\to{}^{239}_{94}Pu+e^-+\bar\upsilon$. …

Concept and Intuition

An isotope of the original nucleus has the same atomic number $Z$ (same element) but generally a different mass number $A$. In radioactive decay: an $\alpha$ particle emission reduces $Z$ by 2 and $A$ by 4; a $\beta^-$ particle emission (electron emission, a neutron converting to a proton) increases $Z$ by 1 while leaving $A$ unchanged.

Step-by-Step Solution

1. Let the decay chain involve $n_\alpha$ alpha particles and $n_\beta$ beta particles.
2. Net change in atomic number: $\Delta Z = -2n_\alpha + n_\beta$.
3. For the final nucleus to be an isotope (same $Z$, so $\Delta Z=0$, but with $A$ reduced due to the alpha decays, so genuinely a different isotope, not the identical nuclide): $n_\beta = 2n_\alpha$. …

Concept and Intuition

In a decay series, mass number (A) changes only because of $\alpha$-emission — each $\alpha$ is a $_2^4\mathrm{He}$ nucleus, so it reduces A by 4. $\beta^-$-emission (an electron ejected when a neutron converts to a proton inside the nucleus) changes Z but leaves A unchanged. So the total mass-number drop alone tells us how many alphas were emitted; once that is known, the atomic-number drop tells us how many betas were needed to make the numbers balance.

Step-by-Step Solution

1. Mass-number change: $238 - 206 = 32$. Since each $\alpha$ reduces A by 4: $n_\alpha = 32/4 = 8$.
2. Atomic-number change: $92 - 82 = 10$ (a net decrease of 10 over the whole series). …

Concept and Intuition

Nuclear reactions release or absorb energy according to the mass difference between reactants and products ($Q = \Delta m\, c^2$). Equivalently, since total energy is conserved, this same $Q$ shows up as the difference between the total kinetic energy carried away by the products and the total kinetic energy the reactants had — this equivalence is exactly what statement (A) states. The other statements test whether you can distinguish related-but-different ideas: mass defect (binding energy) vs. constituent mass, isotopes vs. isotones, and fission vs. fusion.

Step-by-Step Solution

1. (A) By definition, $Q = (\text{KE of products}) - (\text{KE of reactants})$, which equals $[\sum m_{\text{reactants}} - \sum m_{\text{products}}]c^2$. This is the standard textbook definition — true.
2. (B) Because of the mass defect (binding energy), a nucleus's mass is always less than the sum of the masses of its free constituent nucleons, not higher — false.
3. (C) Nuclides with the same number of protons (atomic number $Z$) are isotopes; nuclides with the same number of neutrons are isotones. The statement swaps these — false. …

Concept and Intuition

In a nuclear reaction, the energy released, $Q$, equals the increase in total binding energy going from reactants to products — because a more tightly bound (higher total BE) final configuration corresponds to a lower rest-mass total, and that mass difference is released as kinetic energy/radiation by $E=mc^2$. So instead of tracking individual masses, we can directly use $Q = \sum(\text{BE})_{products} - \sum(\text{BE})_{reactants}$, taking care to multiply each binding energy per nucleon by the number of nucleons in that nucleus to get the total binding energy.

Step-by-Step Solution

1. Reaction: ${}_3\text{Li}^7 + {}_1\text{H}^1 \rightarrow {}_2\text{He}^4 + {}_2\text{He}^4 + Q$.
2. Total BE of ${}_3\text{Li}^7$ (7 nucleons, 5.60 MeV/nucleon): $7 \times 5.60 = 39.2$ MeV.
3. Total BE of ${}_1\text{H}^1$: a single proton has no binding energy (there's nothing to bind to itself), so BE $= 0$. …

Concept and Intuition

In any nuclear reaction (or decay), the total mass number $A$ (nucleons) and the total charge $Z$ (protons, counting $\beta^-$ as charge $-1$ and mass number $0$, and $\gamma$ as carrying neither charge nor mass number) must each be conserved separately. Checking these two simple bookkeeping sums is the fastest way to spot an impossible reaction.

Step-by-Step Solution

1. (A) $_5B^{10}+{}_2He^4 \to {}_7N^{13}+{}_1H^1$: mass numbers $10+4=14=13+1$ ✓, but charges $5+2=7$ on the left vs $7+1=8$ on the right ✗ — charge not conserved, impossible as written (the real reaction actually produces a neutron, $_7N^{13}+{}_0n^1$).
2. (B) $_{11}Na^{24}+{}_1H^1 \to {}_{10}Ne^{20}+{}_2He^4$: charges $11+1=12=10+2$ ✓, but mass numbers $24+1=25 \ne 20+4=24$ ✗ — mass number not conserved, impossible. …