Q.Answer the following questions:

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. …