Atoms and Nuclei MCQs with Answers – Part 6 (NEET Physics)

Explanation: \(\textbf{Identify the ground-state reaction reference:}\) The supplied ground-state reaction energy is: \[ Q_0=190\,\text{MeV} \] This represents the difference between the complete initial and final rest energies when both daughter nuclei occupy their ground states. \(\textbf{Account for the excited daughter masses:}\) The actual first fragment contains excitation energy: \[ E_1^*=6\,\text{MeV} \] The second contains: \[ E_2^*=4\,\text{MeV} \] Their combined nuclear excitation energy is: \[ E_{\mathrm{exc,total}}=E_1^*+E_2^* \] \[ E_{\mathrm{exc,total}}=6+4 \] \[ E_{\mathrm{exc,total}}=10\,\text{MeV} \] \(\textbf{Determine the immediate fission energy release:}\) Excited daughter nuclei have greater rest energy than the corresponding ground-state daughters. Therefore, less energy is available immediately after their formation: \[ Q_{\mathrm{prompt}}=Q_0-E_{\mathrm{exc,total}} \] \[ Q_{\mathrm{prompt}}=190-10 \] \[ Q_{\mathrm{prompt}}=180\,\text{MeV} \] \(\textbf{Calculate subsequent photon energy:}\) The excited fragments return to their ground states. Neglecting recoil corrections, their combined emitted photon energy is: \[ E_{\gamma,\mathrm{total}}=E_1^*+E_2^* \] \[ E_{\gamma,\mathrm{total}}=10\,\text{MeV} \] \(\textbf{Recover the complete reaction energy:}\) \[ Q_{\mathrm{complete}}=Q_{\mathrm{prompt}}+E_{\gamma,\mathrm{total}} \] \[ Q_{\mathrm{complete}}=180+10 \] \[ Q_{\mathrm{complete}}=190\,\text{MeV} \] \(\textbf{Interpret the two stages:}\) The nuclear excitation energy is retained temporarily within the daughter fragments. It is not lost from the complete reaction energy budget. When the fragments de-excite, this stored energy becomes radiation. The energy partition changes between immediate fragment formation and complete de-excitation. However, the total energy released between the specified initial ground state and the final ground-state products is fixed by their rest-mass difference. \(\textbf{Final answer:}\) \[ \boxed{Q_{\mathrm{prompt}}=180\,\text{MeV},\quad E_{\gamma,\mathrm{total}}=10\,\text{MeV},\quad Q_{\mathrm{complete}}=190\,\text{MeV}} \]

Explanation: Process P begins with helium-3 and a free neutron. The free neutron contributes zero nuclear binding energy. The initial combined nuclear binding energy is therefore \(7.7\,\text{MeV}\). The helium-4 product has binding energy \(28.3\,\text{MeV}\). Thus, the binding energy increases by \(20.6\,\text{MeV}\). Statement I is correct because the process can release this energy. Process Q reverses the nuclear transformation. Its final configuration has smaller combined binding energy than its initial helium-4 nucleus. Consequently, the specified separation requires an energy input rather than releasing energy. Statement II is correct under the stated threshold approximation. Statement III is incorrect because the photon's energy originates from the change between initial and final nuclear rest energies. The photon is an outgoing carrier of energy, not the physical source of the nuclear binding-energy difference. Statement IV is correct because interchanging the initial and final states reverses the sign of their rest-energy difference. A reaction and its exact reverse cannot both have positive reaction energy when the nuclear states are identical in the two descriptions. The process classification and reaction-energy sign must be inferred from the complete nuclear configurations, not merely from the presence or absence of a photon. Thus, Statements I, II and IV are correct.

Explanation: \(\textbf{Apply momentum conservation to fusion products:}\) The total initial momentum is negligible. The final helium-4 nucleus and neutron therefore have equal and opposite momenta: \[ \vec p_{\mathrm{He}}=-\vec p_n \] Consequently: \[ p_{\mathrm{He}}=p_n \] \(\textbf{Relate their kinetic energies:}\) For non-relativistic motion: \[ K=\frac{p^2}{2m} \] At equal momentum magnitudes: \[ \frac{K_n}{K_{\mathrm{He}}}=\frac{m_{\mathrm{He}}}{m_n} \] \[ \frac{K_n}{K_{\mathrm{He}}}=\frac41 \] \(\textbf{Use the complete fusion energy budget:}\) \[ K_{\mathrm{He}}+K_n=17.6\,\text{MeV} \] Substituting \(K_n=4K_{\mathrm{He}}\): \[ 5K_{\mathrm{He}}=17.6 \] \[ K_{\mathrm{He}}=3.52\,\text{MeV} \] \(\textbf{Use the helium nucleus as the Rutherford projectile:}\) The helium-4 nucleus has charge: \[ q_{\mathrm{He}}=+2e \] The stationary gold nucleus has charge: \[ q_{\mathrm{Au}}=+79e \] At the head-on Coulomb turning point: \[ K_{\mathrm{He}}=\frac{k(2e)(79e)}{d} \] \(\textbf{Calculate the closest-approach distance:}\) Using the supplied Coulomb conversion: \[ d=\frac{2(79)(1.44)}{K_{\mathrm{He}}}\,\text{fm} \] Substituting: \[ d=\frac{2(79)(1.44)}{3.52}\,\text{fm} \] \[ d=\frac{227.52}{3.52}\,\text{fm} \] \[ d\approx64.6\,\text{fm} \] \(\textbf{Physical interpretation:}\) The fusion reaction energy is shared unequally between the helium nucleus and neutron because their outgoing momentum magnitudes are equal. The helium nucleus carries only \(3.52\,\text{MeV}\), not the entire \(17.6\,\text{MeV}\). Its kinetic energy then determines how close it can approach the positively charged gold nucleus under the separate Rutherford-scattering conditions. The result connects energy partition in a nuclear reaction with electrostatic scattering of one reaction product. \(\textbf{Final answer:}\) \[ \boxed{K_{\mathrm{He}}=3.52\,\text{MeV},\qquad d_{\min}\approx64.6\,\text{fm}} \]

Explanation: \(\textbf{Identify the shared limiting reactant:}\) Both permitted fusion reactions require one deuterium nucleus. There are exactly five deuterium nuclei. Therefore, the total number of fusion events must be five: \[ N_P+N_Q=5 \] \(\textbf{Apply the separate reactant limits:}\) Only three tritium nuclei are available: \[ N_P\leq3 \] Only four helium-3 nuclei are available: \[ N_Q\leq4 \] The reaction counts must also be non-negative integers. \(\textbf{Construct the total energy expression:}\) Reaction P releases: \[ Q_P=17.6\,\text{MeV} \] Reaction Q releases: \[ Q_Q=18.4\,\text{MeV} \] Therefore: \[ E_{\mathrm{total}}=17.6N_P+18.4N_Q \] \(\textbf{Eliminate one reaction count:}\) From: \[ N_P+N_Q=5 \] we obtain: \[ N_P=5-N_Q \] Thus: \[ E_{\mathrm{total}}=17.6(5-N_Q)+18.4N_Q \] Simplifying: \[ E_{\mathrm{total}}=88.0+0.8N_Q \] \(\textbf{Apply the energy-maximisation condition:}\) The energy increases by \(0.8\,\text{MeV}\) whenever one feasible P reaction is replaced by a Q reaction. Therefore, the greatest energy is obtained by using as many Q reactions as possible. The helium-3 supply allows: \[ N_Q=4 \] Hence: \[ N_P=5-4 \] \[ N_P=1 \] \(\textbf{Calculate the maximum released energy:}\) \[ E_{\max}=1(17.6)+4(18.4) \] \[ E_{\max}=17.6+73.6 \] \[ E_{\max}=91.2\,\text{MeV} \] \(\textbf{Check reactant availability:}\) The process consumes five deuterium nuclei, one tritium nucleus and four helium-3 nuclei. It does not exceed any available population. Two tritium nuclei remain unused. \(\textbf{Physical interpretation:}\) The reaction with greater energy release per event is preferred when maximising total energy, but its use is limited by the availability of its specific reactants. The correct allocation must therefore satisfy nuclear reactant counts as well as the energy objective. The problem illustrates why maximising the number of reactions and maximising released energy are related but distinct tasks. \(\textbf{Final answer:}\) \[ \boxed{N_P=1,\qquad N_Q=4,\qquad E_{\max}=91.2\,\text{MeV}} \]

Explanation: \(\textbf{Interpret the sign of the reaction energy:}\) The specified nuclear transformation has: \[ Q=-2.5\,\text{MeV} \] A negative reaction energy means that the complete final nuclear rest-energy sum is greater than the initial rest-energy sum. The difference must be supplied from the kinetic energy of the initial particles or another permitted energy source. \(\textbf{Write the energy balance:}\) With no outgoing radiation: \[ K_f=K_i+Q \] A physically possible outcome requires: \[ K_f\geq0 \] \(\textbf{Test experiment P:}\) The initial neutron kinetic energy is: \[ K_{i,P}=1.0\,\text{MeV} \] Therefore: \[ K_{f,P}=1.0-2.5 \] \[ K_{f,P}=-1.5\,\text{MeV} \] A negative total outgoing kinetic energy is impossible. Thus, experiment P lacks sufficient initial energy. \(\textbf{Test experiment Q:}\) The initial kinetic energy is: \[ K_{i,Q}=3.0\,\text{MeV} \] Therefore: \[ K_{f,Q}=3.0-2.5 \] \[ K_{f,Q}=0.5\,\text{MeV} \] This is non-negative. Hence, experiment Q satisfies the simplified energetic requirement. \(\textbf{Distinguish reaction energy from electrostatic approach:}\) A neutron has zero electric charge and therefore does not encounter the Coulomb repulsion faced by an approaching positively charged nucleus. However, removing a Coulomb barrier does not change the nuclear rest-energy difference. A negative-\(Q\) reaction can still require sufficient incident kinetic energy. Conversely, satisfying the energy requirement does not guarantee that the nuclear interaction will occur. \(\textbf{Physical interpretation:}\) The \(Q\)-value measures the net rest-energy change of a specified reaction. The initial kinetic energy determines whether that increase can be supplied. The elementary energy-balance condition is necessary for the reaction to occur, but reaction dynamics and exact momentum constraints may impose further requirements. \(\textbf{Final answer:}\) \[ K_{f,P}=-1.5\,\text{MeV}\ \text{(forbidden)},\qquad K_{f,Q}=0.5\,\text{MeV} \]

Explanation: \(\textbf{Determine the common initial binding energy:}\) Both channels begin with two deuterium nuclei. Therefore: \[ B_i=2B_D \] \[ B_i=2(2.2) \] \[ B_i=4.4\,\text{MeV} \] \(\textbf{Calculate channel P's energy release:}\) Channel P produces helium-4. The photon has zero nuclear binding energy. Thus: \[ B_{f,P}=28.3\,\text{MeV} \] The reaction energy is: \[ Q_P=B_{f,P}-B_i \] \[ Q_P=28.3-4.4 \] \[ Q_P=+23.9\,\text{MeV} \] \(\textbf{Calculate channel Q's energy release:}\) Channel Q produces a tritium nucleus and one free proton. The free proton contributes zero nuclear binding energy. Therefore: \[ B_{f,Q}=8.5\,\text{MeV} \] Thus: \[ Q_Q=B_{f,Q}-B_i \] \[ Q_Q=8.5-4.4 \] \[ Q_Q=+4.1\,\text{MeV} \] \(\textbf{Compare the two channels:}\) \[ Q_P-Q_Q=23.9-4.1 \] \[ Q_P-Q_Q=19.8\,\text{MeV} \] \(\textbf{Verify charge and nucleon conservation:}\) Both channels begin with: \[ A_i=4,\qquad Z_i=2 \] Channel P produces helium-4 with the same total nuclear composition. Channel Q produces tritium and a proton: \[ A_f=3+1=4 \] \[ Z_f=1+1=2 \] Both equations are balanced. \(\textbf{Interpret the different final nuclear configurations:}\) Identical reactants do not require identical reaction energies when alternative final configurations are possible. The helium-4 product in P has much greater binding energy than the tritium product in Q. The photon in P can carry energy released through the nuclear rearrangement. Its presence is not an additional source of energy. The calculation determines reaction-energy differences, not the probabilities of the two channels. \(\textbf{Final answer:}\) \[ \boxed{Q_P=23.9\,\text{MeV},\qquad Q_Q=4.1\,\text{MeV},\qquad Q_P-Q_Q=19.8\,\text{MeV}} \]

Explanation: The reaction energy and other stated energy contributions can be combined using conservation of total energy. If the initial total kinetic energy is known and all outgoing carriers are included, subtracting the non-fragment outgoing energies determines the fragments' combined kinetic energy. Statement I is correct. However, determining their separate kinetic energies requires more than their masses and combined kinetic energy. The inverse-mass kinetic-energy relation follows when two non-relativistic particles have equal momentum magnitudes. For a two-body decay from rest, momentum conservation directly establishes that condition. When additional particles are emitted, the fragments need not have equal and opposite momenta. Their combined momentum must balance the vector sum of all other outgoing particle momenta. Therefore, Statement II is incorrect. Statement III is correct because zero initial total momentum together with zero net momentum of the other outgoing particles requires the two fragments' momenta to sum to zero. Their momenta are then equal in magnitude and opposite in direction. Statement IV is correct because momentum is a vector. Knowing the energies of emitted particles does not generally establish how their momentum vectors are oriented. Two particles with the same individual energies may have zero or non-zero combined momentum depending on their directions. Thus, energy conservation can determine a total kinetic-energy budget without uniquely determining its division between the heavy fragments. Statements I, III and IV are correct.

Explanation: \(\textbf{Identify the ground-state reaction reference:}\) The direct ground-state reaction has: \[ Q_0=+12.0\,\text{MeV} \] This is the difference between the initial ground-state nuclear rest energy and the final ground-state nuclear rest energies. \(\textbf{Account for photon absorption:}\) The incoming photon supplies: \[ E_\gamma=5.0\,\text{MeV} \] The nucleus absorbs it and occupies an excited state. Its rest energy is now greater by: \[ E^*=5.0\,\text{MeV} \] \(\textbf{Calculate the reaction energy from the excited state:}\) The excited nucleus has greater initial rest energy than the ground-state reference. Therefore: \[ Q^*=Q_0+E^* \] \[ Q^*=12.0+5.0 \] \[ Q^*=17.0\,\text{MeV} \] \(\textbf{Check the complete energy balance:}\) An equivalent calculation can begin before the photon is absorbed. The initial ground-state nucleus is accompanied by a photon carrying \(5.0\,\text{MeV}\). Therefore, the energy available beyond the specified final rest energies is: \[ E_{\mathrm{available}}=Q_0+E_\gamma \] \[ E_{\mathrm{available}}=12.0+5.0 \] \[ E_{\mathrm{available}}=17.0\,\text{MeV} \] \(\textbf{Avoid double-counting the photon:}\) The absorbed photon increases the nucleus's excitation energy. That excitation is already included when the reaction energy is calculated from the excited nuclear state. It must not be added a second time. \(\textbf{Physical interpretation:}\) An absorbed photon may supply nuclear excitation energy that becomes available during a subsequent transformation. The reaction energy from the excited state differs from the ground-state reference because the initial nuclear state differs. However, both approaches to the complete energy accounting give the same result. The calculation depends on identifying the actual initial state and ensuring that the absorbed photon's energy is counted exactly once. \(\textbf{Final answer:}\) \[ \boxed{Q^*=17.0\,\text{MeV},\qquad E_{\mathrm{available}}=17.0\,\text{MeV}} \]

Explanation: Process P describes capture of one free neutron by a nucleus. The initial target contains \(A\) nucleons and \(Z\) protons. After neutron capture, the bound nuclear product contains \(A+1\) nucleons. The neutron carries zero electric charge, so the atomic number remains \(Z\). Statement I is correct. Statement II is incorrect because increasing mass number does not require increasing proton number. The added neutron contributes one nucleon while leaving the proton count unchanged. Process Q is the exact reverse nuclear transformation under the specified state conditions. Therefore, the initial and final nuclear rest-energy configurations are interchanged. The reaction-energy difference changes sign when the reaction direction is reversed. Statement III is correct. If neutron capture in P releases energy, the reverse separation in Q must require energy to restore the original separated nuclear configuration. Statement IV is therefore correct. The photon in P is a carrier of the energy released by nuclear capture. In Q, an incident photon may supply the energy needed for separation. The two descriptions are consistent with conservation of nucleon number, electric charge and total energy. The nuclear rest-energy balance, rather than the mere appearance of a photon, determines which reaction is exothermic. Thus, Statements I, III and IV are correct.

Explanation: \(\textbf{Combine initial kinetic energy with nuclear reaction energy:}\) The nuclear reaction releases: \[ Q=17.6\,\text{MeV} \] The initial centre-of-mass kinetic energy is: \[ K_i=2.4\,\text{MeV} \] Conservation of energy gives: \[ K_f=K_i+Q \] Substituting: \[ K_f=2.4+17.6 \] \[ K_f=20.0\,\text{MeV} \] \(\textbf{Apply momentum conservation:}\) The total initial momentum is zero. Only two final particles are produced. Therefore: \[ \vec p_n+\vec p_{\mathrm{He}}=0 \] Their momentum magnitudes are equal: \[ p_n=p_{\mathrm{He}} \] \(\textbf{Relate the individual kinetic energies:}\) For equal momentum magnitudes: \[ \frac{K_n}{K_{\mathrm{He}}}=\frac{m_{\mathrm{He}}}{m_n} \] Using the approximate masses: \[ \frac{K_n}{K_{\mathrm{He}}}=\frac41 \] Therefore: \[ K_n=4K_{\mathrm{He}} \] \(\textbf{Use the complete outgoing kinetic energy:}\) \[ K_n+K_{\mathrm{He}}=20.0\,\text{MeV} \] Substituting: \[ 4K_{\mathrm{He}}+K_{\mathrm{He}}=20.0 \] \[ 5K_{\mathrm{He}}=20.0 \] \[ K_{\mathrm{He}}=4.0\,\text{MeV} \] \(\textbf{Recover the neutron kinetic energy:}\) \[ K_n=4(4.0) \] \[ K_n=16.0\,\text{MeV} \] \(\textbf{Physical interpretation:}\) The final kinetic-energy sum exceeds the reaction energy because the reactants already possess kinetic energy. The released nuclear energy and the initial kinetic energy must both appear in the complete energy balance. The individual product energies are determined by momentum conservation. The lighter neutron carries the larger kinetic-energy share because the two products have equal momentum magnitudes. The energy partition depends on the actual total outgoing kinetic energy, not only on the reaction \(Q\)-value. \(\textbf{Final answer:}\) \[ \boxed{K_f=20.0\,\text{MeV},\qquad K_n=16.0\,\text{MeV}} \]

Explanation: \(\textbf{Establish the relevant energy comparison:}\) A nuclear reaction's energy is determined by the difference between the combined final and initial total binding energies. For conserved proton and neutron totals: \[ Q=B_f-B_i \] \(\textbf{Calculate the forward reaction's initial binding:}\) The initial nucleus P has: \[ B_i=192\,\text{MeV} \] \(\textbf{Calculate the combined final binding:}\) The two daughter nuclei have: \[ B_Q=98\,\text{MeV} \] \[ B_R=95\,\text{MeV} \] Therefore: \[ B_f=98+95 \] \[ B_f=193\,\text{MeV} \] \(\textbf{Determine the forward reaction energy:}\) \[ Q_{\mathrm{forward}}=193-192 \] \[ Q_{\mathrm{forward}}=+1\,\text{MeV} \] The forward splitting reaction is exothermic. \(\textbf{Determine the exact reverse reaction energy:}\) For the reverse process, the initial and final configurations exchange roles. Therefore: \[ Q_{\mathrm{reverse}}=-Q_{\mathrm{forward}} \] \[ Q_{\mathrm{reverse}}=-1\,\text{MeV} \] The reverse fusion requires a net energy input. \(\textbf{Check using average binding energy per nucleon:}\) For P: \[ \overline B_P=\frac{192}{24} \] \[ \overline B_P=8.00\,\text{MeV/nucleon} \] For Q: \[ \overline B_Q=\frac{98}{12} \] \[ \overline B_Q\approx8.167\,\text{MeV/nucleon} \] For R: \[ \overline B_R=\frac{95}{12} \] \[ \overline B_R\approx7.917\,\text{MeV/nucleon} \] The weighted combined final binding energy is greater than the parent's total binding energy by \(1\,\text{MeV}\). \(\textbf{Physical interpretation:}\) The positive binding energies of individual nuclei do not imply that every transformation involving them releases energy. The direction of energy release depends on which complete nuclear configuration is more strongly bound. The exact reverse of an exothermic nuclear reaction is endothermic when the initial and final nuclear states are otherwise identical. The investigator's argument confuses nuclear binding with the net change in binding. \(\textbf{Final answer:}\) \[ \boxed{Q_{\mathrm{forward}}=+1\,\text{MeV},\qquad Q_{\mathrm{reverse}}=-1\,\text{MeV}} \]

Explanation: The Coulomb potential energy is proportional to the product of the nuclear charge numbers. When \(U\) is plotted against \(1/r\), the proportionality becomes a straight-line relation. Its slope is proportional to \(Z_1Z_2\). The charge-number product of Q is four times that of P. Therefore, Statement I is correct. For a classical head-on approach from large separation, the turning point occurs when the initial relative kinetic energy has become Coulomb potential energy. At fixed kinetic energy, increasing the charge-number product increases the separation at which the turning-point condition is reached. Therefore, Statement II is incorrect and Statement III is correct. A greater Coulomb interaction does not mean that the nuclei necessarily undergo fusion or release more energy once fusion occurs. The Coulomb potential describes their electrostatic approach. The net nuclear reaction energy depends on the difference between the complete initial and final nuclear rest energies. Statement IV is therefore incorrect. The graph slopes provide information about the electrostatic interaction strength and classical approach condition. They do not supply the binding-energy differences needed to calculate a nuclear fusion \(Q\)-value. Thus, Statements I and III are correct.

Explanation: \(\textbf{Determine the common nuclear radius:}\) The nuclear-radius law is: \[ R=R_0A^{1/3} \] Both nuclei have mass number: \[ A=64 \] Therefore: \[ R_P=R_Q=1.2(64)^{1/3} \] \[ R_P=R_Q=1.2(4) \] \[ R_P=R_Q=4.8\,\text{fm} \] \(\textbf{Establish the Rutherford turning-point condition:}\) At the classical head-on turning point, the alpha particle's initial kinetic energy has become electrostatic potential energy. Thus: \[ K=\frac{1}{4\pi\varepsilon_0}\frac{(2e)(Ze)}{d} \] Rearranging: \[ d=\frac{2Z(1.44)}{K}\,\text{fm} \] \(\textbf{Calculate P's turning-point distance:}\) \[ d_P=\frac{2(28)(1.44)}{18.0} \] \[ d_P=\frac{80.64}{18.0} \] \[ d_P=4.48\,\text{fm} \] \(\textbf{Calculate Q's turning-point distance:}\) \[ d_Q=\frac{2(32)(1.44)}{18.0} \] \[ d_Q=\frac{92.16}{18.0} \] \[ d_Q=5.12\,\text{fm} \] \(\textbf{Compare the turning points with nuclear size:}\) For P: \[ d_P=4.48\,\text{fm}\lt R_P=4.8\,\text{fm} \] For Q: \[ d_Q=5.12\,\text{fm}\gt R_Q=4.8\,\text{fm} \] \(\textbf{Interpret the model limitation:}\) The pure-Coulomb calculation for P places the turning point inside the estimated nuclear region. Therefore, the assumption that only electrostatic repulsion acts throughout that approach is not reliably justified. For Q, the calculated turning point remains outside the estimated nuclear surface. The difference arises because nuclear size depends on mass number, while the Coulomb interaction depends on nuclear charge. \(\textbf{Physical interpretation:}\) Isobars have equal mass numbers and hence equal radii under the stated radius law. They need not produce identical alpha-particle scattering behaviour because their proton numbers can differ. The experiment therefore separates two nuclear properties that are sometimes incorrectly treated as interchangeable. \(\textbf{Final answer:}\) \[ \boxed{R_P=R_Q=4.8\,\text{fm},\qquad d_P=4.48\,\text{fm},\qquad d_Q=5.12\,\text{fm}} \]

Explanation: \(\textbf{Identify the nuclear composition changes:}\) The three nuclei are isotopes, so their proton numbers are identical. Their mass numbers increase by one from P to Q and from Q to R. Each transformation can therefore be represented as capture of one free neutron. The free neutron contributes zero nuclear binding energy before capture. \(\textbf{Calculate the first capture energy:}\) For: \[ P+n\rightarrow Q+\gamma \] the reaction energy is: \[ Q_1=B_Q-B_P \] Substituting: \[ Q_1=167.8-160.0 \] \[ Q_1=+7.8\,\text{MeV} \] \(\textbf{Calculate the second capture energy:}\) For: \[ Q+n\rightarrow R+\gamma \] we obtain: \[ Q_2=B_R-B_Q \] \[ Q_2=175.2-167.8 \] \[ Q_2=+7.4\,\text{MeV} \] \(\textbf{Evaluate P's binding energy per nucleon:}\) \[ \overline B_P=\frac{160.0}{20} \] \[ \overline B_P=8.000\,\text{MeV/nucleon} \] \(\textbf{Evaluate Q's binding energy per nucleon:}\) \[ \overline B_Q=\frac{167.8}{21} \] \[ \overline B_Q\approx7.9905\,\text{MeV/nucleon} \] \(\textbf{Evaluate R's binding energy per nucleon:}\) \[ \overline B_R=\frac{175.2}{22} \] \[ \overline B_R\approx7.9636\,\text{MeV/nucleon} \] Therefore: \[ \overline B_P\gt \overline B_Q\gt \overline B_R \] \(\textbf{Resolve the apparent contradiction:}\) Both captures have positive reaction energies because the final nuclei possess greater total binding energies. However, each neutron capture increases the number of bound nucleons. The total binding energy increases by less than would be required to maintain the original average binding energy per nucleon. Consequently, the average decreases even while the total binding energy increases. \(\textbf{Physical interpretation:}\) Energy release in neutron capture is governed by the increase in total nuclear binding energy. It does not require the final nucleus to have greater average binding energy per nucleon than the parent. The distinction is important when comparing neighbouring isotopes, whose nucleon counts are different. \(\textbf{Final answer:}\) \[ \boxed{Q_1=+7.8\,\text{MeV},\quad Q_2=+7.4\,\text{MeV},\quad \overline B_P\gt \overline B_Q\gt \overline B_R} \]

Explanation: Isotopes have the same proton number but different neutron numbers. P and Q both have atomic number 18, while their mass numbers are 40 and 42. Therefore, P and Q are isotopes. Isobars have equal mass numbers but different atomic numbers. Q and R both have mass number 42, but their atomic numbers are 18 and 20. They are consequently isobars. Statement I is correct. To identify isotones, the neutron numbers must be compared. P contains \(40-18=22\) neutrons. R contains \(42-20=22\) neutrons. Thus, P and R are isotones, making Statement II correct. A neutral Q atom contains 18 electrons because its atomic number is 18. S has atomic number 20 and would contain 20 electrons when neutral. After losing two electrons, it contains 18 electrons. Statement III is therefore correct. Statement IV is incorrect under the standard nuclear-radius law. P and S have equal mass number 40. Their predicted nuclear radii are therefore equal when the same value of \(R_0\) is used. Nuclear radius in this approximation depends on mass number, not directly on atomic number. The complete comparison distinguishes nuclear composition, extranuclear electron count and nuclear size. Thus, Statements I, II and III are correct.

Explanation: \(\textbf{Construct the instrument's measurement relation:}\) The true nuclear radius is: \[ R=R_0A^{1/3} \] The displayed radius contains a common additive error: \[ R_{\mathrm{displayed}}=R_0A^{1/3}+\delta \] \(\textbf{Apply the measurement for mass number eight:}\) For: \[ A=8 \] we have: \[ A^{1/3}=2 \] Thus: \[ 3.4=2R_0+\delta \] \(\textbf{Apply the measurement for mass number twenty-seven:}\) For: \[ A=27 \] we have: \[ A^{1/3}=3 \] Therefore: \[ 4.6=3R_0+\delta \] \(\textbf{Eliminate the common calibration error:}\) Subtracting the first equation from the second: \[ 4.6-3.4=(3R_0+\delta)-(2R_0+\delta) \] \[ 1.2=R_0 \] Therefore: \[ R_0=1.2\,\text{fm} \] \(\textbf{Recover the zero error:}\) Using the first measurement: \[ 3.4=2(1.2)+\delta \] \[ \delta=3.4-2.4 \] \[ \delta=1.0\,\text{fm} \] \(\textbf{Predict the true radius for }A=64\textbf{:}\) \[ R_{64}=1.2(64)^{1/3} \] \[ R_{64}=1.2(4) \] \[ R_{64}=4.8\,\text{fm} \] \(\textbf{Predict the instrument's displayed reading:}\) \[ R_{\mathrm{displayed},64}=R_{64}+\delta \] \[ R_{\mathrm{displayed},64}=4.8+1.0 \] \[ R_{\mathrm{displayed},64}=5.8\,\text{fm} \] \(\textbf{Physical interpretation:}\) A constant additive error shifts every radius reading by the same amount. Taking the difference between two readings removes that error, allowing the radius constant to be inferred. The true nuclear radius and the displayed reading are distinct quantities. The instrument will report \(5.8\,\text{fm}\), although the nucleus's actual radius under the model is \(4.8\,\text{fm}\). \(\textbf{Final answer:}\) \[ \boxed{R_0=1.2\,\text{fm},\qquad\delta=1.0\,\text{fm},\qquad R_{\mathrm{displayed},64}=5.8\,\text{fm}} \]

Explanation: \(\textbf{Calculate the parent nucleus's total binding energy:}\) The initial mass number is: \[ A_i=100 \] The average binding energy is: \[ \overline B_i=8.00\,\text{MeV/nucleon} \] Therefore: \[ B_i=A_i\overline B_i \] \[ B_i=100(8.00) \] \[ B_i=800\,\text{MeV} \] \(\textbf{Use the measured reaction energy:}\) For conserved total proton and neutron numbers: \[ Q=B_f-B_i \] The reaction releases: \[ Q=+16.0\,\text{MeV} \] Thus: \[ B_f=B_i+Q \] \[ B_f=800+16 \] \[ B_f=816\,\text{MeV} \] \(\textbf{Determine the number of bound final nucleons:}\) The average binding energy per bound nucleon in the fragments is: \[ \overline B_f=8.50\,\text{MeV/nucleon} \] Therefore: \[ B_f=A_{\mathrm{bound}}\overline B_f \] Rearranging: \[ A_{\mathrm{bound}}=\frac{B_f}{\overline B_f} \] Substituting: \[ A_{\mathrm{bound}}=\frac{816}{8.50} \] \[ A_{\mathrm{bound}}=96 \] \(\textbf{Apply mass-number conservation:}\) The initial system contains 100 nucleons. The bound fragments contain 96 nucleons. The remaining nucleons must therefore appear as free neutrons under the stated product conditions. Thus: \[ x=100-96 \] \[ x=4 \] \(\textbf{Verify the energy balance independently:}\) The fragments' combined binding energy is: \[ B_f=96(8.50) \] \[ B_f=816\,\text{MeV} \] Therefore: \[ Q=816-800 \] \[ Q=+16\,\text{MeV} \] This agrees with the measured energy release. \(\textbf{Physical interpretation:}\) The final average binding energy per bound nucleon cannot be multiplied by the parent's full mass number when free neutrons are emitted. Only nucleons remaining in bound fragments contribute to that final average. The reaction energy fixes the total final binding energy, which then determines the number of bound nucleons. Mass-number conservation supplies the emitted neutron count. The calculation is an inverse use of nuclear binding and reaction-energy data. \(\textbf{Final answer:}\) \[ \boxed{x=4,\qquad B_f=816\,\text{MeV}} \]

Explanation: Fusion requires the reactant nuclei to approach sufficiently closely for the attractive nuclear interaction to become important. The approach of two nuclei depends on their motion relative to one another. It is not determined solely by the sum of their laboratory-frame kinetic energies. In arrangement P, the nuclei initially travel in the same direction with identical velocities. Their initial relative velocity is zero. Therefore, their common motion does not, by itself, bring the nuclei closer together. The kinetic energy associated with uniform motion of the pair's centre of mass does not provide additional energy for overcoming their mutual Coulomb repulsion. The nuclei may subsequently move apart under repulsion, but this does not create an approaching collision. In arrangement Q, the two nuclei move directly towards one another. Their relative velocity is non-zero and directed towards decreasing separation. The approaching motion allows their relative kinetic energy to be converted into increasing electrostatic potential energy. Whether they reach the nuclear interaction region depends on the available relative energy and the repulsive interaction. The equal laboratory kinetic-energy sums therefore do not imply equivalent collision conditions. This distinction is important in interpreting the high kinetic energies associated with thermonuclear fusion. The relative motion of reacting nuclei is relevant to overcoming electrostatic repulsion, while the net fusion reaction energy depends on the initial and final nuclear configurations. Thus, arrangement Q is the one initially providing an approaching collision.

Explanation: \(\textbf{Convert equal sample masses into nucleus counts:}\) The mass of one nucleus is approximately proportional to its mass number. Therefore: \[ m_P:m_Q=8:64 \] \[ m_P:m_Q=1:8 \] The total masses of the two species in the collection are equal: \[ N_Pm_P=N_Qm_Q \] Hence: \[ \frac{N_P}{N_Q}=\frac{m_Q}{m_P} \] \[ \frac{N_P}{N_Q}=8 \] Thus: \[ N_P:N_Q=8:1 \] \(\textbf{Calculate the individual nuclear radii:}\) For P: \[ R_P=R_0(8)^{1/3} \] \[ R_P=2R_0 \] For Q: \[ R_Q=R_0(64)^{1/3} \] \[ R_Q=4R_0 \] \(\textbf{Determine the number-weighted average radius:}\) Let the collection contain eight P nuclei for every one Q nucleus. Then: \[ \overline R=\frac{8R_P+R_Q}{9} \] Substituting: \[ \overline R=\frac{8(2R_0)+4R_0}{9} \] \[ \overline R=\frac{20}{9}R_0 \] \(\textbf{Calculate total binding energy in one representative group:}\) Eight P nuclei contain: \[ A_{P,\mathrm{total}}=8(8) \] \[ A_{P,\mathrm{total}}=64 \] nucleons. One Q nucleus also contains: \[ A_Q=64 \] nucleons. The binding-energy contributions are: \[ B_{P,\mathrm{total}}=64(7.0) \] \[ B_{P,\mathrm{total}}=448\,\text{MeV} \] and: \[ B_Q=64(8.0) \] \[ B_Q=512\,\text{MeV} \] \(\textbf{Determine the collection's binding energy per nucleon:}\) The combined binding energy is: \[ B_{\mathrm{total}}=448+512 \] \[ B_{\mathrm{total}}=960\,\text{MeV} \] The total nucleon count is: \[ A_{\mathrm{total}}=64+64 \] \[ A_{\mathrm{total}}=128 \] Therefore: \[ \overline B_{\mathrm{collection}}=\frac{960}{128} \] \[ \overline B_{\mathrm{collection}}=7.5\,\text{MeV/nucleon} \] \(\textbf{Physical interpretation:}\) Equal sample masses do not imply equal numbers of nuclei when their mass numbers differ. The number-weighted mean radius is dominated by the more numerous lighter nuclei. However, equal approximate sample masses imply equal total nucleon contributions under the stated mass approximation. The collection's average binding energy per nucleon consequently gives equal weight to the two species' per-nucleon values. The appropriate averaging weights depend on whether the quantity is measured per nucleus or per nucleon. \(\textbf{Final answer:}\) \[ \boxed{N_P:N_Q=8:1,\qquad\overline R=\frac{20}{9}R_0,\qquad\overline B=7.5\,\text{MeV/nucleon}} \]

Explanation: \(\textbf{Determine the neutron separation energy to ground-state Y:}\) The parent binding energy is: \[ B_X=200\,\text{MeV} \] The ground-state daughter binding energy is: \[ B_{Y,g}=192\,\text{MeV} \] The neutron separation energy is: \[ S_{n,g}=B_X-B_{Y,g} \] \[ S_{n,g}=200-192 \] \[ S_{n,g}=8\,\text{MeV} \] \(\textbf{Identify the daughter excitation energy:}\) The excited-state daughter has binding energy: \[ B_{Y,e}=190\,\text{MeV} \] The difference in nuclear binding is: \[ E^*=B_{Y,g}-B_{Y,e} \] \[ E^*=192-190 \] \[ E^*=2\,\text{MeV} \] The excited state therefore lies \(2\,\text{MeV}\) above the ground state in rest energy. \(\textbf{Determine the threshold to excited-state Y:}\) Producing an excited daughter requires both neutron separation and the additional excitation energy. Thus: \[ S_{n,e}=S_{n,g}+E^* \] \[ S_{n,e}=8+2 \] \[ S_{n,e}=10\,\text{MeV} \] \(\textbf{Test the supplied photon energy:}\) The incident photon carries: \[ E_\gamma=9.0\,\text{MeV} \] For ground-state Y: \[ 9.0\gt 8.0 \] The channel is energetically accessible. The excess kinetic energy is: \[ K_{\mathrm{excess},g}=9.0-8.0 \] \[ K_{\mathrm{excess},g}=1.0\,\text{MeV} \] For excited-state Y: \[ 9.0\lt 10.0 \] The photon cannot supply the required energy. \(\textbf{Physical interpretation:}\) Removing a neutron can leave the daughter nucleus in different internal energy states. An excited daughter requires a greater energy input than the same daughter in its ground state. The binding energy of the excited state is smaller because its nuclear rest energy is higher. A photon can therefore be energetic enough to open one neutron-emission channel while leaving a more highly excited product channel inaccessible. The calculation establishes energetic accessibility rather than guaranteeing the occurrence of either reaction. \(\textbf{Final answer:}\) \[ \boxed{S_{n,g}=8\,\text{MeV},\qquad S_{n,e}=10\,\text{MeV},\qquad K_{\mathrm{excess},g}=1\,\text{MeV}} \]