Atoms and Nuclei MCQs with Answers – Part 5 (NEET Physics)

Explanation: \(\textbf{Convert the sample mass:}\) The initial sample mass is: \[ m=1.0\,\text{mg} \] \[ m=1.0\times10^{-6}\,\text{kg} \] \(\textbf{Calculate the mass of one nucleus:}\) The approximate mass of one nucleus is: \[ M_N=100u \] Therefore: \[ M_N=100(1.66\times10^{-27}) \] \[ M_N=1.66\times10^{-25}\,\text{kg} \] \(\textbf{Determine the initial number of nuclei:}\) \[ N=\frac{m}{M_N} \] Substituting: \[ N=\frac{1.0\times10^{-6}}{1.66\times10^{-25}} \] \[ N\approx6.024\times10^{18} \] \(\textbf{Apply the reacting fraction:}\) Only one-quarter of the nuclei react: \[ N_{\mathrm{reacted}}=0.25N \] \[ N_{\mathrm{reacted}}\approx1.506\times10^{18} \] \(\textbf{Convert energy released per nucleus:}\) Each reaction releases: \[ Q=200\,\text{MeV} \] In joules: \[ Q=200(1.60\times10^{-13}) \] \[ Q=3.20\times10^{-11}\,\text{J} \] \(\textbf{Calculate total nuclear energy release:}\) \[ E_{\mathrm{nuclear}}=N_{\mathrm{reacted}}Q \] \[ E_{\mathrm{nuclear}}=(1.506\times10^{18})(3.20\times10^{-11}) \] \[ E_{\mathrm{nuclear}}\approx4.819\times10^7\,\text{J} \] \(\textbf{Apply electrical conversion efficiency:}\) The electrical energy produced is: \[ E_{\mathrm{electric}}=0.40E_{\mathrm{nuclear}} \] \[ E_{\mathrm{electric}}\approx0.40(4.819\times10^7) \] \[ E_{\mathrm{electric}}\approx1.928\times10^7\,\text{J} \] \(\textbf{Calculate average useful power:}\) The experiment lasts: \[ t=10\,\text{s} \] Therefore: \[ P=\frac{E_{\mathrm{electric}}}{t} \] \[ P=\frac{1.928\times10^7}{10} \] \[ P\approx1.928\times10^6\,\text{W} \] \[ P\approx1.93\,\text{MW} \] \(\textbf{Physical interpretation:}\) The number of reacting nuclei determines the total nuclear energy release. The efficiency determines how much of that energy becomes useful electrical energy. The experiment duration then determines average electrical power. These are three separate quantities, and none should be substituted for another. A small sample can contain an enormous number of nuclei, making the aggregate released energy substantial even when only a fraction undergoes the specified transformation. \(\textbf{Final answer:}\) \[ \boxed{E_{\mathrm{nuclear}}\approx4.82\times10^7\,\text{J},\qquad P_{\mathrm{electric}}\approx1.93\,\text{MW}} \]

Explanation: \(\textbf{Determine the initial binding energy:}\) The parent contains: \[ A_i=80 \] Its average binding energy is: \[ \overline B_i=8.00\,\text{MeV/nucleon} \] Therefore: \[ B_i=80(8.00) \] \[ B_i=640\,\text{MeV} \] \(\textbf{Calculate P's binding contribution:}\) P contains 30 nucleons. Its binding energy is: \[ B_P=30(8.20) \] \[ B_P=246\,\text{MeV} \] \(\textbf{Represent Q's unknown contribution:}\) Q contains 50 nucleons. Its binding energy is: \[ B_Q=50b \] Therefore, the combined final binding energy is: \[ B_f=246+50b \] \(\textbf{Apply the positive-reaction-energy condition:}\) For conserved constituent nucleon totals: \[ Q_{\mathrm{reaction}}=B_f-B_i \] The reaction releases energy when: \[ Q_{\mathrm{reaction}}\gt 0 \] Thus: \[ 246+50b-640\gt 0 \] \[ 50b\gt 394 \] \[ b\gt \frac{394}{50} \] \[ b\gt 7.88\,\text{MeV/nucleon} \] \(\textbf{Check the threshold interpretation:}\) At exactly: \[ b=7.88\,\text{MeV/nucleon} \] the combined final binding energy equals the initial binding energy. The reaction energy would then be zero. A strictly positive reaction energy requires \(b\) to exceed this value. \(\textbf{Physical interpretation:}\) One final nucleus already has a higher average binding energy than the parent. Consequently, the other product need not also have a higher average binding energy for the complete reaction to release energy. The relevant requirement is that the products' weighted total binding energy exceed the parent's total binding energy. This illustrates why an energy-release condition can be expressed as a threshold on an unknown product's binding energy per nucleon. \(\textbf{Final answer:}\) \[ \boxed{b\gt 7.88\,\text{MeV/nucleon}} \]

Explanation: Statement I is correct under the specified condition. If the parent nucleus has the greatest average binding energy per nucleon and every final bound fragment has a smaller value, then the final combined binding energy cannot exceed the initial total when the full nucleon content is conserved. The conclusion follows because the final total is a sum of mass-number-weighted binding energies. Statement II is also correct. A free neutron is still a nucleon and must be included in mass-number conservation. However, a free neutron is not part of a bound nucleus and has zero nuclear binding energy relative to the free-nucleon reference. Therefore, it contributes to the nucleon count but not to the final sum of bound-nucleus binding energies. Statement III is incorrect. Positive reaction energy means that the specified final configuration has lower combined rest energy than the corresponding initial configuration. It does not independently establish a large reaction probability or rapid reaction rate. A reaction may encounter an electrostatic barrier or other physical constraints even when its energy balance is favourable. Statement IV is correct because average binding energies must be converted into total binding energies using the appropriate mass numbers. Complete reactant and product information is then needed to compare initial and final totals. A graph showing only one nucleus's average binding energy cannot determine the energy release of an unspecified transformation. Thus, Statements I, II and IV are correct.

Explanation: \(\textbf{Interpret the forward energy requirement:}\) The formation process requires a net input of: \[ 15\,\text{MeV} \] Therefore, its reaction energy is negative: \[ Q_{\mathrm{forward}}=-15\,\text{MeV} \] \(\textbf{Relate reaction energy to total binding energy:}\) For a reaction conserving proton and neutron totals: \[ Q=B_f-B_i \] In the forward direction, the final system is the single heavy nucleus. The initial system is the pair of separated bound fragments. Therefore: \[ B_{\mathrm{heavy}}-B_{\mathrm{fragments}}=-15\,\text{MeV} \] Rearranging: \[ B_{\mathrm{fragments}}-B_{\mathrm{heavy}}=+15\,\text{MeV} \] The separated bound fragments have greater combined nuclear binding energy. \(\textbf{Reverse the transformation:}\) The reverse process begins with the heavy nucleus and ends with the same pair of fragments. Thus: \[ Q_{\mathrm{reverse}}=B_{\mathrm{fragments}}-B_{\mathrm{heavy}} \] \[ Q_{\mathrm{reverse}}=+15\,\text{MeV} \] \(\textbf{Check the sign relationship:}\) Because the initial and final configurations exchange roles: \[ Q_{\mathrm{reverse}}=-Q_{\mathrm{forward}} \] \[ Q_{\mathrm{reverse}}=-(-15) \] \[ Q_{\mathrm{reverse}}=+15\,\text{MeV} \] \(\textbf{Physical interpretation:}\) The reaction energies of exactly reversed transformations have equal magnitudes and opposite signs. The net binding-energy change is also reversed. A positive binding energy for each individual nucleus does not imply that every reaction involving those nuclei releases energy. The comparison depends on the total binding-energy difference between the complete configurations. This result concerns reaction energetics and does not independently determine the ease or mechanism of either direction. \(\textbf{Final answer:}\) \[ \boxed{Q_{\mathrm{forward}}=-15\,\text{MeV},\qquad Q_{\mathrm{reverse}}=+15\,\text{MeV}} \]

Explanation: \(\textbf{Establish the common reaction-energy convention:}\) For each process, proton and neutron totals are conserved. Therefore: \[ Q_{\mathrm{reaction}}=B_f-B_i \] A positive result indicates energy release. A negative result indicates net energy input. \(\textbf{Evaluate process P:}\) The initial binding energy is: \[ B_{i,P}=12(7.5) \] \[ B_{i,P}=90\,\text{MeV} \] The final products contain three four-nucleon nuclei. Therefore: \[ B_{f,P}=3[4(7.0)] \] \[ B_{f,P}=84\,\text{MeV} \] The reaction energy is: \[ Q_P=84-90 \] \[ Q_P=-6\,\text{MeV} \] Process P requires energy. \(\textbf{Evaluate process Q:}\) The initial binding energy is: \[ B_{i,Q}=100(8.0) \] \[ B_{i,Q}=800\,\text{MeV} \] The final configuration contains two nuclei of mass number 50. Therefore: \[ B_{f,Q}=2[50(8.3)] \] \[ B_{f,Q}=830\,\text{MeV} \] The reaction energy is: \[ Q_Q=830-800 \] \[ Q_Q=+30\,\text{MeV} \] Process Q releases energy. \(\textbf{Evaluate process R:}\) The initial system contains two nuclei having ten nucleons each. Its combined binding energy is: \[ B_{i,R}=2[10(7.0)] \] \[ B_{i,R}=140\,\text{MeV} \] The final nucleus has: \[ B_{f,R}=20(7.2) \] \[ B_{f,R}=144\,\text{MeV} \] Therefore: \[ Q_R=144-140 \] \[ Q_R=+4\,\text{MeV} \] \(\textbf{Rank the three reaction energies:}\) The calculated values are: \[ Q_P=-6\,\text{MeV} \] \[ Q_R=+4\,\text{MeV} \] \[ Q_Q=+30\,\text{MeV} \] Therefore: \[ Q_P\lt Q_R\lt Q_Q \] \(\textbf{Physical interpretation:}\) Process P is endothermic because its products have smaller combined binding energy. Processes R and Q are exothermic because their final configurations are more strongly bound in total. Q releases more energy than R, even though the two processes involve different nuclear mass scales. The sign and magnitude of reaction energy depend on total binding-energy differences, not merely on whether a process joins or separates nuclei. \(\textbf{Final answer:}\) \[ Q_P=-6\,\text{MeV},\qquad Q_R=+4\,\text{MeV},\qquad Q_Q=+30\,\text{MeV} \]

Explanation: \(\textbf{Identify the initial nucleon count:}\) The incident neutron contributes one nucleon. Therefore: \[ A_i=235+1 \] \[ A_i=236 \] \(\textbf{Apply mass-number conservation:}\) The products contain barium-144, an unknown krypton isotope and three free neutrons. Thus: \[ 236=144+A+3 \] Rearranging: \[ A=236-144-3 \] \[ A=89 \] \(\textbf{Independently verify electric-charge conservation:}\) The initial nuclear charge number is: \[ Z_i=92 \] The final charge numbers are: \[ Z_f=56+36+3(0) \] \[ Z_f=92 \] Therefore, electric charge is conserved. \(\textbf{Write the completed nuclear equation:}\) \[ {}^{235}_{92}\mathrm U+{}^{1}_{0}\mathrm n \rightarrow{}^{144}_{56}\mathrm{Ba} +{}^{89}_{36}\mathrm{Kr} +3{}^{1}_{0}\mathrm n+Q \] \(\textbf{Interpret positive reaction energy:}\) For a balanced nuclear reaction: \[ Q=(M_i-M_f)c^2 \] Positive reaction energy corresponds to: \[ M_i\gt M_f \] Equivalently, when the total numbers of protons and neutrons are conserved: \[ Q=B_f-B_i\gt 0 \] The complete final nuclear configuration is therefore more strongly bound in total. \(\textbf{Physical interpretation:}\) The incident neutron changes the initial nucleon count from 235 to 236. The emitted neutrons must be included in the final mass-number balance even though they carry no electric charge. A positive reaction energy does not mean that nucleons have been created or destroyed. It indicates that some initial rest energy becomes available as kinetic energy or radiation. The particular fragment pair is one possible fission channel, not the only possible product combination for uranium-235 fission. \(\textbf{Final answer:}\) \[ \boxed{A=89,\qquad x=3,\qquad Q\gt 0\Longleftrightarrow B_f\gt B_i} \]

Explanation: Both events begin with uranium-235 and one incident neutron. The initial total mass number is therefore 236, while the initial total nuclear charge number is 92. For event P, the product mass numbers add to \(144+89+3=236\). Their charge numbers add to \(56+36=92\). Thus, event P satisfies both conservation conditions. For event Q, the mass numbers add to \(140+93+3=236\). The charge numbers add to \(54+38=92\). Event Q therefore satisfies the same two conservation conditions. Statement I is correct. The two balanced reactions contain different fission-fragment pairs. Statement II is correct because the daughter combination in neutron-induced fission is not necessarily unique. Statement III is incorrect because equality of emitted neutron counts does not guarantee equality of the total final rest masses. The two fragment pairs may have different combined binding energies. Their corresponding reaction energies may therefore differ. Statement IV is correct because free neutrons released during fission may encounter and induce fission in other suitable nuclei. Whether they actually do so depends on the conditions in the surrounding material. Thus, Statements I, II and IV are correct.

Explanation: \(\textbf{Check the complete reaction balance:}\) The initial mass-number total is: \[ A_i=235+1 \] \[ A_i=236 \] The final mass-number total is: \[ A_f=142+92+2 \] \[ A_f=236 \] The charge numbers satisfy: \[ 92=56+36 \] Thus, the total numbers of protons and neutrons are conserved. \(\textbf{Determine the initial binding-energy sum:}\) The initial uranium nucleus has binding energy: \[ B_{\mathrm U}=1780\,\text{MeV} \] The incident neutron is free and has zero nuclear binding energy. Therefore: \[ B_i=1780\,\text{MeV} \] \(\textbf{Determine the final binding-energy sum:}\) The barium and krypton fragments have combined binding energy: \[ B_f=1180+770 \] \[ B_f=1950\,\text{MeV} \] The two free neutrons contribute zero nuclear binding energy. \(\textbf{Calculate the fission reaction energy:}\) For the balanced reaction: \[ Q=B_f-B_i \] Substituting: \[ Q=1950-1780 \] \[ Q=+170\,\text{MeV} \] \(\textbf{Calculate the energy release per initial nucleon:}\) The complete initial system contains 236 nucleons. Therefore: \[ q=\frac{Q}{236} \] \[ q=\frac{170}{236} \] \[ q\approx0.7203\,\text{MeV/nucleon} \] \(\textbf{Interpret the positive result:}\) The products have greater combined nuclear binding energy than the original uranium nucleus. Their combined rest energy is therefore smaller than that of the complete initial system. The released energy can appear as fragment kinetic energy, emitted-neutron kinetic energy or radiation. The incident neutron must be included when calculating energy released per participating initial nucleon. It must not be assigned an additional nuclear binding-energy contribution while initially free. \(\textbf{Final answer:}\) \[ \boxed{Q=+170\,\text{MeV},\qquad q\approx0.720\,\text{MeV/nucleon}} \]

Explanation: \(\textbf{Check whether neutral atomic masses are consistent:}\) A neutral uranium atom contains 92 electrons. The neutral barium and krypton atoms together contain: \[ 56+36=92 \] electrons. Thus, the electron rest-mass contributions balance between the initial and final neutral atomic mass sums. The incident and emitted neutrons contain no electrons. \(\textbf{Calculate the initial mass sum:}\) The initial system contains one uranium atom and one neutron. Therefore: \[ M_i=M({}^{235}\mathrm U)+m_n \] \[ M_i=235.044u+1.009u \] \[ M_i=236.053u \] \(\textbf{Calculate the final mass sum:}\) The final system contains the two neutral atomic species and three neutrons. Thus: \[ M_f=M({}^{144}\mathrm{Ba})+M({}^{89}\mathrm{Kr})+3m_n \] Substituting: \[ M_f=143.923+88.918+3(1.009) \] \[ M_f=143.923+88.918+3.027 \] \[ M_f=235.868u \] \(\textbf{Determine the mass decrease:}\) \[ \Delta M=M_i-M_f \] \[ \Delta M=236.053-235.868 \] \[ \Delta M=0.185u \] \(\textbf{Convert the mass decrease into released energy:}\) \[ Q=\Delta Mc^2 \] \[ Q=0.185(931.5) \] \[ Q=172.3275\,\text{MeV} \] \[ Q\approx172.3\,\text{MeV} \] \(\textbf{Physical interpretation:}\) The positive result is consistent with an energy-releasing fission reaction. The initial and final sums contain the same number of electrons when the neutral atomic-mass convention is used. Therefore, an additional correction involving 92 electron masses would count the electron contribution incorrectly. The mass-number balance is a separate requirement from equality of rest masses. Nucleon number is conserved, while the decrease in combined rest mass corresponds to energy released during the transformation. \(\textbf{Final answer:}\) \[ \boxed{\Delta M=0.185u,\qquad Q\approx172.3\,\text{MeV}} \]

Explanation: \(\textbf{Distinguish reaction energy from total available energy:}\) The fission reaction energy is: \[ Q=+180\,\text{MeV} \] The incident neutron also carries initial kinetic energy: \[ K_i=2\,\text{MeV} \] Thus, the total energy available beyond the final rest energies includes both quantities. \(\textbf{Apply energy conservation:}\) For the complete process: \[ M_ic^2+K_i=M_fc^2+K_f+E_\gamma \] Using: \[ Q=(M_i-M_f)c^2 \] we obtain: \[ K_f+E_\gamma=K_i+Q \] \(\textbf{Calculate total available outgoing energy:}\) \[ E_{\mathrm{available}}=2+180 \] \[ E_{\mathrm{available}}=182\,\text{MeV} \] \(\textbf{Account for electromagnetic radiation:}\) The radiation carries: \[ E_\gamma=12\,\text{MeV} \] Therefore, the total material-particle kinetic energy is: \[ K_f=182-12 \] \[ K_f=170\,\text{MeV} \] \(\textbf{Separate neutron and fragment kinetic energies:}\) The emitted neutrons together carry: \[ K_n=20\,\text{MeV} \] Therefore: \[ K_{\mathrm{fragments}}=K_f-K_n \] \[ K_{\mathrm{fragments}}=170-20 \] \[ K_{\mathrm{fragments}}=150\,\text{MeV} \] \(\textbf{Check the complete energy budget:}\) The total outgoing energy beyond the final rest masses is: \[ 150+20+12=182\,\text{MeV} \] This equals the sum of reaction energy and initial kinetic energy. \(\textbf{Physical interpretation:}\) A fission reaction's positive \(Q\)-value does not have to appear entirely as kinetic energy of the heavy fragments. Some energy may be carried by free neutrons and radiation. The incident projectile's initial kinetic energy must also be included in the complete energy balance. The result is the combined kinetic energy of the fragments, not the kinetic energy of each fragment separately. \(\textbf{Final answer:}\) \[ \boxed{K_{\mathrm{fragments}}=150\,\text{MeV}} \]

Explanation: \(\textbf{Fix the conserved initial nucleon count:}\) The initial system contains uranium-235 and one neutron. Therefore: \[ A_i=235+1 \] \[ A_i=236 \] \(\textbf{Check the product charge numbers:}\) The xenon fragment has atomic number 54. The strontium fragment has atomic number 38. Thus: \[ 54+38=92 \] The final nuclear charge equals the original uranium nuclear charge. \(\textbf{Determine event P's unknown fragment mass number:}\) Let the strontium mass number in P be \(A_P\). Two neutrons are emitted. Therefore: \[ 236=140+A_P+2 \] \[ A_P=236-142 \] \[ A_P=94 \] The second fragment is: \[ {}^{94}_{38}\mathrm{Sr} \] \(\textbf{Determine event Q's unknown fragment mass number:}\) Let the strontium mass number be \(A_Q\). Three neutrons are emitted: \[ 236=140+A_Q+3 \] \[ A_Q=236-143 \] \[ A_Q=93 \] The second fragment is: \[ {}^{93}_{38}\mathrm{Sr} \] \(\textbf{Interpret the changed neutron emission:}\) Both events begin with the same total number of nucleons. Both also produce a xenon fragment having the same mass number. Consequently, emitting one additional free neutron leaves one fewer nucleon for the accompanying strontium fragment. The element remains strontium because its proton number is unchanged. The isotope changes because its neutron number differs. \(\textbf{Final answer:}\) \[ \boxed{P:\ {}^{94}_{38}\mathrm{Sr},\qquad Q:\ {}^{93}_{38}\mathrm{Sr}} \]

Explanation: \(\textbf{Identify the effective continuation rule:}\) Each fission releases three neutrons. However, only two neutrons from each event initiate further fissions. Therefore, each event produces exactly two fission events in the next generation. The third neutron does not contribute to the continuation of the chain. \(\textbf{Count generation 0:}\) The chain begins with one event: \[ N_0=1 \] \(\textbf{Count generation 1:}\) Each generation-0 event causes two events: \[ N_1=2N_0 \] \[ N_1=2 \] \(\textbf{Count generation 2:}\) \[ N_2=2N_1 \] \[ N_2=2(2) \] \[ N_2=4 \] \(\textbf{Count generation 3:}\) \[ N_3=2N_2 \] \[ N_3=2(4) \] \[ N_3=8 \] \(\textbf{Calculate the cumulative fission count:}\) \[ N_{\mathrm{total}}=N_0+N_1+N_2+N_3 \] \[ N_{\mathrm{total}}=1+2+4+8 \] \[ N_{\mathrm{total}}=15 \] \(\textbf{Calculate the total energy release:}\) The energy released per event is: \[ Q_f=200\,\text{MeV} \] Therefore: \[ E_{\mathrm{total}}=N_{\mathrm{total}}Q_f \] \[ E_{\mathrm{total}}=15(200) \] \[ E_{\mathrm{total}}=3000\,\text{MeV} \] \(\textbf{Interpret the finite-generation model:}\) The calculation counts four generations, numbered 0 through 3. It does not assume that every emitted neutron causes fission. The three-neutron emission count and the two-neutron effective continuation count have different roles. The model is an idealised description of the multiplication of fission events over a finite number of generations. The calculation does not establish that an actual sample can sustain unlimited multiplication. \(\textbf{Final answer:}\) \[ \boxed{N_3=8,\qquad N_{\mathrm{total}}=15,\qquad E_{\mathrm{total}}=3000\,\text{MeV}} \]

Explanation: The Assertion is true because the emission of free neutrons makes further fission events possible but does not guarantee that they occur. A continuing chain reaction requires neutrons from one generation to initiate suitable reactions in subsequent nuclei. The observation of three emitted neutrons establishes the neutron yield of one event. It does not establish how those neutrons interact after leaving the original fission fragments. Some neutrons may escape the material without encountering another suitable nucleus. Other neutrons may be absorbed in processes that do not produce fission. Consequently, the effective number of subsequent fission events can be smaller than the emitted neutron count. The Reason correctly identifies these possible outcomes. A material may therefore contain nuclei capable of fission without automatically maintaining a chain reaction. The reaction energy of an individual fission event is also distinct from the conditions required for continuation of successive events. A favourable nuclear energy balance does not, by itself, determine the probability that every emitted neutron causes another fission. Thus, both Assertion and Reason are true, and the Reason correctly explains the Assertion.

Explanation: \(\textbf{Identify the conserved initial configuration:}\) Both channels begin with uranium-235 and one free neutron. Therefore: \[ A_i=236 \] The initial bound nucleus has binding energy: \[ B_i=1780\,\text{MeV} \] The incident neutron contributes zero nuclear binding energy. \(\textbf{Check the final nucleon balance:}\) For channel P: \[ 233+3=236 \] For channel Q: \[ 234+2=236 \] Therefore, both final systems contain the correct total number of nucleons. The proton-number balance is specified to hold. \(\textbf{Calculate channel P's reaction energy:}\) The final bound fragments have combined binding energy: \[ B_{f,P}=1930\,\text{MeV} \] The emitted neutrons are free. Therefore: \[ Q_P=B_{f,P}-B_i \] \[ Q_P=1930-1780 \] \[ Q_P=+150\,\text{MeV} \] \(\textbf{Calculate channel Q's reaction energy:}\) For Q: \[ B_{f,Q}=1915\,\text{MeV} \] Therefore: \[ Q_Q=1915-1780 \] \[ Q_Q=+135\,\text{MeV} \] \(\textbf{Compare the two channels:}\) \[ Q_P-Q_Q=150-135 \] \[ Q_P-Q_Q=15\,\text{MeV} \] Thus, channel P releases \(15\,\text{MeV}\) more. \(\textbf{Interpret the neutron-count difference:}\) The free neutrons contribute to nucleon-number conservation. Their rest masses are already accounted for through the common free-nucleon reference. The number of emitted neutrons alone does not determine the fission reaction energy. The combined binding energies of the corresponding fragments must also be known. The analyst's conclusion is therefore unsupported by neutron count and contradicted by the supplied binding-energy data. \(\textbf{Final answer:}\) \[ \boxed{Q_P=+150\,\text{MeV},\qquad Q_Q=+135\,\text{MeV}} \]

Explanation: Statement I is incorrect because the energy balance of a fission reaction depends on the entire change from the initial nuclear configuration to the final configuration. Although separating a bound nucleus into free nucleons requires energy, fission does not ordinarily leave all nucleons free. Instead, it produces new bound nuclear fragments. The binding energy of those fragments can exceed the binding energy of the original heavy nucleus. Statement II is therefore correct. When the final combined binding energy is greater, the final rest energy is smaller and the difference can be released. Statement III is also correct. Nucleon number is conserved in the elementary nuclear fission reactions considered here. Any free neutrons emitted during the event are part of the final nucleon-number balance. Statement IV is incorrect because emitted free neutrons carry some of the parent's original nucleons. The mass numbers of the bound fragments therefore add to the initial total minus the number of free emitted neutrons. The reaction also conserves electric charge, which is carried by the charged nuclear fragments. The qualitative binding-energy curve helps explain why suitable heavy-nucleus splitting processes can release energy. It does not establish that every conceivable splitting channel is exothermic. Thus, Statements II and III are correct.

Explanation: \(\textbf{Recover the complete initial nucleon count:}\) The final products include xenon-139, strontium-94 and three neutrons. Therefore: \[ A_f=139+94+3 \] \[ A_f=236 \] The compound system immediately before fission must contain 236 nucleons. \(\textbf{Recover the nuclear charge:}\) The two heavy fragments have charge numbers 54 and 38. The free neutrons have charge number zero. Therefore: \[ Z_f=54+38+3(0) \] \[ Z_f=92 \] The compound nucleus has atomic number 92. \(\textbf{Infer the original target composition:}\) The compound nucleus forms after absorption of one neutron. Neutron absorption increases mass number by one but leaves atomic number unchanged. Thus: \[ A_{\mathrm{target}}=236-1 \] \[ A_{\mathrm{target}}=235 \] And: \[ Z_{\mathrm{target}}=92 \] \(\textbf{Identify the element:}\) Atomic number 92 corresponds to uranium. Therefore, the target is: \[ {}^{235}_{92}\mathrm U \] After neutron absorption, the compound nucleus is: \[ {}^{236}_{92}\mathrm U \] \(\textbf{Check the complete reaction:}\) \[ {}^{235}_{92}\mathrm U+{}^{1}_{0}\mathrm n \rightarrow{}^{139}_{54}\mathrm{Xe} +{}^{94}_{38}\mathrm{Sr} +3{}^{1}_{0}\mathrm n+Q \] The mass numbers balance: \[ 235+1=139+94+3 \] The charge numbers also balance: \[ 92=54+38 \] \(\textbf{Physical interpretation:}\) The detected products permit the unknown parent to be inferred by reversing the conservation equations. The absorbed neutron must be distinguished from the neutrons emitted after fission. Although the compound nucleus contains 236 nucleons, the original target contains only 235. The element is identified from proton number, not mass number. \(\textbf{Final answer:}\) \[ \text{Target }{}^{235}_{92}\mathrm U,\qquad\text{compound nucleus }{}^{236}_{92}\mathrm U \]

Explanation: \(\textbf{Convert energy released per fission:}\) Each event releases: \[ Q_f=200\,\text{MeV} \] Using the supplied energy conversion: \[ Q_f=200(1.60\times10^{-13}) \] \[ Q_f=3.20\times10^{-11}\,\text{J} \] \(\textbf{Apply the measured event count:}\) The number of fissions is: \[ N=5.0\times10^{12} \] Therefore: \[ E_{\mathrm{total}}=NQ_f \] Substituting: \[ E_{\mathrm{total}}=(5.0\times10^{12})(3.20\times10^{-11}) \] \[ E_{\mathrm{total}}=160\,\text{J} \] \(\textbf{Calculate the detector's measured energy:}\) The detector records: \[ 30\% \] of the released energy. Therefore: \[ E_{\mathrm{detected}}=0.30E_{\mathrm{total}} \] \[ E_{\mathrm{detected}}=0.30(160) \] \[ E_{\mathrm{detected}}=48\,\text{J} \] \(\textbf{Determine the unrecorded portion:}\) The remaining energy is: \[ E_{\mathrm{other}}=160-48 \] \[ E_{\mathrm{other}}=112\,\text{J} \] \(\textbf{Check the complete energy partition:}\) \[ E_{\mathrm{detected}}+E_{\mathrm{other}}=48+112 \] \[ E_{\mathrm{detected}}+E_{\mathrm{other}}=160\,\text{J} \] \(\textbf{Physical interpretation:}\) Energy per fission must be multiplied by the total number of actual fission events. The detector's response fraction is then applied to the aggregate nuclear energy. The fact that only part of the released energy is recorded does not reduce the reaction energy of an individual fission. The remaining energy is still part of the physical energy budget. The total energy released, detected energy and energy per nuclear event describe distinct quantities. \(\textbf{Final answer:}\) \[ \boxed{E_{\mathrm{total}}=160\,\text{J},\qquad E_{\mathrm{detected}}=48\,\text{J}} \]

Explanation: \(\textbf{Establish the conservation conditions:}\) The initial compound nucleus contains 236 nucleons and 92 protons. Every valid final configuration must therefore satisfy: \[ \sum A_f=236 \] \[ \sum Z_f=92 \] Free neutrons contribute to mass number but not electric charge. \(\textbf{Test configuration P:}\) The mass-number total is: \[ 140+93+3=236 \] The charge-number total is: \[ 54+38=92 \] Therefore, P satisfies both conservation conditions. \(\textbf{Test configuration Q:}\) The mass-number total is: \[ 144+90+2=236 \] The charge-number total is: \[ 56+36=92 \] Therefore, Q is also balanced. \(\textbf{Test configuration R:}\) The mass-number total is: \[ 140+94+3=237 \] The charge-number total is: \[ 54+38=92 \] Although electric charge is conserved, the final nucleon count exceeds the initial count by one. Therefore, R is inconsistent with the stated reaction. \(\textbf{Test configuration S:}\) The mass-number total is: \[ 144+89+3=236 \] The charge-number total is: \[ 56+36=92 \] Therefore, S satisfies both conditions. \(\textbf{Physical interpretation:}\) A valid elementary nuclear reaction must conserve both mass number and electric charge. Satisfying only one conservation law is insufficient. Different fragment pairs and neutron counts may be compatible with the same initial compound nucleus. The conservation equations determine whether a proposed channel is balanced, but do not independently guarantee that it will occur or specify its probability. Thus, configurations P, Q and S are balanced, while R is not.

Explanation: \(\textbf{Distinguish neutron production from effective continuation:}\) Each fission emits three neutrons. However, only neutrons that actually initiate another fission contribute to the next generation's event count. The effective continuation count differs among the three surroundings. \(\textbf{Analyse surroundings P:}\) The chain starts with: \[ N_0=1 \] No emitted neutron causes another fission. Therefore: \[ N_1=0 \] With no first-generation events, there are no later fissions produced through the specified chain. Thus: \[ P:\quad1,0,0,0 \] \(\textbf{Analyse surroundings Q:}\) Exactly one neutron from each fission produces a further event. Therefore, every event is replaced by one event in the next generation. The sequence is: \[ Q:\quad1,1,1,1 \] The event count remains constant from generation to generation under the idealised rule. \(\textbf{Analyse surroundings R:}\) Exactly two neutrons from every fission produce further events. Therefore: \[ N_{j+1}=2N_j \] Starting from: \[ N_0=1 \] successive generations contain: \[ N_1=2 \] \[ N_2=4 \] \[ N_3=8 \] Hence: \[ R:\quad1,2,4,8 \] \(\textbf{Physical interpretation:}\) All three surroundings have the same neutron yield per individual fission. Nevertheless, their chain behaviours differ because the emitted neutrons have different subsequent outcomes. In P the chain stops, in Q the event count remains constant, and in R the event count grows through the specified generations. Neutron emission is therefore a necessary enabling feature of these neutron-induced chains but does not determine their continuation by itself. The finite sequences are idealised and do not imply that any physical material can sustain unlimited multiplication.

Explanation: \(\textbf{Recover the complete final rest mass:}\) The reaction energy is: \[ Q=(M_i-M_f)c^2 \] Therefore: \[ M_f=M_i-\frac{Q}{c^2} \] \(\textbf{Convert reaction energy into a mass difference:}\) \[ \Delta M=\frac{180}{931.5}u \] \[ \Delta M\approx0.193237u \] \(\textbf{Determine the final mass sum:}\) \[ M_f=236.050000u-0.193237u \] \[ M_f\approx235.856763u \] This includes both nuclear fragments and all three emitted neutrons. \(\textbf{Calculate the emitted neutrons' combined mass:}\) \[ M_{3n}=3m_n \] \[ M_{3n}=3(1.008665u) \] \[ M_{3n}=3.025995u \] \(\textbf{Determine the bound-fragment mass sum:}\) Let \(M_{\mathrm{fragments}}\) denote the sum of the two bound nuclear masses. Then: \[ M_f=M_{\mathrm{fragments}}+M_{3n} \] Therefore: \[ M_{\mathrm{fragments}}=M_f-M_{3n} \] Substituting: \[ M_{\mathrm{fragments}}=235.856763-3.025995 \] \[ M_{\mathrm{fragments}}\approx232.830768u \] \[ M_{\mathrm{fragments}}\approx232.8308u \] \(\textbf{Interpret the result:}\) The measured reaction energy determines the rest-mass sum of every final product. However, the problem asks only for the combined mass of the two bound fragments. The masses of the three free neutrons must therefore be subtracted from the complete final mass. The positive reaction energy means that the final complete rest-mass sum is smaller than the initial rest-mass sum. It does not mean that the emitted neutrons have zero rest mass. Free neutrons have zero nuclear binding energy relative to the free-nucleon reference, but each still possesses its ordinary rest mass. \(\textbf{Final answer:}\) \[ \boxed{M_{\mathrm{fragments}}\approx232.8308u} \]