Atoms and Nuclei MCQs with Answers – Part 3 (NEET Physics)

Explanation: \(\textbf{Identify the mass convention:}\) The carbon-12 mass definition refers to the complete neutral atom. A neutral carbon atom contains six extranuclear electrons because its atomic number is six. The bare nucleus does not include these electrons. \(\textbf{Stage 1 — Calculate the total electron rest mass:}\) The number of electrons is: \[ N_e=6 \] The supplied electron mass is: \[ m_e=0.000549u \] Therefore: \[ M_e=6m_e \] \[ M_e=6(0.000549u) \] \[ M_e=0.003294u \] \(\textbf{Stage 2 — Estimate nuclear mass:}\) Neglecting electronic binding-energy corrections: \[ M_{\text{atom}}\approx M_{\text{nucleus}}+M_e \] Rearranging: \[ M_{\text{nucleus}}\approx M_{\text{atom}}-M_e \] Substituting: \[ M_{\text{nucleus}}\approx12u-0.003294u \] \[ M_{\text{nucleus}}\approx11.996706u \] \(\textbf{Stage 3 — Calculate the specified percentage:}\) The requested relative difference uses neutral atomic mass as the denominator. \[ p=\frac{M_{\text{atom}}-M_{\text{nucleus}}}{M_{\text{atom}}}\times100\% \] Substituting: \[ p=\frac{0.003294u}{12u}\times100\% \] \[ p=0.0002745\times100\% \] \[ p=0.02745\% \] \(\textbf{Interpret the approximation:}\) The percentage is small because electron masses are much smaller than nucleon masses. However, it is not zero. Thus, the neutral carbon-12 atom has mass exactly \(12u\) by definition, while its bare nucleus does not have mass exactly \(12u\). The calculation concerns the removal of the extranuclear electron mass contribution. It does not assume that each of the twelve nucleons has mass exactly \(1u\). \(\textbf{Final answer:}\) \[ \boxed{(M_{\text{nucleus}},p)\approx(11.996706u,0.02745\%)} \]

Explanation: Both nuclei are isotopes, so they have identical atomic numbers and identical positive nuclear charges. Their mass numbers differ because their neutron counts differ. From the empirical radius law, the radius ratio equals the cube root of the mass-number ratio. Since the mass numbers are 125 and 64, their cube roots are five and four, respectively. Therefore, Q has a radius \(5/4\) times that of P, making Statement I correct. Nuclear volume varies as the cube of radius and therefore directly as mass number under the stated law. Consequently, the volume of Q is \(125/64\) times the volume of P. Average positive nuclear charge density is nuclear charge divided by nuclear volume. The nuclear charges are equal, but Q has the larger volume. Its average positive charge density must therefore be smaller, making Statement II incorrect. Specifically, Q's charge density is \(64/125\) times that of P, so Statement III is correct. Statement IV is also correct because neutral atoms of the same element have the same electron count. The approximately constant nuclear matter density does not imply constant positive charge density across isotopes. Neutrons contribute to nuclear mass and volume without adding positive electric charge. Thus, Statements I, III and IV are correct.

Explanation: Graph P plots nuclear radius against the cube root of mass number. The empirical radius law directly states that \(R\) is proportional to \(A^{1/3}\). Therefore, Graph P should be a straight line through the origin with slope \(R_0\). It matches description 2. Graph Q plots mean nuclear matter density against mass number. Nuclear mass is approximately proportional to \(A\). The spherical nuclear volume is proportional to \(R^3\), which is also proportional to \(A\) under the radius law. Hence, their ratio is approximately independent of mass number. Graph Q should therefore be approximately horizontal, matching description 1. Graph R plots nuclear volume against mass number. Substituting \(R=R_0A^{1/3}\) into the spherical-volume formula gives \(V=(4/3)\pi R_0^3A\). Therefore, Graph R is a straight line through the origin with slope \((4/3)\pi R_0^3\). It matches description 3. Both P and R are increasing straight lines, but their horizontal variables and slopes are different. A radius law that is linear in \(A^{1/3}\) produces a volume law linear in \(A\). The approximate constancy of nuclear density is a consequence of these relations, not an independent statement that all nuclei have equal volumes.

Explanation: \(\textbf{Identify the isotope fractions:}\) The lighter-isotope abundance is: \[ f_L=0.70 \] The heavier-isotope abundance is: \[ f_H=0.30 \] \(\textbf{Stage 1 — Calculate the original average neutral atomic mass:}\) \[ \overline M_A=f_LM_L+f_HM_H \] Substituting: \[ \overline M_A=0.70(39.960u)+0.30(41.960u) \] \[ \overline M_A=27.972u+12.588u \] \[ \overline M_A=40.560u \] \(\textbf{Stage 2 — Apply the double-ionisation mass correction:}\) Every atom loses two electrons. Therefore, the mass removed from each atom is approximately: \[ \Delta m_e=2m_e \] \[ \Delta m_e=2(0.00055u) \] \[ \Delta m_e=0.00110u \] Because every atom loses the same number of electrons, the average ionic mass decreases by this same amount. \[ \overline M_{2+}=\overline M_A-2m_e \] \[ \overline M_{2+}=40.560u-0.00110u \] \[ \overline M_{2+}=40.55890u \] \(\textbf{Stage 3 — Calculate the neutron number of each isotope:}\) Both isotopes have atomic number: \[ Z=20 \] For the lighter isotope: \[ N_L=40-20 \] \[ N_L=20 \] For the heavier isotope: \[ N_H=42-20 \] \[ N_H=22 \] \(\textbf{Stage 4 — Calculate average neutron number:}\) \[ \overline N=f_LN_L+f_HN_H \] \[ \overline N=0.70(20)+0.30(22) \] \[ \overline N=14.0+6.6 \] \[ \overline N=20.6 \] \(\textbf{Physical interpretation:}\) Double ionisation changes the electron count and measured ionic mass but does not change nuclear neutron numbers. Because both isotopes belong to the same element, they lose the same number of electrons when converted to the specified charge state. The average ionic mass can therefore be found by correcting the average neutral atomic mass. The average neutron number remains a property of the isotope mixture and is unaffected by electron removal. \(\textbf{Final answer:}\) \[ \boxed{(\overline M_{2+},\overline N)=(40.55890u,20.6)} \]

Explanation: The radius measurements can be tested against the empirical relation \(R=R_0A^{1/3}\). For nucleus P, the cube root of mass number 27 is three, giving an inferred \(R_0\) of \(3.6/3=1.2\,\text{fm}\). For Q, the cube root of mass number 64 is four, giving \(R_0=4.8/4=1.2\,\text{fm}\). For R, the cube root of mass number 125 is five, giving \(R_0=6.0/5=1.2\,\text{fm}\). Therefore, all three measurements support a common nuclear-radius parameter. The associated nuclear volumes are proportional to their mass numbers. If nuclear masses are approximately proportional to mass numbers, this also supports approximately equal nuclear matter densities. However, positive nuclear charge density is a different quantity. It depends on the number of protons divided by nuclear volume. The atomic numbers of P, Q and R have not been supplied. Therefore, their nuclear positive charges are unknown. Equal matter densities do not guarantee equal positive charge densities because the proton fractions \(Z/A\) may differ. The analyst's first conclusion is supported by the data, whereas the second requires additional information about nuclear composition.

Explanation: \(\textbf{Identify the incident particles:}\) The beryllium nucleus has: \[ A_{\mathrm{Be}}=9,\qquad Z_{\mathrm{Be}}=4 \] The alpha particle has: \[ A_\alpha=4,\qquad Z_\alpha=2 \] \(\textbf{Stage 1 — Apply conservation of electric charge:}\) The total initial nuclear charge number is: \[ Z_{\text{initial}}=4+2 \] \[ Z_{\text{initial}}=6 \] The carbon product has atomic number six. Therefore, the remaining particle must have zero electric charge: \[ Z_X=6-6 \] \[ Z_X=0 \] \(\textbf{Stage 2 — Apply conservation of nucleon number:}\) The total initial mass number is: \[ A_{\text{initial}}=9+4 \] \[ A_{\text{initial}}=13 \] A neutron has mass number one. Therefore, the carbon product must have: \[ A_C=13-1 \] \[ A_C=12 \] \(\textbf{Stage 3 — Write the balanced nuclear reaction:}\) \[ {}^{9}_{4}\mathrm{Be}+{}^{4}_{2}\mathrm{He} \rightarrow{}^{12}_{6}\mathrm{C}+{}^{1}_{0}\mathrm{n} \] \(\textbf{Check both conserved quantities:}\) For nucleon number: \[ 9+4=12+1 \] For nuclear charge number: \[ 4+2=6+0 \] Both conditions are satisfied. \(\textbf{Physical interpretation:}\) The emitted neutral particle must carry one nucleon without contributing positive electric charge. A neutron satisfies these properties. A photon carries energy and momentum but does not have a nucleon number of one. A proton carries positive charge and therefore cannot be the neutral particle specified in the experiment. The conservation analysis is consistent with the interpretation of the penetrating radiation as neutrons. \(\textbf{Final answer:}\) \[ {}^{12}_{6}\mathrm{C}\ \text{and}\ {}^{1}_{0}\mathrm{n} \]

Explanation: The Assertion is true because the absence of electric deflection establishes only that the radiation has no detectable net electric charge. An electrically neutral neutron would not undergo direct electric-force deflection. However, an electrically neutral photon would also not be deflected by an ordinary electrostatic field in this experiment. Therefore, neutrality alone cannot distinguish neutrons from electromagnetic radiation. The Reason is true because neutrons have nonzero rest mass comparable to the proton mass, whereas photons have zero rest mass. Their energy and momentum transfers to target particles must consequently be analysed using the appropriate physical relations. The observation of energetic protons emerging from paraffin provides additional information about momentum and energy transfer. Chadwick's interpretation showed that the observed recoil behaviour was consistent with collisions involving massive neutral particles. It was not necessary to imagine that the emitted radiation had acquired a positive charge merely because it produced proton recoil. The proton's charge belongs to the target particle knocked out of the material. The Reason therefore explains why an electric-field observation must be supplemented by collision evidence. Both statements correctly distinguish particle neutrality from particle identification.

Explanation: \(\textbf{Identify the invariant nuclear quantity:}\) Isotones have equal neutron numbers. The nuclear composition relation is: \[ A=Z+N \] For a family with fixed neutron number: \[ A=Z+\text{constant} \] This predicts a straight-line graph of mass number against atomic number. \(\textbf{Stage 1 — Determine the graph slope:}\) Using the two plotted points: \[ m=\frac{\Delta A}{\Delta Z} \] \[ m=\frac{47-40}{25-18} \] \[ m=\frac{7}{7} \] \[ m=1 \] \(\textbf{Stage 2 — Determine the intercept:}\) The graph equation is: \[ A=Z+N \] Therefore, the intercept on the mass-number axis is the common neutron number. Using the first point: \[ N=A-Z \] \[ N=40-18 \] \[ N=22 \] Thus: \[ A=Z+22 \] The intercept is 22. \(\textbf{Stage 3 — Determine the ion's electron count:}\) The second point represents a nucleus having: \[ Z=25 \] Its neutral atom contains 25 electrons. A triply positive ion has lost three electrons. Therefore: \[ N_e=25-3 \] \[ N_e=22 \] \(\textbf{Physical interpretation:}\) The graph slope reflects the one-to-one increase in mass number when proton number increases while neutron number remains constant. The intercept identifies the common neutron population. The electron count, however, is determined by atomic number and ionic charge. Its numerical equality with the neutron number in this particular ion is coincidental, not a general property of isotones. \(\textbf{Final answer:}\) \[ \boxed{(m,\text{intercept},N_e)=(1,22,22)} \]

Explanation: \(\textbf{Distinguish neutron fraction from atomic abundance:}\) The given fraction refers to the total number of neutrons, not the total number of atoms. An atom of the heavier isotope contributes more neutrons than an atom of the lighter isotope. Therefore, the neutron fraction cannot be used directly as the atomic abundance. \(\textbf{Stage 1 — Determine the neutron counts:}\) For neon-20: \[ N_{20}=20-10 \] \[ N_{20}=10 \] For neon-22: \[ N_{22}=22-10 \] \[ N_{22}=12 \] \(\textbf{Stage 2 — Construct the neutron-population relation:}\) Let \(x\) be the fraction of neon-22 atoms. Then the neon-20 fraction is \(1-x\). The fraction of all neutrons belonging to neon-22 nuclei is: \[ f_n=\frac{12x}{10(1-x)+12x} \] The measured value is: \[ \frac{12x}{10(1-x)+12x}=\frac{12}{17} \] Simplifying: \[ \frac{12x}{10+2x}=\frac{12}{17} \] Cross-multiplying: \[ 204x=120+24x \] \[ 180x=120 \] \[ x=\frac{2}{3} \] \(\textbf{Stage 3 — Calculate average atomic mass:}\) \[ \overline m=(1-x)m_{20}+xm_{22} \] Substituting: \[ \overline m=\frac{1}{3}(19.99u)+\frac{2}{3}(21.99u) \] \[ \overline m=\frac{19.99+43.98}{3}u \] \[ \overline m=\frac{63.97}{3}u \] \[ \overline m\approx21.323u \] \(\textbf{Consistency check:}\) Two-thirds of the atoms are neon-22, but they account for a larger fraction of the neutrons. This is expected because each neon-22 nucleus contains twelve neutrons rather than ten. The average atomic mass is closer to the heavier isotopic mass, consistent with its greater abundance. \(\textbf{Final answer:}\) \[ \boxed{\left(f_{22},\overline m\right)=\left(\frac{2}{3},21.32u\right)} \]

Explanation: \(\textbf{Infer the proton number of P:}\) A singly negative ion contains one electron more than the corresponding neutral atom. Therefore: \[ N_{e,P}=Z_P+1 \] Given: \[ 27=Z_P+1 \] Hence: \[ Z_P=26 \] \(\textbf{Determine P's mass number:}\) The neutron number of P is 32. Therefore: \[ A_P=Z_P+N_P \] \[ A_P=26+32 \] \[ A_P=58 \] \(\textbf{Infer the proton number of Q:}\) A doubly positive ion has lost two electrons. Therefore: \[ N_{e,Q}=Z_Q-2 \] Given: \[ 28=Z_Q-2 \] Hence: \[ Z_Q=30 \] \(\textbf{Apply the isobar condition:}\) Since Q is an isobar of P: \[ A_Q=A_P \] \[ A_Q=58 \] Its neutron number is: \[ N_Q=A_Q-Z_Q \] \[ N_Q=58-30 \] \[ N_Q=28 \] \(\textbf{Classify their nuclear relationship:}\) The nuclei have equal mass numbers, making them isobars. Their atomic numbers differ, so they are not isotopes. Their neutron numbers also differ, so they are not isotones. \(\textbf{Physical interpretation:}\) The ionic electron counts must first be corrected for the different charge states. Only then can the nuclear proton numbers be determined. The equal mass numbers do not require equal proton or neutron numbers. The result demonstrates that information about extranuclear electrons can be combined with nuclear composition constraints without confusing ionic charge with nuclear charge. \(\textbf{Final answer:}\) \[ \boxed{(A_P,A_Q,N_Q)=(58,58,28)} \] The nuclei are isobars only.

Explanation: \(\textbf{Identify the relevant atomic relationship:}\) The atoms are isotopes of the same element. Therefore, their atomic numbers are identical. Each neutral atom consequently contains the same number of electrons. \(\textbf{Stage 1 — Express atomic and nuclear masses:}\) Neglecting the stated electronic binding-energy corrections: \[ M_{A1}\approx M_{N1}+Zm_e \] \[ M_{A2}\approx M_{N2}+Zm_e \] Here \(Z\) is the common atomic number. \(\textbf{Stage 2 — Subtract the two mass equations:}\) \[ M_{A2}-M_{A1} =(M_{N2}+Zm_e)-(M_{N1}+Zm_e) \] The equal electron-mass terms cancel. Therefore: \[ M_{A2}-M_{A1}\approx M_{N2}-M_{N1} \] \(\textbf{Stage 3 — Substitute measured atomic masses:}\) \[ M_{N2}-M_{N1}\approx41.9595u-39.9620u \] \[ M_{N2}-M_{N1}\approx1.9975u \] \(\textbf{Interpret the result:}\) The nuclear mass difference is approximately equal to the measured neutral atomic mass difference. It is unnecessary to subtract the full electron mass contribution from both atoms before comparing them because those equal contributions cancel automatically. However, this cancellation does not occur in the same form when comparing neutral atoms of different atomic numbers. The conclusion also does not imply that the mass difference equals exactly two atomic mass units. Actual nuclear and atomic masses are measured quantities, not merely integer mass-number labels. \(\textbf{Final answer:}\) \[ \boxed{M_{N2}-M_{N1}\approx1.9975u} \]

Explanation: \(\textbf{Connect nuclear size with measured density:}\) The approximate nuclear mass can be inferred from: \[ M=\rho V \] The measured radius determines the volume when the nucleus is treated as a sphere. \(\textbf{Stage 1 — Convert nuclear radius into SI units:}\) \[ R=6.0\,\text{fm} \] \[ R=6.0\times10^{-15}\,\text{m} \] \(\textbf{Stage 2 — Calculate spherical nuclear volume:}\) \[ V=\frac{4}{3}\pi R^3 \] Substituting: \[ V=\frac{4}{3}(3.14)(6.0\times10^{-15})^3 \] First evaluate the radius cube: \[ R^3=216\times10^{-45}\,\text{m}^3 \] \[ R^3=2.16\times10^{-43}\,\text{m}^3 \] Therefore: \[ V\approx4.187(2.16\times10^{-43}) \] \[ V\approx9.04\times10^{-43}\,\text{m}^3 \] \(\textbf{Stage 3 — Determine the nuclear mass:}\) \[ M=\rho V \] \[ M=(2.3\times10^{17})(9.04\times10^{-43})\,\text{kg} \] \[ M\approx2.079\times10^{-25}\,\text{kg} \] \[ M\approx2.08\times10^{-25}\,\text{kg} \] \(\textbf{Stage 4 — Determine the average mass per nucleon:}\) The nucleus contains 125 nucleons. Therefore: \[ \overline m_N=\frac{M}{A} \] \[ \overline m_N=\frac{2.079\times10^{-25}}{125}\,\text{kg} \] \[ \overline m_N\approx1.66\times10^{-27}\,\text{kg} \] \(\textbf{Physical check:}\) The inferred average mass per nucleon is close to the usual atomic mass unit expressed in kilograms. This is consistent with the nuclear mass being approximately proportional to nucleon number. The calculation is an estimate because nuclear radius, average density and spherical shape are approximate physical descriptions. The agreement does not establish that all individual nucleons have exactly the same mass. \(\textbf{Final answer:}\) \[ \boxed{(M,\overline m_N)\approx(2.08\times10^{-25},1.66\times10^{-27})\,\text{kg}} \]

Explanation: \(\textbf{Interpret the measurement correctly:}\) The experiment gives an allowed interval for nuclear radius rather than a single exact value. A candidate is acceptable only if its predicted radius lies inside that interval. \(\textbf{Stage 1 — Calculate the candidate radii:}\) The empirical relation is: \[ R=1.2A^{1/3}\,\text{fm} \] For \(A=27\): \[ R_{27}=1.2(27)^{1/3} \] \[ R_{27}=1.2(3)=3.6\,\text{fm} \] This is below the allowed range. For \(A=64\): \[ R_{64}=1.2(64)^{1/3} \] \[ R_{64}=1.2(4)=4.8\,\text{fm} \] This lies inside the allowed range. For \(A=81\): \[ R_{81}=1.2(81)^{1/3} \] \[ R_{81}\approx1.2(4.327) \] \[ R_{81}\approx5.19\,\text{fm} \] This exceeds the maximum allowed radius of \(5.1\,\text{fm}\). For \(A=125\): \[ R_{125}=1.2(125)^{1/3} \] \[ R_{125}=1.2(5)=6.0\,\text{fm} \] This is also outside the interval. \(\textbf{Stage 2 — Determine the compatible mass number:}\) Only the candidate having: \[ A=64 \] is consistent with the radius measurement. \(\textbf{Stage 3 — Determine its neutron number:}\) The atomic number is: \[ Z=30 \] Therefore: \[ N=A-Z \] \[ N=64-30 \] \[ N=34 \] \(\textbf{Physical interpretation:}\) An approximate radius measurement can distinguish among candidate nuclear mass numbers when their predicted sizes are sufficiently separated. The experimental interval must be respected rather than replacing it by its midpoint alone. Once a compatible mass number is identified, the known atomic number determines the neutron number. The inference depends on accepting the specified empirical radius relation within the precision of the experiment. \(\textbf{Final answer:}\) \[ \boxed{(A,Z,N)=(64,30,34)} \]

Explanation: \(\textbf{Identify the fixed and changing quantities:}\) Both atoms have the same effective atomic radius. Their nuclear mass numbers differ. Therefore, the relevant change in each geometric fraction arises from the nuclear radius. \(\textbf{Stage 1 — Compare nuclear radii:}\) The empirical radius law gives: \[ R=R_0A^{1/3} \] Hence: \[ \frac{R_Q}{R_P}=\left(\frac{A_Q}{A_P}\right)^{1/3} \] Substituting: \[ \frac{R_Q}{R_P}=\left(\frac{216}{27}\right)^{1/3} \] \[ \frac{R_Q}{R_P}=8^{1/3} \] \[ \frac{R_Q}{R_P}=2 \] \(\textbf{Stage 2 — Compare geometric area fractions:}\) Since both atomic radii are equal: \[ \frac{f_{S,Q}}{f_{S,P}}=\left(\frac{R_Q}{R_P}\right)^2 \] Therefore: \[ \frac{f_{S,Q}}{f_{S,P}}=2^2 \] \[ \frac{f_{S,Q}}{f_{S,P}}=4 \] \(\textbf{Stage 3 — Compare nuclear volume fractions:}\) Again using equal atomic radii: \[ \frac{f_{V,Q}}{f_{V,P}}=\left(\frac{R_Q}{R_P}\right)^3 \] Therefore: \[ \frac{f_{V,Q}}{f_{V,P}}=2^3 \] \[ \frac{f_{V,Q}}{f_{V,P}}=8 \] \(\textbf{Physical interpretation:}\) A twofold increase in nuclear radius produces a fourfold increase in geometric cross-sectional area and an eightfold increase in volume. These ratios describe geometric nuclear fractions inside the two idealised atoms. They do not directly give the probabilities of large-angle Rutherford scattering. Actual scattering probabilities depend on Coulomb interaction, incident energy, impact parameter and experimental conditions. The distinction between geometric size and electrostatic scattering behaviour is important when interpreting nuclear dimensions. \(\textbf{Final answer:}\) \[ \boxed{\left(\frac{f_{S,Q}}{f_{S,P}},\frac{f_{V,Q}}{f_{V,P}}\right)=(4,8)} \]

Explanation: \(\textbf{Distinguish proton and neutron densities:}\) Average proton number density is proton count divided by nuclear volume. Average neutron number density is neutron count divided by nuclear volume. The two isotopes have equal proton counts but different neutron counts and nuclear volumes. \(\textbf{Stage 1 — Compare nuclear volumes:}\) The empirical radius law gives: \[ R\propto A^{1/3} \] For spherical nuclei: \[ V\propto R^3 \] Therefore: \[ V\propto A \] Consequently: \[ \frac{V_Q}{V_P}=\frac{A_Q}{A_P} \] \[ \frac{V_Q}{V_P}=\frac{125}{64} \] \(\textbf{Stage 2 — Compare average proton number densities:}\) Both nuclei contain: \[ Z=20 \] protons. Therefore: \[ \rho_p=\frac{Z}{V} \] Taking the ratio: \[ \frac{\rho_{p,Q}}{\rho_{p,P}} =\frac{Z/V_Q}{Z/V_P} \] \[ \frac{\rho_{p,Q}}{\rho_{p,P}}=\frac{V_P}{V_Q} \] \[ \frac{\rho_{p,Q}}{\rho_{p,P}}=\frac{64}{125} \] \(\textbf{Stage 3 — Determine the neutron counts:}\) For P: \[ N_P=A_P-Z \] \[ N_P=64-20 \] \[ N_P=44 \] For Q: \[ N_Q=125-20 \] \[ N_Q=105 \] \(\textbf{Stage 4 — Compare neutron number densities:}\) \[ \rho_n=\frac{N}{V} \] Therefore: \[ \frac{\rho_{n,Q}}{\rho_{n,P}} =\frac{N_Q/V_Q}{N_P/V_P} \] \[ \frac{\rho_{n,Q}}{\rho_{n,P}} =\frac{N_Q}{N_P}\frac{V_P}{V_Q} \] Substituting: \[ \frac{\rho_{n,Q}}{\rho_{n,P}} =\frac{105}{44}\times\frac{64}{125} \] \[ \frac{\rho_{n,Q}}{\rho_{n,P}} =\frac{6720}{5500} \] \[ \frac{\rho_{n,Q}}{\rho_{n,P}}=\frac{336}{275} \] \(\textbf{Physical interpretation:}\) The heavier isotope has a larger nuclear volume. Its unchanged total proton count is distributed over that larger volume, so its average proton number density is smaller. Its neutron number, however, increases substantially. The increase in neutron count more than compensates for the increase in volume, producing a larger average neutron number density. The empirical law predicts approximately constant total nucleon density, not individually constant proton and neutron densities for every isotope. \(\textbf{Final answer:}\) \[ \boxed{\left(\frac{\rho_{p,Q}}{\rho_{p,P}},\frac{\rho_{n,Q}}{\rho_{n,P}}\right)=\left(\frac{64}{125},\frac{336}{275}\right)} \]

Explanation: \(\textbf{Distinguish equal masses from equal atom counts:}\) The laboratory mixes equal total masses, not equal numbers of atoms. Because the lighter isotope has less mass per atom, an equal-mass portion contains more lighter atoms. \(\textbf{Stage 1 — Express the atom counts:}\) Let the total mass of each pure sample be \(M\). The lighter isotope has atomic mass: \[ m_L=10.00u \] The heavier isotope has atomic mass: \[ m_H=11.00u \] Their atom counts are proportional to total sample mass divided by individual atomic mass. Therefore: \[ N_L=\frac{M}{10u} \] \[ N_H=\frac{M}{11u} \] \(\textbf{Stage 2 — Determine the atom-count ratio:}\) \[ \frac{N_L}{N_H} =\frac{M/(10u)}{M/(11u)} \] \[ \frac{N_L}{N_H}=\frac{11}{10} \] Thus: \[ N_L:N_H=11:10 \] \(\textbf{Stage 3 — Calculate the lighter-isotope atom fraction:}\) \[ f_L=\frac{N_L}{N_L+N_H} \] Using the atom-count ratio: \[ f_L=\frac{11}{11+10} \] \[ f_L=\frac{11}{21} \] The heavier-isotope fraction is: \[ f_H=\frac{10}{21} \] \(\textbf{Stage 4 — Determine neutron numbers:}\) The atomic number is: \[ Z=5 \] For the lighter isotope: \[ N_{n,L}=10-5=5 \] For the heavier isotope: \[ N_{n,H}=11-5=6 \] \(\textbf{Stage 5 — Calculate the average neutron number:}\) \[ \overline N_n=f_LN_{n,L}+f_HN_{n,H} \] \[ \overline N_n=\frac{11}{21}(5)+\frac{10}{21}(6) \] \[ \overline N_n=\frac{55+60}{21} \] \[ \overline N_n=\frac{115}{21} \] \(\textbf{Physical interpretation:}\) Equal sample masses do not imply equal isotopic abundances by number. The lighter isotope contributes more atoms because each of its atoms has less mass. The neutron average must therefore be weighted using the actual atom counts, not the equal sample-mass fractions. The resulting average is slightly smaller than 5.5 because lighter-isotope atoms are more numerous. \(\textbf{Final answer:}\) \[ \boxed{\left(f_L,\overline N_n\right)=\left(\frac{11}{21},\frac{115}{21}\right)} \]

Explanation: The removed particle is a neutron, which carries no electric charge. Removing it decreases the total nucleon count by one without decreasing the proton number. Therefore, the daughter nucleus has mass number \(A-1\) and atomic number \(Z\), making Statement I correct. According to the empirical radius relation, radius increases with the cube root of mass number. Since the daughter's mass number is smaller, its predicted radius is also smaller. Statement II is therefore correct. Under the same law, nuclear volume varies directly with mass number. Approximate nuclear mass also varies directly with nucleon number. Consequently, their ratio remains approximately constant, so Statement III is correct. Statement IV is incorrect because nuclear positive charge depends on proton count, not total nucleon count. No proton has been removed. The daughter nucleus therefore retains nuclear charge \(+Ze\). The process changes the isotope because neutron number changes while atomic number remains fixed. The stated radius and density conclusions describe the hypothetical daughter under the empirical model. They do not imply that removing a neutron from every real nucleus is energetically possible.

Explanation: \(\textbf{Relate nuclear volume to nucleon number:}\) For a spherical nucleus: \[ V=\frac{4}{3}\pi R^3 \] Using the empirical radius law: \[ R=R_0A^{1/3} \] we obtain: \[ V=\frac{4}{3}\pi R_0^3A \] \(\textbf{Stage 1 — Determine average volume per nucleon:}\) The volume per nucleon is: \[ v_N=\frac{V}{A} \] Therefore: \[ v_N=\frac{4}{3}\pi R_0^3 \] Substituting: \[ v_N=\frac{4}{3}(3.14)(1.2\,\text{fm})^3 \] First calculate: \[ (1.2)^3=1.728 \] Therefore: \[ v_N=\frac{4}{3}(3.14)(1.728)\,\text{fm}^3 \] \[ v_N\approx7.23\,\text{fm}^3 \] \(\textbf{Stage 2 — Calculate the hypothetical cube volume:}\) The cube has side: \[ \ell=10\,\text{fm} \] Its volume is: \[ V_{\text{cube}}=\ell^3 \] \[ V_{\text{cube}}=(10\,\text{fm})^3 \] \[ V_{\text{cube}}=1000\,\text{fm}^3 \] \(\textbf{Stage 3 — Estimate the corresponding nucleon count:}\) For uniform matter at this average density: \[ N_{\text{equivalent}}=\frac{V_{\text{cube}}}{v_N} \] Substituting: \[ N_{\text{equivalent}}=\frac{1000}{7.23} \] \[ N_{\text{equivalent}}\approx138 \] \(\textbf{Physical interpretation:}\) The result represents an equivalent average nucleon count in a hypothetical volume of uniformly distributed nuclear matter. It does not mean that individual nucleons are rigid spheres each occupying precisely \(7.23\,\text{fm}^3\). The estimate arises from the macroscopic density implied by the empirical nuclear-radius law. The average nucleon number density is approximately constant because nuclear volume increases in proportion to \(A\). Thus, the same average density can be applied to the stated hypothetical cube. \(\textbf{Final answer:}\) \[ \boxed{\left(v_N,N_{\text{equivalent}}\right)\approx(7.23\,\text{fm}^3,138)} \]

Explanation: \(\textbf{Distinguish nuclear masses from ionic masses:}\) Both neutral atoms have the same atomic number. Consequently, both initially contain twelve electrons. Their different mass numbers do not change the number of electrons in a neutral atom. \(\textbf{Stage 1 — Determine the nuclear mass difference:}\) For a neutral atom: \[ M_A\approx M_N+Zm_e \] The electron contributions are equal for both isotopes. Therefore, they cancel when the atomic masses are subtracted. \[ \Delta M_N\approx M_{A,H}-M_{A,L} \] Substituting: \[ \Delta M_N\approx25.98260u-23.98500u \] \[ \Delta M_N\approx1.99760u \] \(\textbf{Stage 2 — Determine the lighter ionic mass:}\) The lighter isotope forms a singly positive ion. It loses one electron. Therefore: \[ M_{L,+}\approx M_{A,L}-m_e \] \[ M_{L,+}\approx23.98500u-0.00055u \] \[ M_{L,+}\approx23.98445u \] \(\textbf{Stage 3 — Determine the heavier ionic mass:}\) The heavier isotope forms a triply positive ion. It loses three electrons. Therefore: \[ M_{H,3+}\approx M_{A,H}-3m_e \] \[ M_{H,3+}\approx25.98260u-3(0.00055u) \] \[ M_{H,3+}\approx25.98095u \] \(\textbf{Stage 4 — Compare the ionic masses:}\) \[ \Delta M_{\text{ions}}=M_{H,3+}-M_{L,+} \] \[ \Delta M_{\text{ions}}=25.98095u-23.98445u \] \[ \Delta M_{\text{ions}}=1.99650u \] \(\textbf{Independent check:}\) The heavier ion has lost two more electrons than the lighter ion. Therefore, its ionic mass difference must be smaller than the corresponding neutral atomic mass difference by: \[ 2m_e=0.00110u \] Indeed: \[ 1.99760u-0.00110u=1.99650u \] \(\textbf{Physical interpretation:}\) When comparing neutral isotopes of one element, equal electron contributions cancel from the nuclear mass difference. When comparing ions in different charge states, they do not cancel in the same way. The distinction is essential whenever the electron-mass correction is comparable to the precision of the measured mass difference. \(\textbf{Final answer:}\) \[ \boxed{(\Delta M_N,\Delta M_{\text{ions}})=(1.99760u,1.99650u)} \]

Explanation: Dataset P concerns total nuclear volume and total nucleon number. If nuclear mass increases approximately in proportion to mass number, while nuclear volume also increases in proportion to mass number, their ratio remains nearly constant. Therefore, Dataset P supports approximately constant average nuclear matter density, making Statement I correct. Dataset Q concerns the fraction of nucleons that are protons. Nuclear positive charge is contributed only by protons, whereas both protons and neutrons contribute to nucleon number and nuclear mass. If proton fraction decreases while total nucleon density stays approximately constant, average positive charge density can decrease. Therefore, Statement II is correct. Statement III is incorrect because the radius law and approximate matter-density constancy impose no requirement that proton and neutron counts be equal. Nuclei can have different proton-to-neutron ratios while satisfying the same approximate radius law. Statement IV is correct because the two datasets describe related but distinct physical properties. Total nucleon density measures nucleons per unit nuclear volume. Positive charge density measures proton charge per unit nuclear volume. Increasing neutron number while maintaining proton number can enlarge nuclear volume without proportionately increasing positive charge. Consequently, the observations are mutually compatible rather than contradictory. The distinction is especially important for interpreting isotope comparisons.