Atoms and Nuclei MCQs with Answers – Part 7 (NEET Physics)

Explanation: The positively charged alpha particle experiences electrostatic interaction with the positive charge concentrated in an atomic nucleus. The nuclear charge is determined by the number of protons in that nucleus. Removing extranuclear electrons changes the net electric charge of the atom but does not change its proton number. Therefore, ionisation does not increase the nuclear charge. It also does not remove protons or neutrons from the nucleus. The nuclear mass number and nuclear radius remain unchanged under the specified conditions. In a real scattering experiment, extranuclear electrons can influence the electric field experienced by a projectile at sufficiently large distances. However, the question explicitly considers energetic large-angle scattering in which electronic screening is neglected. Under this approximation, the close-encounter repulsive interaction depends on the same nuclear charge in both samples. The target-nucleus counts and incident-beam conditions are also stipulated to be equal. Consequently, the elementary nuclear-scattering prediction is the same for the two samples. The investigator has confused the net positive ionic charge with the positive charge of the nucleus itself. The distinction is important because the atom's electronic configuration and its nuclear composition are separate physical properties. Therefore, the proposed increase in large-angle nuclear scattering is not supported by the stated model.

Explanation: \(\textbf{Identify the governing ionisation-energy relation:}\) For a hydrogen-like ion: \[ I_n=\frac{13.6Z^2}{n^2}\,\text{eV} \] The relevant quantity is the ratio: \[ \frac Zn \] rather than nuclear charge number alone. \(\textbf{Evaluate system P:}\) For P: \[ Z_P=1,\qquad n_P=2 \] Therefore: \[ I_P=\frac{13.6(1)^2}{2^2} \] \[ I_P=\frac{13.6}{4} \] \[ I_P=3.40\,\text{eV} \] \(\textbf{Evaluate system Q:}\) For Q: \[ Z_Q=2,\qquad n_Q=4 \] Thus: \[ I_Q=\frac{13.6(2)^2}{4^2} \] \[ I_Q=\frac{13.6(4)}{16} \] \[ I_Q=3.40\,\text{eV} \] \(\textbf{Evaluate system R:}\) For R: \[ Z_R=3,\qquad n_R=6 \] Therefore: \[ I_R=\frac{13.6(3)^2}{6^2} \] \[ I_R=\frac{13.6(9)}{36} \] \[ I_R=3.40\,\text{eV} \] \(\textbf{Compare the incident photon with the thresholds:}\) Every system receives: \[ E_\gamma=4.00\,\text{eV} \] Since: \[ 4.00\gt 3.40 \] single-photon ionisation is energetically allowed in all three systems. \(\textbf{Calculate the photoelectron kinetic energy:}\) Under the stipulated approximation: \[ K_e=E_\gamma-I_n \] Thus, in every case: \[ K_e=4.00-3.40 \] \[ K_e=0.60\,\text{eV} \] \(\textbf{Physical interpretation:}\) Different hydrogen-like ions can have the same binding energy in appropriately chosen stationary states. Increasing nuclear charge strengthens the Coulomb attraction, but a correspondingly higher principal quantum number can compensate in the Bohr energy formula. The three systems have different nuclear charges and orbital radii, yet identical initial ionisation thresholds. Identical photons therefore produce equal excess photoelectron kinetic energies. The equality follows from the common value of \(Z^2/n^2\), not from identical atomic structures. \(\textbf{Final answer:}\) \[ \boxed{I_P=I_Q=I_R=3.40\,\text{eV},\qquad K_{e,P}=K_{e,Q}=K_{e,R}=0.60\,\text{eV}} \]

Explanation: \(\textbf{Write the neutron-separation mass relation:}\) The neutron-removal process is: \[ X\rightarrow Y+n \] Its required energy is: \[ S_n=[M_Y+m_n-M_X]c^2 \] Rearranging: \[ M_Yc^2=M_Xc^2-m_nc^2+S_n \] \(\textbf{Write the proton-separation mass relation:}\) The proton-removal process is: \[ X\rightarrow Z+p \] Therefore: \[ S_p=[M_Z+m_p-M_X]c^2 \] Rearranging: \[ M_Zc^2=M_Xc^2-m_pc^2+S_p \] \(\textbf{Subtract the daughter rest energies:}\) The parent rest-energy term is common to both expressions. Therefore: \[ (M_Y-M_Z)c^2 =-m_nc^2+S_n+m_pc^2-S_p \] Collecting terms: \[ (M_Y-M_Z)c^2 =(S_n-S_p)-(m_n-m_p)c^2 \] \(\textbf{Insert the measured separation energies:}\) \[ S_n=9.0\,\text{MeV} \] \[ S_p=7.0\,\text{MeV} \] Thus: \[ S_n-S_p=2.0\,\text{MeV} \] Using the supplied free-nucleon mass difference: \[ (M_Y-M_Z)c^2=2.0-1.293 \] \[ (M_Y-M_Z)c^2=0.707\,\text{MeV} \] \(\textbf{Interpret the sign:}\) The mass difference is positive: \[ M_Y\gt M_Z \] Therefore, the neutron-removal daughter Y has greater bare nuclear mass than the proton-removal daughter Z. \(\textbf{Check the underlying composition:}\) The two daughters have the same mass number because each process removes one nucleon. However, neutron removal leaves the parent's proton number unchanged. Proton removal reduces the proton number by one. The two daughters are consequently different nuclear species, and their masses need not be equal. \(\textbf{Physical interpretation:}\) Comparing the separation energies alone would suggest a \(2.0\,\text{MeV}\) difference, but that ignores the different rest masses of the removed nucleons. The neutron is heavier than the proton. The correct daughter-mass comparison must include both the separation-energy difference and the free-nucleon mass difference. This provides an inverse way to infer relative nuclear masses from measured separation energies. \(\textbf{Final answer:}\) \[ \boxed{(M_Y-M_Z)c^2=+0.707\,\text{MeV}} \]

Explanation: \(\textbf{Identify the ground-state reaction reference:}\) The supplied reaction energy: \[ Q_0=+150\,\text{MeV} \] refers to the transformation between the ground-state compound nucleus and the specified ground-state fragments. It does not include any nuclear excitation energy. \(\textbf{Account for the actual initial state:}\) The compound nucleus initially contains excitation energy: \[ E_i^*=8.0\,\text{MeV} \] Its rest energy is therefore \(8.0\,\text{MeV}\) greater than that of the corresponding ground-state compound nucleus. This additional initial energy increases the amount potentially available during the transformation. \(\textbf{Account for the excited daughter:}\) One fragment is produced with excitation energy: \[ E_f^*=5.0\,\text{MeV} \] Its excited-state rest energy is greater than its ground-state rest energy by that amount. Energy retained as fragment excitation is not immediately available as fragment kinetic energy or prompt radiation. \(\textbf{Calculate the immediate reaction energy:}\) The reaction energy between the specified actual nuclear states is: \[ Q_{\mathrm{immediate}}=Q_0+E_i^*-E_f^* \] Substituting: \[ Q_{\mathrm{immediate}}=150+8-5 \] \[ Q_{\mathrm{immediate}}=153\,\text{MeV} \] \(\textbf{Determine the later photon energy:}\) The excited daughter subsequently reaches its ground state. It releases its stored excitation energy: \[ E_{\gamma,\mathrm{later}}=E_f^* \] \[ E_{\gamma,\mathrm{later}}=5.0\,\text{MeV} \] \(\textbf{Calculate the complete energy release:}\) \[ Q_{\mathrm{complete}} =Q_{\mathrm{immediate}}+E_{\gamma,\mathrm{later}} \] \[ Q_{\mathrm{complete}}=153+5 \] \[ Q_{\mathrm{complete}}=158\,\text{MeV} \] \(\textbf{Check against the original ground-state reference:}\) The complete final fragments are in their ground states. The actual initial compound nucleus has \(8.0\,\text{MeV}\) more rest energy than the ground-state reference. Therefore: \[ Q_{\mathrm{complete}}=Q_0+E_i^* \] \[ Q_{\mathrm{complete}}=150+8 \] \[ Q_{\mathrm{complete}}=158\,\text{MeV} \] The two methods agree. \(\textbf{Physical interpretation:}\) Excitation of the parent increases the energy available to the reaction. Excitation retained in a daughter temporarily reduces the prompt energy available outside nuclear rest energy. When the daughter de-excites, the stored energy is subsequently released. The complete accounting depends on the actual initial and final nuclear states and must not count the same excitation energy twice. \(\textbf{Final answer:}\) \[ \boxed{Q_{\mathrm{immediate}}=153\,\text{MeV},\qquad E_{\gamma,\mathrm{later}}=5\,\text{MeV},\qquad Q_{\mathrm{complete}}=158\,\text{MeV}} \]

Explanation: \(\textbf{Check nuclear composition conservation:}\) The parent P contains six protons and six neutrons. Together, Q and R contain: \[ Z_{Q+R}=3+3 \] \[ Z_{Q+R}=6 \] and: \[ N_{Q+R}=3+3 \] \[ N_{Q+R}=6 \] Thus, the splitting and combining processes conserve the constituent proton and neutron counts. \(\textbf{Calculate the initial binding energy of X:}\) Process X begins with P: \[ B_{i,X}=92\,\text{MeV} \] \(\textbf{Calculate the final binding energy of X:}\) The daughter nuclei have combined binding energy: \[ B_{f,X}=B_Q+B_R \] \[ B_{f,X}=30+65 \] \[ B_{f,X}=95\,\text{MeV} \] \(\textbf{Calculate the splitting reaction energy:}\) \[ Q_X=B_{f,X}-B_{i,X} \] \[ Q_X=95-92 \] \[ Q_X=+3\,\text{MeV} \] \(\textbf{Calculate the reverse reaction energy:}\) Process Y reverses the initial and final configurations. Therefore: \[ Q_Y=-Q_X \] \[ Q_Y=-3\,\text{MeV} \] \(\textbf{Interpret the calculation:}\) The splitting process releases \(3\,\text{MeV}\), while the reverse combining process requires \(3\,\text{MeV}\). The sign follows from comparing the complete total binding energies rather than noting that individual nuclei are bound. The positive total binding-energy difference shows that process X releases energy, whereas its reverse requires energy input. A positive binding energy for each fragment does not by itself determine whether their formation is energetically favourable. The difference between the complete initial and final binding energies is decisive. \(\textbf{Physical interpretation:}\) The same nucleons can form different bound nuclear configurations. One configuration may have greater total binding energy and therefore lower total rest energy. Transforming towards that configuration releases energy, while returning to the less strongly bound configuration requires energy. This distinction applies to both nuclear splitting and nuclear combining processes.

Explanation: \(\textbf{Calculate the initially emitted neutron population:}\) The experiment begins with: \[ N_0=20 \] fissions. Each releases three neutrons. Therefore: \[ N_{n,\mathrm{emitted}}=20(3) \] \[ N_{n,\mathrm{emitted}}=60 \] \(\textbf{Determine neutrons escaping the assembly:}\) One-fourth escape: \[ N_{\mathrm{escape}}=\frac14(60) \] \[ N_{\mathrm{escape}}=15 \] These neutrons cause no additional fissions. \(\textbf{Determine non-fission absorptions:}\) Another one-fourth are absorbed without fission: \[ N_{\mathrm{absorption}}=\frac14(60) \] \[ N_{\mathrm{absorption}}=15 \] These also do not contribute to the additional fission population. \(\textbf{Count neutrons inducing additional fission:}\) One-half induce fission: \[ N_{\mathrm{successful}}=\frac12(60) \] \[ N_{\mathrm{successful}}=30 \] Every successful neutron induces one fission. Therefore: \[ N_{\mathrm{additional}}=30 \] \(\textbf{Check neutron accounting:}\) \[ N_{\mathrm{escape}}+N_{\mathrm{absorption}}+N_{\mathrm{successful}} =15+15+30 \] \[ N_{\mathrm{total}}=60 \] Every neutron produced by the initial fissions has been assigned an outcome. \(\textbf{Determine the total fission count:}\) \[ N_f=N_0+N_{\mathrm{additional}} \] \[ N_f=20+30 \] \[ N_f=50 \] \(\textbf{Calculate the total fission energy:}\) Each fission releases: \[ Q_f=200\,\text{MeV} \] Therefore: \[ E=N_fQ_f \] \[ E=50(200) \] \[ E=10000\,\text{MeV} \] \(\textbf{Physical interpretation:}\) The number of emitted neutrons is greater than the number that successfully induces fission. Both neutron escape and absorption without fission reduce the population available for subsequent reactions. The observation includes one additional generation but excludes further multiplication by neutrons emitted in that generation. Therefore, the energy calculation uses only the initial and explicitly induced fissions. The problem separates emitted-neutron count, successful-neutron count and total nuclear energy release. \(\textbf{Final answer:}\) \[ \boxed{N_f=50,\qquad E=10000\,\text{MeV}} \]

Explanation: The hydrogen atom begins in the same stationary state and ends in the same stationary state in both pathways. The complete atomic energy decrease is therefore the same. In pathway P, this energy is emitted in a single photon. In pathway Q, it is distributed among three photons corresponding to consecutive transitions. The sum of those three photon energies equals the direct \(4\to1\) transition energy. Statement I is correct. However, photon momentum is a vector. The direction of each emitted photon matters when calculating total photon momentum. The three photons in Q need not travel in the same direction. Therefore, equal combined photon energies do not guarantee equal combined momentum vectors. Statement II is incorrect. Pathway P contains only the transition \(4\to1\), so it produces one photon. Pathway Q contains three transitions and therefore produces three photons. Statement III is correct. For each photon, energy and frequency are related by \(E=hf\). Since Planck's constant is the same for all the photons, equality of total emitted energy implies equality of the corresponding sums of photon frequencies. Thus, the frequencies of the three cascade photons add to the frequency of the direct-transition photon. Statement IV is correct. The result shows that conservation of atomic transition energy constrains the total radiated energy but does not uniquely determine photon count or the vector distribution of emitted momenta. Statements I, III and IV are correct.

Explanation: Experiment P is Rutherford's alpha-particle scattering experiment. The passage of most alpha particles with little deflection indicates that much of the atomic volume does not contain a strongly repulsive concentration of positive charge. The rare large-angle and backward scattering events require a strong repulsive interaction during sufficiently close encounters. A small, massive, positively charged nucleus provides a consistent explanation. Statement I is therefore correct. However, the scattering observations do not establish that a classically revolving electron can remain in a stable orbit indefinitely. A revolving electron has centripetal acceleration even when its speed is constant. According to classical electromagnetic theory, an accelerated electric charge would radiate energy. The resulting energy loss creates a stability problem for the classical planetary description. Statement II is incorrect. Experiment Q concerns discrete photon energies produced by hydrogen. Since photon energy satisfies \(E_\gamma=h\nu\), distinct spectral lines correspond to distinct atomic energy differences. Statement III is correct. Bohr's model introduces permitted stationary states in which electrons do not continuously radiate. It associates radiation with transitions between those states. This modification addresses the classical atomic-stability problem and the discrete hydrogen spectrum. It does not require rejecting Rutherford's conclusion that most positive charge and atomic mass are concentrated in a small nucleus. Statement IV is correct. The two experiments therefore constrain different aspects of a satisfactory atomic model: nuclear structure and electronic energy behaviour. Statements I, III and IV are correct.

Explanation: \(\textbf{Compare the permitted orbital radii:}\) The initial and final principal quantum numbers are: \[ n_i=4,\qquad n_f=2 \] The radius law gives: \[ r_n=n^2a_0 \] Therefore: \[ r_i=16a_0 \] \[ r_f=4a_0 \] The radius ratio is: \[ \frac{r_f}{r_i}=\frac4{16} \] \[ \frac{r_f}{r_i}=\frac14 \] \(\textbf{Determine the change in electrostatic force:}\) The electrostatic force magnitude satisfies: \[ F_n=\frac{ke^2}{r_n^2} \] For the same nuclear and electronic charges: \[ \frac{F_f}{F_i}=\left(\frac{r_i}{r_f}\right)^2 \] Substituting: \[ \frac{F_f}{F_i}=\left(\frac{16}{4}\right)^2 \] \[ \frac{F_f}{F_i}=16 \] \(\textbf{Compare electron orbital speeds:}\) Using: \[ v_n=\frac{v_1}{n} \] we obtain: \[ \frac{v_f}{v_i}=\frac{n_i}{n_f} \] \[ \frac{v_f}{v_i}=\frac42 \] \[ \frac{v_f}{v_i}=2 \] The electron moves faster in the smaller orbit. \(\textbf{Determine the change in revolution period:}\) The period satisfies: \[ T=\frac{2\pi r}{v} \] Therefore: \[ \frac{T_i}{T_f}=\frac{r_i}{r_f}\frac{v_f}{v_i} \] Substituting: \[ \frac{T_i}{T_f}=4(2) \] \[ \frac{T_i}{T_f}=8 \] \(\textbf{Calculate the photon energy:}\) The initial atomic energy is: \[ E_4=-\frac{13.6}{16} \] \[ E_4=-0.85\,\text{eV} \] The final atomic energy is: \[ E_2=-\frac{13.6}{4} \] \[ E_2=-3.40\,\text{eV} \] The emitted photon carries: \[ E_\gamma=E_4-E_2 \] \[ E_\gamma=-0.85-(-3.40) \] \[ E_\gamma=2.55\,\text{eV} \] \(\textbf{Interpret the simultaneous changes:}\) As the electron moves to a smaller permitted orbit, the attractive force becomes stronger. Its orbital speed increases, while the time needed to complete one revolution decreases. Nevertheless, the atom's total energy decreases because its electrostatic potential energy becomes sufficiently more negative. The emitted photon's energy comes from the difference between stationary-state total energies, not directly from the change in revolution frequency. \(\textbf{Final answer:}\) \[ \boxed{\frac{F_f}{F_i}=16,\qquad\frac{T_i}{T_f}=8,\qquad E_\gamma=2.55\,\text{eV}} \]

Explanation: \(\textbf{Determine the photon emitted by helium:}\) For \(\mathrm{He}^{+}\): \[ Z=2 \] The emission transition is: \[ 4\to2 \] Therefore: \[ E_\gamma=13.6Z^2\left(\frac1{2^2}-\frac1{4^2}\right) \] Substituting: \[ E_\gamma=13.6(4)\left(\frac14-\frac1{16}\right) \] \[ E_\gamma=54.4\left(\frac3{16}\right) \] \[ E_\gamma=10.20\,\text{eV} \] \(\textbf{Test the photon against ground-state hydrogen:}\) For ordinary hydrogen: \[ Z=1 \] The excitation energy from \(n=1\) to \(n=2\) is: \[ \Delta E_{1\to2,\mathrm H}=13.6\left(1-\frac14\right) \] \[ \Delta E_{1\to2,\mathrm H}=10.20\,\text{eV} \] The incident photon exactly matches this energy difference. Therefore, a hydrogen atom initially in \(n=1\) can absorb the photon and reach \(n=2\). \(\textbf{Test the photon against ground-state helium ions:}\) For \(\mathrm{He}^{+}\), the smallest bound-bound excitation from the ground state is \(1\to2\). Its energy is: \[ \Delta E_{1\to2,\mathrm{He}^{+}} =13.6(2)^2\left(1-\frac14\right) \] \[ \Delta E_{1\to2,\mathrm{He}^{+}}=40.8\,\text{eV} \] Since: \[ 10.20\lt 40.8 \] the emitted photon cannot excite a ground-state helium ion to its first excited state. Every higher bound-bound excitation from that ground state requires still more energy. \(\textbf{Explain the spectral coincidence:}\) The \(4\to2\) transition in \(\mathrm{He}^{+}\) and the \(2\to1\) transition in hydrogen have the same photon energy. They therefore have the same vacuum wavelength. However, the absorbing gas must possess an initially occupied lower level separated from an available higher level by that same energy. Ground-state hydrogen meets this condition, whereas ground-state \(\mathrm{He}^{+}\) does not. \(\textbf{Physical interpretation:}\) Equal photon wavelengths from different atomic species do not imply identical energy-level structures. The nuclear charge changes the scale of hydrogen-like energy differences. Whether a photon is absorbed also depends on the occupied initial state of the receiving atom. This problem combines spectral coincidence with state-dependent absorption rather than identifying the emitting species from wavelength alone. \(\textbf{Final answer:}\) \[ E_\gamma=10.20\,\text{eV},\qquad \mathrm H:\ 1\to2\ \text{allowed},\qquad \mathrm{He}^{+}:\ \text{no ground-state bound-bound absorption} \]

Explanation: The two alpha particles have the same mass and the same impact-parameter magnitude. Their kinetic energies differ by a factor of two. For a non-relativistic particle, momentum magnitude is related to kinetic energy by: \[ p=\sqrt{2mK} \] Therefore, doubling kinetic energy increases momentum magnitude by a factor of \(\sqrt2\). The incident angular-momentum magnitude about the target centre is: \[ L=pb \] Since \(b\) is unchanged, Q has \(\sqrt2\) times P's initial angular-momentum magnitude. Statement I is correct. Increasing incident kinetic energy makes the repulsive Coulomb interaction less effective at turning the trajectory through a large angle at the same impact parameter. Consequently, the higher-energy particle undergoes a smaller scattering deflection under the stipulated conditions. Statement II is incorrect. The Coulomb force is conservative. The target is fixed, and neither radiation nor any other energy-loss process is included. At sufficiently large initial and final separations, the electrostatic potential energy is negligible at both locations. Each alpha particle therefore regains its own original kinetic energy after scattering. Statement III is correct. The impact parameter is not the same as the distance of closest approach. The latter depends on the incident kinetic energy, angular momentum and Coulomb interaction. Although \(b\) is unchanged, doubling the incident energy changes the trajectory and its closest-approach distance. Statement IV is incorrect. The comparison distinguishes the geometric specification of a trajectory from its energy-dependent Coulomb deflection. Thus, Statements I and III are correct.

Explanation: \(\textbf{Calculate the direct-transition energy:}\) The energy levels are: \[ E_2=-3.400\,\text{eV} \] \[ E_3=-\frac{13.6}{9}\,\text{eV} \] \[ E_4=-0.850\,\text{eV} \] For pathway P: \[ E_{\gamma,P}=E_4-E_2 \] \[ E_{\gamma,P}=-0.850-(-3.400) \] \[ E_{\gamma,P}=2.550\,\text{eV} \] \(\textbf{Determine the two cascade photon energies:}\) For \(4\to3\): \[ E_{4\to3}=13.6\left(\frac19-\frac1{16}\right) \] \[ E_{4\to3}\approx0.6611\,\text{eV} \] For \(3\to2\): \[ E_{3\to2}=13.6\left(\frac14-\frac19\right) \] \[ E_{3\to2}\approx1.8889\,\text{eV} \] Their sum is: \[ E_{\gamma,Q}=0.6611+1.8889 \] \[ E_{\gamma,Q}=2.5500\,\text{eV} \] \(\textbf{Identify the absorption conditions of the second gas:}\) The absorbing hydrogen atoms initially occupy \(n=2\). They may absorb photons that match upward transitions beginning at that level. For \(2\to3\): \[ E_{2\to3}=1.8889\,\text{eV} \] For \(2\to4\): \[ E_{2\to4}=2.5500\,\text{eV} \] \(\textbf{Test pathway P's photon:}\) The direct \(4\to2\) photon carries \(2.55\,\text{eV}\). It matches absorption through \(2\to4\). Therefore, it can be absorbed. \(\textbf{Test each photon from pathway Q:}\) The \(3\to2\) photon carries approximately \(1.8889\,\text{eV}\). It matches absorption through \(2\to3\). The \(4\to3\) photon carries approximately \(0.6611\,\text{eV}\). This is smaller than the lowest upward excitation gap from \(n=2\), so it cannot drive any bound-bound absorption from that initial state. \(\textbf{Physical interpretation:}\) Both emission pathways have the same total radiated energy because their initial and final atomic states coincide. Nevertheless, their individual photon energies differ. A receiving gas responds to individual photon energies and its own initially populated levels, not simply to the sum of the energy in a cascade. Consequently, equal total emitted energies need not produce equal absorption patterns. \(\textbf{Final answer:}\) \[ E_{\gamma,P}=E_{\gamma,Q}=2.55\,\text{eV},\qquad 2\to4\ \text{and}\ 2\to3\ \text{absorption permitted} \]

Explanation: \(\textbf{Establish the initial nuclear and electronic composition:}\) Uranium has atomic number: \[ Z_U=92 \] The original uranium atom is electrically neutral. It therefore contains: \[ N_{e,i}=92 \] extranuclear electrons. The incident neutron carries no electric charge and introduces no extranuclear electron. \(\textbf{Check nuclear charge conservation:}\) The product nuclear charge numbers are: \[ Z_{\mathrm{Sr}}=38,\qquad Z_{\mathrm{Xe}}=54 \] Their sum is: \[ Z_f=38+54 \] \[ Z_f=92 \] Thus, the fission process conserves nuclear electric charge. \(\textbf{Check mass-number conservation:}\) Initially: \[ A_i=235+1 \] \[ A_i=236 \] Finally: \[ A_f=94+140+2 \] \[ A_f=236 \] The two emitted neutrons account for the nucleons absent from the bound fragments. \(\textbf{Determine electrons retained by the strontium ion:}\) Neutral strontium contains 38 electrons. The specified ion is doubly positive, so it has lost two: \[ N_{e,\mathrm{Sr}^{2+}}=38-2 \] \[ N_{e,\mathrm{Sr}^{2+}}=36 \] \(\textbf{Determine electrons retained by the xenon ion:}\) Neutral xenon contains 54 electrons. The specified ion is singly positive: \[ N_{e,\mathrm{Xe}^{+}}=54-1 \] \[ N_{e,\mathrm{Xe}^{+}}=53 \] \(\textbf{Determine the total bound electron count:}\) \[ N_{e,\mathrm{bound}}=36+53 \] \[ N_{e,\mathrm{bound}}=89 \] \(\textbf{Determine the remaining free electrons:}\) The complete system initially contains 92 electrons. Therefore: \[ N_{e,\mathrm{free}}=92-89 \] \[ N_{e,\mathrm{free}}=3 \] \(\textbf{Check total electric charge:}\) The two positive ions carry a combined charge of: \[ (+2e)+(+e)=+3e \] The three free electrons carry: \[ 3(-e)=-3e \] The complete electronic and ionic charge remains zero. \(\textbf{Physical interpretation:}\) Nuclear fission rearranges the original protons and neutrons among nuclear fragments and emitted particles. The electronic states of the resulting atomic species can change separately. A nuclear equation need not show spectator electrons, but a complete accounting of the final ionic system must conserve electric charge. The additional three free electrons are not three additional nucleons and do not alter the balanced nuclear equation. \(\textbf{Final answer:}\) \[ \boxed{N_{e,\mathrm{free}}=3,\qquad N_{e,\mathrm{bound}}=89} \]

Explanation: \(\textbf{Infer the principal quantum number from radius:}\) The measured radius is: \[ r=16a_0 \] For hydrogen: \[ r_n=n^2a_0 \] Therefore: \[ 16a_0=n^2a_0 \] \[ n^2=16 \] \[ n=4 \] \(\textbf{Independently infer the state from kinetic energy:}\) The reported kinetic energy is: \[ K=0.850\,\text{eV} \] The Bohr relation is: \[ K_n=\frac{13.6}{n^2}\,\text{eV} \] Therefore: \[ 0.850=\frac{13.6}{n^2} \] Rearranging: \[ n^2=\frac{13.6}{0.850} \] \[ n^2=16 \] \[ n=4 \] \(\textbf{Check mutual consistency of the first two readings:}\) Both the radius and kinetic-energy readings independently identify the fourth stationary orbit. They are therefore mutually compatible. The remaining measurement must correspond to that same principal quantum number. \(\textbf{Determine the required angular momentum:}\) Bohr's angular-momentum postulate gives: \[ L_n=\frac{nh}{2\pi} \] For \(n=4\): \[ L_{\mathrm{required}}=\frac{4h}{2\pi} \] The reported angular momentum is: \[ L_{\mathrm{reported}}=\frac{3h}{2\pi} \] This belongs to \(n=3\), not \(n=4\). \(\textbf{Check the alternative interpretation:}\) If the angular-momentum reading were accepted, the electron would occupy \(n=3\). Its permitted radius would then be: \[ r_3=9a_0 \] Its kinetic energy would be: \[ K_3=\frac{13.6}{9} \] \[ K_3\approx1.511\,\text{eV} \] Both other readings would then require correction. That contradicts the stipulation that exactly one reading is incorrect. \(\textbf{Physical interpretation:}\) The three Bohr quantities are not independently adjustable properties of a stationary hydrogen orbit. Radius, kinetic energy and angular momentum are all determined by the same integer principal quantum number. Two consistent independent measurements can therefore identify which third reading is incompatible with the model. This is a model-consistency test, not merely a calculation of an orbital quantity. \(\textbf{Final answer:}\) \[ \boxed{n=4,\qquad L_{\mathrm{correct}}=\frac{4h}{2\pi}} \]

Explanation: \(\textbf{Determine the conserved mechanical energy:}\) Initially, the alpha particle is very far from the nucleus. The Coulomb potential energy is negligible there. Therefore: \[ E_{\mathrm{total}}=K_\infty \] \[ E_{\mathrm{total}}=12.0\,\text{MeV} \] \(\textbf{Identify the turning-point potential energy:}\) At the head-on turning point \(r=d\), the alpha particle momentarily stops. Thus: \[ K(d)=0 \] Conservation of mechanical energy gives: \[ U(d)=12.0\,\text{MeV} \] \(\textbf{Use the inverse-separation potential relation:}\) The Coulomb potential energy is proportional to \(1/r\). Therefore: \[ U(r)=U(d)\frac dr \] This relation determines the potential energy at any specified multiple of the turning-point separation. \(\textbf{Calculate the potential energy at P:}\) Point P is at: \[ r_P=4d \] Therefore: \[ U_P=12.0\left(\frac d{4d}\right) \] \[ U_P=3.0\,\text{MeV} \] \(\textbf{Calculate the kinetic energy at P:}\) \[ K_P=E_{\mathrm{total}}-U_P \] \[ K_P=12.0-3.0 \] \[ K_P=9.0\,\text{MeV} \] \(\textbf{Calculate the potential energy at Q:}\) Point Q is at: \[ r_Q=2d \] Thus: \[ U_Q=12.0\left(\frac d{2d}\right) \] \[ U_Q=6.0\,\text{MeV} \] \(\textbf{Calculate the kinetic energy at Q:}\) \[ K_Q=12.0-6.0 \] \[ K_Q=6.0\,\text{MeV} \] \(\textbf{Compare the speeds:}\) For the same non-relativistic alpha particle: \[ K=\frac12mv^2 \] Therefore: \[ \frac{v_Q}{v_P}=\sqrt{\frac{K_Q}{K_P}} \] Substituting: \[ \frac{v_Q}{v_P}=\sqrt{\frac69} \] \[ \frac{v_Q}{v_P}=\sqrt{\frac23} \] \(\textbf{Physical interpretation:}\) The alpha particle loses kinetic energy as it approaches the positively charged nucleus. The lost kinetic energy becomes increasing positive Coulomb potential energy. Its speed decreases continuously until it reaches the turning point. The kinetic energies at intermediate separations cannot be found by assuming that all the initial energy has already become potential energy. The speed ratio must also be obtained from the square root of the kinetic-energy ratio, not from the kinetic-energy ratio itself. \(\textbf{Final answer:}\) \[ \boxed{K_P=9.0\,\text{MeV},\qquad K_Q=6.0\,\text{MeV},\qquad \frac{v_Q}{v_P}=\sqrt{\frac23}} \]

Explanation: \(\textbf{Express nuclear volume as a function of mass number:}\) For a spherical nucleus: \[ V=\frac{4\pi R^3}{3} \] Using: \[ R=R_0A^{1/3} \] gives: \[ V=\frac{4\pi R_0^3}{3}A \] Define: \[ V_0=\frac{4\pi R_0^3}{3} \] Then: \[ V=V_0A \] \(\textbf{Recover the average mass number from the volume measurement:}\) The measured mean volume is: \[ \overline V=20V_0 \] Since volume is proportional to mass number: \[ \overline A=20 \] \(\textbf{Infer the isotope number fractions:}\) Let the number fraction of Q be \(f\). The number fraction of P is \(1-f\). Therefore: \[ \overline A=(1-f)A_P+fA_Q \] Substituting: \[ 20=(1-f)(8)+27f \] \[ 20=8+19f \] \[ 19f=12 \] \[ f=\frac{12}{19} \] Thus: \[ f_Q=\frac{12}{19},\qquad f_P=\frac7{19} \] \(\textbf{Calculate the separate nuclear radii:}\) For P: \[ R_P=R_0(8)^{1/3} \] \[ R_P=2R_0 \] For Q: \[ R_Q=R_0(27)^{1/3} \] \[ R_Q=3R_0 \] \(\textbf{Determine the actual number-weighted mean radius:}\) \[ \overline R=f_PR_P+f_QR_Q \] Substituting: \[ \overline R=\frac7{19}(2R_0)+\frac{12}{19}(3R_0) \] \[ \overline R=\frac{14+36}{19}R_0 \] \[ \overline R=\frac{50}{19}R_0 \] \(\textbf{Determine the mass fraction contributed by Q:}\) Nuclear mass is approximately proportional to mass number. Therefore: \[ w_Q=\frac{f_QA_Q}{f_PA_P+f_QA_Q} \] Substituting: \[ w_Q=\frac{(12/19)(27)}{(7/19)(8)+(12/19)(27)} \] \[ w_Q=\frac{324}{56+324} \] \[ w_Q=\frac{324}{380} \] \[ w_Q=\frac{81}{95} \] \(\textbf{Interpret the different averages:}\) The mean nuclear volume directly determines the mean mass number under the given radius law. The mean radius must instead be obtained by averaging the individual cube-root radii. The mass fraction of Q is greater than its number fraction because each Q nucleus contains more nucleons than each P nucleus. The same isotope mixture therefore has different number-weighted and mass-weighted descriptions. \(\textbf{Final answer:}\) \[ \boxed{f_Q=\frac{12}{19},\qquad \overline R=\frac{50}{19}R_0,\qquad w_Q=\frac{81}{95}} \]

Explanation: \(\textbf{Identify the required momentum balance:}\) The parent nucleus is initially stationary. Its initial momentum is therefore: \[ p_i=0 \] The complete final system contains two heavy fragments and one free neutron. All three momenta must be included in conservation of momentum. \(\textbf{Determine the unknown fragment momentum:}\) Take rightward momentum as positive. The final momentum sum is: \[ p_P+p_Q+p_n=0 \] Substituting: \[ -4p+p_Q+p=0 \] Therefore: \[ p_Q=3p \] The positive sign shows that fragment Q moves to the right. \(\textbf{Compare the magnitudes of heavy-fragment momenta:}\) The measured momenta are: \[ |p_P|=4p \] \[ |p_Q|=3p \] They are not equal. The difference is required to balance the momentum carried by the emitted neutron. \(\textbf{Calculate P's kinetic energy:}\) For a non-relativistic particle: \[ K=\frac{p^2}{2m} \] Fragment P has mass \(2m\). Thus: \[ K_P=\frac{(-4p)^2}{2(2m)} \] \[ K_P=\frac{16p^2}{4m} \] \[ K_P=\frac{4p^2}{m} \] \(\textbf{Calculate Q's kinetic energy:}\) Fragment Q has mass \(3m\). Therefore: \[ K_Q=\frac{(3p)^2}{2(3m)} \] \[ K_Q=\frac{9p^2}{6m} \] \[ K_Q=\frac{3p^2}{2m} \] \(\textbf{Determine the kinetic-energy ratio:}\) \[ \frac{K_P}{K_Q} =\frac{4p^2/m}{3p^2/(2m)} \] \[ \frac{K_P}{K_Q}=\frac83 \] \(\textbf{Check the total final momentum:}\) \[ p_{\mathrm{total}}=-4p+3p+p \] \[ p_{\mathrm{total}}=0 \] The calculated momentum of Q satisfies conservation of momentum. \(\textbf{Physical interpretation:}\) In a two-body fission final state from rest, the two heavy fragments have equal and opposite momenta. That conclusion cannot be applied unchanged when a free neutron carries non-zero momentum. The neutron changes the momentum balance of the heavy fragments. Their separate kinetic energies must be calculated from their actual momentum magnitudes and masses. Momentum conservation must therefore precede the kinetic-energy comparison. \(\textbf{Final answer:}\) \[ \boxed{p_Q=+3p,\qquad\frac{K_P}{K_Q}=\frac83} \]

Explanation: The Assertion compares orbital size, orbital motion and ionisation energy. These quantities depend differently on the principal quantum number. The Bohr radius relation shows that orbital radius increases in proportion to \(n^2\). The revolution period increases in proportion to \(n^3\). Consequently, an electron occupying a sufficiently high stationary state has a much larger orbital radius and a much longer revolution period than one in a low state. The atom's bound-state total energy remains negative for every finite permitted \(n\). However, its magnitude decreases as \(n\) increases. The zero of atomic energy corresponds to a separated electron and nucleus at infinite distance. Therefore, the energy required to ionise hydrogen from the \(n\)th state equals the magnitude of its negative total energy. That ionisation energy tends towards zero as \(n\) increases without bound. The Reason correctly states the limiting behaviour underlying these conclusions. A larger orbit does not mean that the atom is more strongly bound. Instead, the electron becomes progressively less tightly bound even while its orbit grows larger. The limiting state at zero binding energy marks ionisation rather than another ordinary finite-radius bound orbit. Both Assertion and Reason are true, and the Reason correctly explains the Assertion.

Explanation: \(\textbf{Identify the initial energy available:}\) The nuclear excitation energy is: \[ E^*=1.000000\,\text{MeV} \] The excited nucleus initially has zero momentum. It emits one photon and returns to its ground state. \(\textbf{Apply momentum conservation:}\) The initial momentum is zero. Therefore: \[ \vec p_N+\vec p_\gamma=0 \] The recoil momentum magnitude equals the photon momentum magnitude: \[ p_N=p_\gamma=\frac{E_\gamma}{c} \] \(\textbf{Express the nuclear recoil kinetic energy:}\) The ground-state nuclear recoil is non-relativistic: \[ K_N=\frac{p_N^2}{2M} \] Substituting the photon momentum: \[ K_N=\frac{E_\gamma^2}{2Mc^2} \] With the supplied nuclear rest energy: \[ K_N=\frac{E_\gamma^2}{2(10\,000\,\text{MeV})} \] \(\textbf{Construct the complete energy balance:}\) The initial excitation energy is shared between the photon and nuclear recoil: \[ E^*=E_\gamma+K_N \] Therefore: \[ 1.000000=E_\gamma+\frac{E_\gamma^2}{20\,000} \] Here, all energies are expressed in MeV. \(\textbf{Use a justified small-recoil approximation:}\) The nucleus has rest energy much greater than the excitation energy. Therefore, recoil energy is expected to be very small. For a first estimate, use: \[ E_\gamma\approx1.000000\,\text{MeV} \] inside the recoil term. Then: \[ K_N\approx\frac{(1.000000)^2}{20\,000}\,\text{MeV} \] \[ K_N\approx0.000050\,\text{MeV} \] Converting: \[ K_N\approx50\,\text{eV} \] \(\textbf{Calculate the emitted photon energy:}\) From energy conservation: \[ E_\gamma=E^*-K_N \] \[ E_\gamma\approx1.000000-0.000050 \] \[ E_\gamma\approx0.999950\,\text{MeV} \] \(\textbf{Check the approximation:}\) The recoil correction is only about \(5\times10^{-5}\) of the excitation energy. It is therefore consistent to treat it as a small correction. \(\textbf{Physical interpretation:}\) An initially stationary nucleus cannot emit a photon while remaining stationary, because the photon carries momentum. The recoil nucleus carries momentum opposite to the photon. Its small kinetic energy comes from the original nuclear excitation energy. Consequently, the photon carries slightly less energy than the full excitation difference. Neglecting recoil is often useful in elementary nuclear-energy calculations, but it is an approximation rather than an exact statement. \(\textbf{Final answer:}\) \[ \boxed{E_\gamma\approx0.999950\,\text{MeV},\qquad K_N\approx50\,\text{eV}} \]

Explanation: \(\textbf{Identify the initial nuclear composition:}\) The nucleus initially has: \[ Z_i=13,\qquad N_i=14 \] Therefore: \[ A_i=Z_i+N_i \] \[ A_i=13+14 \] \[ A_i=27 \] \(\textbf{Apply neutron capture:}\) One neutron is added. The proton number is unchanged: \[ Z_f=13 \] The neutron number becomes: \[ N_f=14+1 \] \[ N_f=15 \] Thus: \[ A_f=13+15 \] \[ A_f=28 \] \(\textbf{Compare the nuclear volumes:}\) The radius law gives: \[ R=R_0A^{1/3} \] For a spherical nucleus: \[ V=\frac{4\pi R_0^3}{3}A \] Therefore: \[ \frac{V_f}{V_i}=\frac{A_f}{A_i} \] \[ \frac{V_f}{V_i}=\frac{28}{27} \] \(\textbf{Calculate the nuclear charge-density ratio:}\) Average charge density is: \[ \rho_q=\frac{Ze}{V} \] The proton number remains 13, so the nuclear charge is unchanged. Therefore: \[ \frac{\rho_{q,f}}{\rho_{q,i}}=\frac{V_i}{V_f} \] \[ \frac{\rho_{q,f}}{\rho_{q,i}}=\frac{27}{28} \] \(\textbf{Calculate the neutron number-density ratio:}\) Average neutron number density is: \[ n_n=\frac NV \] Thus: \[ \frac{n_{n,f}}{n_{n,i}}=\frac{N_f}{N_i}\frac{V_i}{V_f} \] Substituting: \[ \frac{n_{n,f}}{n_{n,i}}=\frac{15}{14}\frac{27}{28} \] \[ \frac{n_{n,f}}{n_{n,i}}=\frac{405}{392} \] \[ \frac{n_{n,f}}{n_{n,i}}\approx1.0332 \] \(\textbf{Interpret the simultaneous changes:}\) The positive nuclear charge remains constant because neutron capture does not change the proton number. However, the nuclear volume increases, so the average positive charge density decreases. The number of neutrons increases from 14 to 15. This increase is proportionally greater than the increase in volume, so the average neutron number density rises slightly. \(\textbf{Physical interpretation:}\) The approximate constancy of total nuclear matter density does not require constant proton density and constant neutron density separately. Their relative contributions can change when a neutron is captured. This distinction connects nuclear composition with the radius law and the meaning of different kinds of nuclear density. \(\textbf{Final answer:}\) \[ \boxed{\frac{\rho_{q,f}}{\rho_{q,i}}=\frac{27}{28},\qquad \frac{n_{n,f}}{n_{n,i}}=\frac{405}{392}} \]