Atoms and Nuclei MCQs with Answers – Part 1 (NEET Physics)

Explanation: An alpha particle is the nucleus of a helium atom containing two protons and two neutrons. It therefore carries a positive charge of \(+2e\). The gold nucleus also carries positive charge, equal to \(+Ze\), where \(Z=79\) for gold. Consequently, an approaching alpha particle experiences electrostatic repulsion from the gold nucleus. The magnitude of this force depends on the nuclear charge and the instantaneous separation between the two particles. Since the nucleus is much more massive than an alpha particle, its recoil can be neglected in the elementary Rutherford scattering model. Atomic electrons are comparatively light and do not produce the appreciable large-angle deflections observed experimentally. The zinc sulphide screen detects scattered alpha particles through small flashes of light called scintillations. Thus, the experiment probes the concentrated positive charge of the nucleus through the deflection of positively charged alpha particles.

Explanation: The two observations provide complementary information about the internal structure of an atom. The passage of most alpha particles without appreciable deflection indicates that they ordinarily encounter very little concentrated charge along their paths. This is consistent with an atom whose volume is largely empty space. However, the occasional backward scattering requires a very strong repulsive interaction. Such a large repulsive force becomes possible when an alpha particle approaches a highly concentrated positive charge. Rutherford therefore inferred the existence of a very small, positively charged nucleus. The nucleus contains nearly all the mass of the atom, while electrons occupy the surrounding atomic region. If positive charge were spread diffusely throughout the atom, it would not account for the observed rare, very large deflections in the Rutherford model. The rarity of these events also indicates that the strongly repulsive central region occupies only a tiny fraction of the atomic volume. Both observations are therefore necessary to establish the nuclear picture of the atom.

Explanation: The elementary Rutherford scattering analysis considers the Coulomb interaction between an alpha particle and a single target nucleus. A sufficiently thin gold foil makes it reasonable to neglect multiple scattering in the simplest interpretation. When the foil becomes thicker, an alpha particle travels through a greater amount of material. Its probability of undergoing more than one appreciable deflection consequently increases. The final direction recorded by the detector may then result from several successive interactions rather than one isolated encounter. Such multiple scattering makes it harder to relate a measured final angle to the impact parameter of a single collision. Increasing foil thickness does not change the nuclear charge of gold or the identity of the incident alpha particle. Nor does it eliminate the Coulomb repulsion responsible for scattering. Energy loss during passage through the material may introduce another complication. Therefore, the thin-foil condition is important for connecting experimentally observed angular distributions with a simple single-nucleus scattering model.

Explanation: \(\textbf{Given observations:}\) Total number of incident alpha particles: \[ N=80{,}000 \] The particles scattered through angles greater than \(1^\circ\) belong to the second and third intervals. \(\textbf{Number undergoing appreciable scattering:}\) \[ N_{\theta\gt 1^\circ}=102+10=112 \] \(\textbf{Percentage calculation:}\) \[ P=\frac{N_{\theta\gt 1^\circ}}{N}\times100 \] Substituting the observed counts: \[ P=\frac{112}{80{,}000}\times100 \] \[ P=0.14\% \] The percentage undergoing scattering through angles greater than \(90^\circ\) is much smaller. \[ P_{\theta\gt 90^\circ}=\frac{10}{80{,}000}\times100 \] \[ P_{\theta\gt 90^\circ}=0.0125\% \] \(\textbf{Physical interpretation:}\) The relatively small percentage of appreciably scattered particles shows that strong interactions are uncommon. Backward scattering requires an alpha particle to experience particularly strong electrostatic repulsion. This occurs during a very close approach to concentrated positive nuclear charge. Since such encounters are extremely rare, the nucleus must occupy only a small fraction of the atomic volume. The data therefore support a small, dense, positively charged nucleus surrounded by a largely empty atomic region. The required combination is \(0.14\%\) and concentrated central positive charge.

Explanation: The impact parameter is defined using the initial direction of motion, before the incident particle is appreciably deflected. It is the perpendicular distance from the initial velocity line to the centre of the target nucleus. For particle P, that line passes through O, so its impact parameter is zero. However, the gold nucleus and the alpha particle both carry positive charge. Their electrostatic repulsion increases strongly as their separation decreases. During the head-on approach, the alpha particle loses kinetic energy while gaining electrostatic potential energy. It momentarily stops at a finite centre-to-centre separation and then moves back along its original path. Thus, zero impact parameter does not imply zero closest-approach distance. Particle Q has a nonzero impact parameter because its initial velocity line misses the nuclear centre. The impact parameter is not generally identical to the eventual minimum separation, since the path bends under the Coulomb force.

Explanation: The scattering angle depends on how closely an incident alpha particle approaches the repulsive nuclear region. With the incident energy and nuclear charge fixed, a smaller impact parameter produces a larger deflection. Statement I is correct because P has zero impact parameter and approaches the nucleus head-on. Under the stationary-nucleus approximation, it reverses direction along the same line, giving a scattering angle of \(180^\circ\). Statement II is incorrect because R has a larger impact parameter than Q. Consequently, R experiences a smaller, not greater, deflection. Statement III is correct because Q has a finite impact parameter whereas P approaches head-on. Hence, Q has a scattering angle smaller than \(180^\circ\). Statement IV is incorrect because a large impact parameter corresponds to an approach farther from the nuclear centre. The Coulomb interaction bends R's trajectory but does not require it to pass through the nucleus. Therefore, only Statements I and III are correct.

Explanation: \(\textbf{Governing interaction:}\) The alpha particle has charge \(+2e\), while a target nucleus has charge \(+Ze\). The electrostatic force magnitude is: \[ F=\frac{1}{4\pi\varepsilon_0}\frac{(2e)(Ze)}{r^2} \] Hence, for identical incident alpha particles: \[ F\propto\frac{Z}{r^2} \] \(\textbf{Comparison of the two situations:}\) For the first target, \(Z_A=30\) and the separation is \(r_A=r\). For the second target, \(Z_B=60\) and the separation is \(r_B=3r/2\). Therefore: \[ \frac{F_B}{F_A}=\frac{Z_B}{Z_A}\left(\frac{r_A}{r_B}\right)^2 \] \(\textbf{Substitution:}\) \[ \frac{F_B}{F_A}=\frac{60}{30}\left(\frac{r}{3r/2}\right)^2 \] \[ =2\left(\frac{2}{3}\right)^2 \] \[ =2\times\frac{4}{9} \] \(\textbf{Physical check:}\) Doubling the nuclear charge alone would double the force. However, increasing the separation reduces the force by a factor of \(4/9\). Combining both effects gives a force slightly smaller than the original. Since the question asks for a ratio of force magnitudes, the result is dimensionless. \(\textbf{Final answer:}\) \[ \boxed{\frac{F_B}{F_A}=\frac{8}{9}} \]

Explanation: The graph compares scattering at the same incident alpha-particle energy and the same impact parameter. These controlled conditions are important because scattering depends on both the geometry of the encounter and the strength of the electrostatic interaction. A nucleus with greater positive charge exerts stronger repulsion on an approaching alpha particle at a given separation. With the other conditions fixed, this produces a larger change in the particle's direction. The vertical coordinate represents the scattering angle, so a curve appearing higher corresponds to greater deflection. Since P lies above Q at every common nonzero impact parameter, nucleus P has the greater positive charge. The downward trend of both curves is also consistent with Rutherford scattering because a larger impact parameter generally produces a more distant encounter and smaller deflection. A difference in atomic radius alone does not explain the ordering of the pure Coulomb-scattering curves when the relevant nuclear charges are identical. The result should be inferred from the controlled comparison, not merely from the existence of two separate curves. Therefore, nucleus P has the larger nuclear charge.

Explanation: \(\textbf{Given:}\) \[ K=6\,\text{MeV},\qquad Z=50 \] The alpha particle carries charge \(+2e\). \(\textbf{Physical condition:}\) The approach is head-on, so the alpha particle moves directly towards the nuclear centre. At the turning point, its speed becomes momentarily zero in the stationary-nucleus approximation. Thus, its initial kinetic energy has been converted into electrostatic potential energy. \(\textbf{Energy conservation:}\) \[ K=\frac{1}{4\pi\varepsilon_0}\frac{(2e)(Ze)}{d} \] Rearranging: \[ d=\frac{2Z}{K}\left(\frac{e^2}{4\pi\varepsilon_0}\right) \] \(\textbf{Substitution:}\) \[ d=\frac{2(50)(1.44\,\text{MeV\,fm})}{6\,\text{MeV}} \] \[ d=\frac{144\,\text{MeV\,fm}}{6\,\text{MeV}} \] \[ d=24\,\text{fm} \] \(\textbf{Physical check:}\) The energy unit MeV cancels, leaving a distance in femtometres. The factor 2 is essential because the incident alpha particle carries two elementary positive charges. The result represents the centre-to-centre distance at the electrostatic turning point, not necessarily the radius of the target nucleus. \(\textbf{Final answer:}\) \[ \boxed{d=24\,\text{fm}} \]

Explanation: \(\textbf{Given comparisons:}\) \[ Z_A=40,\qquad K_B=2K_A \] \[ d_B=\frac{3}{4}d_A \] \(\textbf{Stage 1 — Determine nuclear charge:}\) For head-on Coulomb scattering: \[ d=\frac{1}{4\pi\varepsilon_0}\frac{2Ze^2}{K} \] Thus, the closest-approach distance is proportional to nuclear charge divided by incident kinetic energy. \[ d\propto\frac{Z}{K} \] Taking the ratio for the two experiments: \[ \frac{d_B}{d_A}=\frac{Z_B}{Z_A}\frac{K_A}{K_B} \] Substituting the supplied conditions: \[ \frac{3}{4}=\frac{Z_B}{40}\times\frac{1}{2} \] \[ Z_B=40\times\frac{3}{4}\times2 \] \[ Z_B=60 \] \(\textbf{Stage 2 — Compare turning-point forces:}\) The Coulomb-force magnitude is: \[ F=\frac{1}{4\pi\varepsilon_0}\frac{2Ze^2}{d^2} \] Therefore: \[ \frac{F_B}{F_A}=\frac{Z_B}{Z_A}\left(\frac{d_A}{d_B}\right)^2 \] \(\textbf{Substitution:}\) \[ \frac{F_B}{F_A}=\frac{60}{40}\left(\frac{4}{3}\right)^2 \] \[ =\frac{3}{2}\times\frac{16}{9} \] \[ =\frac{8}{3} \] \(\textbf{Consistency check:}\) Although the alpha particle in experiment B has twice the initial energy, its closest-approach distance is only three-fourths of that in A because the target charge is also greater. At the shorter turning-point distance, the inverse-square dependence further increases the repulsive force. Hence, the force ratio is greater than the nuclear-charge ratio alone. \(\textbf{Final answer:}\) \[ \boxed{\left(Z_B,\frac{F_B}{F_A}\right)=\left(60,\frac{8}{3}\right)} \]

Explanation: \(\textbf{Initial conditions:}\) \[ K_0=10\,\text{MeV} \] The electrostatic potential energy is taken as zero at infinite separation. At the turning point, the alpha particle momentarily stops. Therefore, its entire initial kinetic energy has become electrostatic potential energy. \(\textbf{Energy at closest approach:}\) \[ K_0=\frac{1}{4\pi\varepsilon_0}\frac{2Ze^2}{d} \] \(\textbf{Stage 1 — Potential energy at }2d\textbf{:}\) The repulsive electrostatic potential energy varies inversely with separation. \[ U(2d)=\frac{1}{4\pi\varepsilon_0}\frac{2Ze^2}{2d} \] Using the turning-point relation: \[ U(2d)=\frac{K_0}{2} \] \[ U(2d)=5\,\text{MeV} \] \(\textbf{Stage 2 — Remaining kinetic energy:}\) Mechanical energy remains conserved throughout the approach. \[ K(2d)+U(2d)=K_0 \] \[ K(2d)=10-5 \] \[ K(2d)=5\,\text{MeV} \] \(\textbf{Stage 3 — Determine speed:}\) Kinetic energy is proportional to the square of speed for the same particle. \[ \frac{K(2d)}{K_0}=\frac{v^2}{v_0^2} \] \[ \frac{5}{10}=\frac{v^2}{v_0^2} \] \[ v^2=\frac{v_0^2}{2} \] \[ v=\frac{v_0}{\sqrt{2}} \] \(\textbf{Physical check:}\) The particle is slowing down as it approaches the repulsive nucleus. However, at \(2d\), it has not reached the turning point, so its speed must remain positive. Half the initial kinetic energy corresponds to a speed reduced by a factor of \(\sqrt{2}\), not by a factor of 2. \(\textbf{Final answer:}\) \[ \boxed{\left(K,v\right)=\left(5\,\text{MeV},\frac{v_0}{\sqrt{2}}\right)} \]

Explanation: \(\textbf{Given:}\) \[ K=7.2\,\text{MeV},\qquad Z=79 \] \[ R_N=7\,\text{fm},\qquad R_\alpha=2\,\text{fm} \] \(\textbf{Stage 1 — Calculate the electrostatic turning point:}\) For a head-on alpha particle: \[ K=\frac{1}{4\pi\varepsilon_0}\frac{2Ze^2}{d} \] Rearranging: \[ d=\frac{2Z}{K}\left(\frac{e^2}{4\pi\varepsilon_0}\right) \] \(\textbf{Substitution:}\) \[ d=\frac{2(79)(1.44)}{7.2}\,\text{fm} \] \[ d=\frac{227.52}{7.2}\,\text{fm} \] \[ d=31.6\,\text{fm} \] \(\textbf{Stage 2 — Test the contact condition:}\) The approximate centre-to-centre separation at physical contact would be the sum of the two effective radii. \[ d_{\text{contact}}=R_N+R_\alpha \] \[ d_{\text{contact}}=7+2 \] \[ d_{\text{contact}}=9\,\text{fm} \] Comparing the two distances: \[ 31.6\,\text{fm}\gt 9\,\text{fm} \] The particle reverses direction before the two objects come into contact. \(\textbf{Physical interpretation:}\) The calculated turning point is an electrostatic distance, not a direct measurement of the nuclear radius. A turning point outside the nuclear surface can provide an upper bound on nuclear dimensions under suitable assumptions. This is why Rutherford scattering can establish that nuclei are very small without requiring the incident alpha particles to strike their surfaces. The numerical result also confirms that the Coulomb turning point occurs well outside the assumed contact distance.

Explanation: \(\textbf{Head-on encounter:}\) Particle P has zero impact parameter, so its motion is directed along the line joining it to the nuclear centre. At its turning point, its speed momentarily becomes zero under the stationary-nucleus approximation. The initial kinetic energy is then entirely stored as repulsive electrostatic potential energy. \[ K_0=\frac{1}{4\pi\varepsilon_0}\frac{2Ze^2}{d_P} \] \(\textbf{Oblique encounter:}\) Particle Q approaches with a nonzero impact parameter and therefore possesses angular momentum about the nuclear centre. The Coulomb force acts along the line joining the particles, so it exerts no torque about that centre. Consequently, Q retains nonzero angular momentum throughout the encounter. At its nearest point, the radial component of velocity becomes zero, but its tangential velocity remains nonzero. Therefore, some kinetic energy is still present. \[ K_0=K_{\text{tangential}}+\frac{1}{4\pi\varepsilon_0}\frac{2Ze^2}{d_Q} \] Since \(K_{\text{tangential}}\gt0\): \[ \frac{1}{4\pi\varepsilon_0}\frac{2Ze^2}{d_Q}\lt K_0 \] Using the head-on turning-point relation: \[ \frac{1}{d_Q}\lt \frac{1}{d_P} \] Thus: \[ d_Q\gt d_P \] \(\textbf{Physical conclusion:}\) The head-on trajectory allows all incident kinetic energy to oppose the Coulomb repulsion at the turning point. For a nonzero impact parameter, some energy remains associated with transverse motion. Therefore, the oblique particle cannot approach as closely as the head-on particle when their initial energies and target nucleus are identical. This distinction also explains why a point of closest approach does not necessarily correspond to a point where the particle is momentarily at rest.

Explanation: Statement I is correct because Rutherford's nuclear model places almost the entire atomic mass in a small nucleus. The electron masses contribute only a small fraction of the total atomic mass. Statement II is correct because a large deflection of a positively charged alpha particle requires a strong repulsive electric field. Such strong deflections are naturally explained by a small region containing concentrated positive charge. Statement III is correct because a gold nucleus has charge \(+79e\). An electrically neutral gold atom must therefore contain 79 electrons with total charge \(-79e\). In Rutherford's model, these electrons occupy the region outside the nucleus. Statement IV is incorrect because a predominance of small deflections does not demonstrate a diffuse distribution of positive charge. Instead, it is consistent with a very small nucleus occupying only a tiny fraction of the atomic volume. The relatively rare large-angle events provide the decisive evidence against a uniformly distributed positive charge. Together, the observations support nuclear concentration of charge and mass along with overall atomic neutrality.

Explanation: \(\textbf{Initial orbit:}\) The electron is attracted towards the proton by the Coulomb force. For a circular orbit: \[ \frac{1}{4\pi\varepsilon_0}\frac{e^2}{R^2}=\frac{mv^2}{R} \] Multiplying both sides by \(R/2\): \[ \frac{1}{2}mv^2=\frac{e^2}{8\pi\varepsilon_0R} \] Thus, the kinetic energy is: \[ K=\frac{e^2}{8\pi\varepsilon_0R} \] The electrostatic potential energy is: \[ U=-\frac{e^2}{4\pi\varepsilon_0R}=-2K \] Therefore, the total mechanical energy is: \[ E=K+U=-K \] Hence: \[ E=-\frac{e^2}{8\pi\varepsilon_0R} \] \(\textbf{Stage 1 — Final total energy:}\) Both the total energy and the kinetic energy vary inversely with orbital radius, but total energy is negative. The radius changes from \(R_i=4r\) to \(R_f=r\). \[ \frac{E_f}{E_i}=\frac{R_i}{R_f} \] \[ \frac{E_f}{-2}=\frac{4r}{r}=4 \] \[ E_f=-8\,\text{eV} \] \(\textbf{Stage 2 — Energy radiated:}\) Energy lost by the orbiting electron is emitted as electromagnetic radiation in the classical description. \[ E_{\text{radiated}}=E_i-E_f \] \[ E_{\text{radiated}}=-2-(-8) \] \[ E_{\text{radiated}}=6\,\text{eV} \] \(\textbf{Stage 3 — Kinetic-energy comparison:}\) Because \(K=-E\): \[ K_i=2\,\text{eV},\qquad K_f=8\,\text{eV} \] \[ \frac{K_f}{K_i}=\frac{8}{2}=4 \] \(\textbf{Physical check:}\) The total mechanical energy becomes more negative as the electron spirals inward. At the same time, the electron's kinetic energy increases because a smaller circular orbit requires a greater orbital speed. There is no contradiction: the decrease in electrostatic potential energy is greater than the increase in kinetic energy. The remaining energy is carried away as radiation. This classical behaviour is precisely what creates the atomic-stability problem for Rutherford's model. \(\textbf{Final answer:}\) \[ \boxed{\left(E_{\text{radiated}},\frac{K_f}{K_i}\right)=\left(6\,\text{eV},4\right)} \]

Explanation: The Assertion is true because Rutherford's model explains the experimental evidence for a small, positively charged nucleus but fails to explain important properties of the surrounding electrons. In the classical model, an electron revolves around the nucleus under electrostatic attraction. Even if its speed were momentarily constant, circular motion involves centripetal acceleration because the direction of velocity continuously changes. Classical electromagnetic theory predicts that an accelerated charged particle emits electromagnetic radiation. The revolving electron would therefore lose mechanical energy. Its orbital radius would progressively decrease, eventually leading to collapse into the nucleus. This contradicts the observed stability of atoms. The Reason is consequently true. The radiation problem also affects the predicted atomic spectrum. In the classical orbital picture, the electron's angular frequency changes continuously as its orbit contracts, and classical theory associates the emitted radiation frequency with the orbital frequency. Hence, the model predicts a continuously changing emission frequency rather than the observed set of discrete spectral lines. The same classical radiation mechanism therefore explains both major limitations stated in the Assertion. Thus, both statements are true, and the Reason correctly explains the Assertion.

Explanation: \(\textbf{What must be compared:}\) The number of particles scattered backwards must be compared relative to the total incident particles in each measurement. Absolute backscattering counts alone are insufficient because the incident particle totals differ. \(\textbf{Measurement P:}\) \[ f_P=\frac{12}{24{,}000} \] \[ f_P=5.0\times10^{-4} \] Expressed as a percentage: \[ P_P=5.0\times10^{-4}\times100=0.05\% \] \(\textbf{Measurement Q:}\) \[ f_Q=\frac{18}{60{,}000} \] \[ f_Q=3.0\times10^{-4} \] Therefore: \[ P_Q=3.0\times10^{-4}\times100=0.03\% \] \(\textbf{Comparison of scattering fractions:}\) \[ \frac{f_P}{f_Q}=\frac{5.0\times10^{-4}}{3.0\times10^{-4}} \] \[ \frac{f_P}{f_Q}=\frac{5}{3} \] \(\textbf{Experimental interpretation:}\) Measurement Q records more backward-scattered particles in absolute number because a larger incident population was examined. However, its fraction of backward-scattered particles is smaller. Such normalization is essential when comparing scattering observations obtained with different beam intensities or incident counts. The data establish a difference in the observed scattering fractions, but they do not independently identify its physical cause. Information about incident energy, foil properties and other experimental conditions would be needed to attribute that difference to a particular factor. \(\textbf{Final answer:}\) \[ \boxed{\frac{f_P}{f_Q}=\frac{5}{3}} \]

Explanation: Two different aspects of the Coulomb-scattering process must be considered. First, at the same impact parameter, the higher-energy alpha particle is less strongly deflected by a given repulsive nuclear field. Its greater incident kinetic energy makes the Coulomb interaction less effective at changing its direction of motion. Thus, beam Q undergoes a smaller scattering angle than beam P for the same nonzero impact parameter. Second, the head-on case must be analysed through energy conservation. At closest approach, the incident kinetic energy has been converted into electrostatic potential energy. The relevant relation is: \[ K=\frac{1}{4\pi\varepsilon_0}\frac{2Ze^2}{d} \] For a fixed target nucleus: \[ d\propto\frac{1}{K} \] Therefore: \[ \frac{d_Q}{d_P}=\frac{K_P}{K_Q} \] \[ \frac{d_Q}{d_P}=\frac{K}{3K}=\frac{1}{3} \] The more energetic alpha particle can penetrate farther into the repulsive Coulomb region before turning back in a head-on encounter. There is no contradiction between approaching more closely in a head-on experiment and being deflected through a smaller angle at a fixed nonzero impact parameter. These are different geometrical conditions. Increasing the incident energy produces both stated effects.

Explanation: The Coulomb interaction between an alpha particle and a positively charged nucleus is repulsive. Consequently, the force on the alpha particle always acts away from the nuclear centre along the line joining the two particles. During the inward journey, its velocity is directed towards the nucleus, opposite to the force. The work done by the repulsive force is therefore negative during approach. The alpha particle loses kinetic energy while its electrostatic potential energy increases. At the turning point, its instantaneous velocity becomes zero in the stationary-nucleus approximation. The force, however, remains finite and directed away from the nucleus. Subsequently, the alpha particle moves outward in the same direction as the repulsive force. The force now performs positive work, increasing the particle's kinetic energy. Thus, a change in the sign of work does not require a reversal of the force itself. It follows from the reversal of the particle's displacement direction relative to the force. Both the Assertion and Reason are true, and the Reason supplies the correct physical explanation.

Explanation: \(\textbf{Momentum before and after scattering:}\) Since nuclear recoil is neglected and the interaction is elastic, each alpha particle has the same kinetic energy before and after scattering. Therefore, the magnitude of its momentum remains unchanged. Let this momentum magnitude be \(p\). The initial and final momentum vectors make an angle \(\theta\). \(\textbf{Magnitude of momentum change:}\) The change in momentum is the vector difference between final and initial momentum. \[ \Delta\vec p=\vec p_f-\vec p_i \] Using the cosine rule: \[ (\Delta p)^2=p^2+p^2-2p^2\cos\theta \] \[ (\Delta p)^2=2p^2(1-\cos\theta) \] Using the half-angle identity: \[ \Delta p=2p\sin\frac{\theta}{2} \] \(\textbf{Particle P:}\) \[ \Delta p_P=2p\sin30^\circ \] \[ \Delta p_P=2p\times\frac{1}{2}=p \] \(\textbf{Particle Q:}\) \[ \Delta p_Q=2p\sin60^\circ \] \[ \Delta p_Q=2p\times\frac{\sqrt{3}}{2} \] \[ \Delta p_Q=\sqrt{3}p \] \(\textbf{Required ratio:}\) \[ \frac{\Delta p_Q}{\Delta p_P}=\frac{\sqrt{3}p}{p} \] \[ \frac{\Delta p_Q}{\Delta p_P}=\sqrt{3} \] \(\textbf{Physical check:}\) A larger scattering angle corresponds to a greater change in momentum direction. Even though both particles retain the same momentum magnitude, their momentum changes are different because momentum is a vector. The nucleus, or the support holding it approximately stationary, receives the corresponding impulse. Hence, conservation of the alpha particle's momentum alone must not be assumed. \(\textbf{Final answer:}\) \[ \boxed{\frac{\Delta p_Q}{\Delta p_P}=\sqrt{3}} \]