Atoms MCQs with Answers – Part 2 (Class 12 Physics)

Explanation: The impact parameter \(b\) describes how close the incoming alpha particle would pass to the nucleus if it continued along its original straight path. It is measured as a perpendicular distance from the nucleus to that initial line of motion. A small \(b\) means the alpha particle passes close to the nucleus and experiences stronger repulsion. A large \(b\) means the particle passes far from the nucleus and suffers only a small deflection. It is a geometrical measure of approach, not a distance measured after the particle has already scattered.

Explanation: The impact parameter tells how close the alpha particle passes to the nucleus. When \(b\) is small, the alpha particle passes closer to the positive nucleus. Coulomb repulsion is then stronger because the separation becomes smaller. A stronger repulsive interaction changes the direction of the alpha particle more. Therefore, small \(b\) corresponds to larger scattering angle, while large \(b\) corresponds to smaller scattering angle.

Explanation: In a head-on approach, the incoming alpha particle is directed exactly toward the centre of the nucleus. The initial straight-line path passes through the nucleus itself. Therefore, the perpendicular distance from the nucleus to this initial line is zero. Since this perpendicular distance is the impact parameter, the head-on case has \(b=0\). This case gives maximum repulsion and is used in the closest-approach calculation.

Explanation: A larger impact parameter means the alpha particle passes farther from the nucleus. At larger separation, the Coulomb repulsion is weaker and produces a smaller change in direction. A smaller impact parameter means closer approach and stronger deflection. Thus the scattering angle has an inverse qualitative relation with \(b\). The relation is not being treated as a simple straight-line law here; the important physical trend is that closer paths scatter more strongly.

Explanation: Path P is the head-on case, so it has \(b=0\) and produces the strongest repulsive effect. Path Q passes close to the nucleus, so it has a small impact parameter and a large but not maximum deflection. Path R passes far from the nucleus, so it experiences weaker repulsion and a small deflection. The scattering angle increases as the impact parameter decreases. This ordering compares closeness of approach, not the length of the path through the foil.

Explanation: The impact parameter is defined using the incoming path the alpha particle would have followed if there were no deflection. It is a perpendicular distance from the nucleus to that initial straight-line path. The final scattered path depends on the repulsive interaction and is not used to define \(b\). This distinction matters because \(b\) describes the initial geometry of the encounter. The scattering angle is an outcome, while the impact parameter is part of the starting condition.

Explanation: Particle P has the smaller impact parameter. This means its initial path passes closer to the nucleus than the path of particle Q. At closer approach, the Coulomb repulsion is stronger and acts more effectively to change the direction of motion. Since the kinetic energy and target nucleus are the same, the difference in deflection is due to the difference in \(b\). The particle with impact parameter \(b\) is deflected more than the one with \(2b\).

Explanation: The impact parameter measures how close the initial path comes to the nucleus. If \(b\) is large, the alpha particle passes far from the nucleus and the repulsive force produces only a small deflection. If \(b\) is small, the particle passes close to the nucleus and the deflection is larger. In a head-on path, the incoming line goes directly through the nucleus, so \(b=0\). All three statements follow from the same geometrical and Coulomb-force reasoning.

Explanation: In a head-on approach, the alpha particle moves against the repulsive electric field of the nucleus. Its kinetic energy decreases as electrostatic potential energy increases. At the closest approach, the alpha particle is momentarily at rest, so its initial kinetic energy is stored as electrostatic potential energy. The charges are \(+2e\) and \(+Ze\), so their product is \(2Ze^2\). Hence \(K=\frac{1}{4\pi\varepsilon_0}\frac{2Ze^2}{r_0}\), with \(r_0\) appearing in the denominator because Coulomb potential energy varies as \(\frac{1}{r}\).

Explanation: \( \textbf{Closest-approach relation:} \) \[ K=\frac{1}{4\pi\varepsilon_0}\frac{2Ze^2}{r_0} \] \( \textbf{Rearranged form:} \) \[ r_0=\frac{1}{4\pi\varepsilon_0}\frac{2Ze^2}{K} \] \( \textbf{Fixed quantities:} \) The target nucleus is the same, so \(Z\), \(e\), and \(\varepsilon_0\) remain fixed. \( \textbf{Proportionality:} \) \[ r_0\propto \frac{1}{K} \] \( \textbf{Energy change:} \) If \(K\) becomes \(2K\), then \(r_0\) becomes \(\frac{r_0}{2}\). \( \textbf{Physical meaning:} \) A more energetic alpha particle gets closer before being stopped by repulsion. \( \textbf{Final answer:} \) The new closest approach is \(\frac{r_0}{2}\).

Explanation: \( \textbf{Energy relation:} \) \[ K=\frac{1}{4\pi\varepsilon_0}\frac{2Ze^2}{r_0} \] \( \textbf{Rearranged form:} \) \[ r_0=\frac{1}{4\pi\varepsilon_0}\frac{2Ze^2}{K} \] \( \textbf{Kinetic energy fixed:} \) \(K\) does not change. \( \textbf{Charge dependence:} \) \[ r_0\propto Z \] \( \textbf{New nuclear charge:} \) \(Z\) becomes \(3Z\). \( \textbf{New distance:} \) \[ r_0'=3r_0 \] \( \textbf{Physical meaning:} \) A larger positive nuclear charge stops the alpha particle at a larger distance for the same incoming energy. \( \textbf{Final answer:} \) The closest approach becomes \(3r_0\).

Explanation: \( \textbf{Given:} \) \(Z=79\), \(K=5.0\,\text{MeV}\), \(k=9.0\times10^9\,\text{N m}^2\text{C}^{-2}\), \(e=1.6\times10^{-19}\,\text{C}\). \( \textbf{Energy conversion:} \) \[ K=5.0\times1.6\times10^{-13}\,\text{J}=8.0\times10^{-13}\,\text{J} \] \( \textbf{Closest-approach relation:} \) \[ K=k\frac{2Ze^2}{r_0} \] \( \textbf{Rearranged form:} \) \[ r_0=k\frac{2Ze^2}{K} \] \( \textbf{Charge factor:} \) \[ 2Z=2(79)=158 \] \( \textbf{Square of charge:} \) \[ e^2=(1.6\times10^{-19})^2=2.56\times10^{-38}\,\text{C}^2 \] \( \textbf{Numerator:} \) \[ (9.0\times10^9)(158)(2.56\times10^{-38})=3.64\times10^{-26} \] \( \textbf{Closest approach:} \) \[ r_0=\frac{3.64\times10^{-26}}{8.0\times10^{-13}}=4.55\times10^{-14}\,\text{m} \] \( \textbf{Reporting:} \) To two significant figures, \(r_0\approx4.6\times10^{-14}\,\text{m}\). \( \textbf{Final answer:} \) The closest approach is approximately \(4.6\times10^{-14}\,\text{m}\).

Explanation: In a head-on scattering event, the alpha particle gets as close to the nucleus as its kinetic energy allows. The closest approach \(r_0\) gives an upper estimate of the scale of the nuclear region because the alpha particle cannot penetrate closer in this simple Coulomb picture. If \(r_0\) is very small, it supports the idea that the nucleus is tiny compared with the atom. This relation belongs to Rutherford scattering, not to direct spectral-line calculation. It connects energy conservation with nuclear-size estimation.

Explanation: The closest approach is determined by energy conservation. The alpha particle’s initial kinetic energy is converted into electrostatic potential energy at the turning point. Since \(r_0\propto \frac{1}{K}\) for the same nucleus, larger kinetic energy means smaller closest approach distance. Physically, the more energetic alpha particle can move closer to the repelling nucleus before stopping. The distance is not independent of energy because the stopping point is set by the balance between kinetic energy and electric potential energy.

Explanation: For head-on scattering, \[ r_0=\frac{1}{4\pi\varepsilon_0}\frac{2Ze^2}{K} \] \( \textbf{Main proportionality:} \) \[ r_0\propto \frac{Z}{K} \] \( \textbf{Row P:} \) If \(K\) doubles and \(Z\) is unchanged, \(r_0\) becomes half, so P is correct. \( \textbf{Row Q:} \) If \(Z\) doubles and \(K\) is unchanged, \(r_0\) doubles, so Q is correct. \( \textbf{Row R:} \) If both \(Z\) and \(K\) double, the ratio \(\frac{Z}{K}\) is unchanged, so R is correct. \( \textbf{Row S:} \) If \(K\) is halved and \(Z\) is unchanged, \(r_0\) should double, not become half. \( \textbf{Final answer:} \) Row S gives the wrong effect.

Explanation: A large scattering angle requires a strong change in the direction of the alpha particle. If the incoming kinetic energy is higher, the particle has greater momentum and is less easily deflected by the same repulsive interaction. Therefore, fewer particles are scattered through a given large angle when the beam energy is increased. This is consistent with the qualitative Rutherford-scattering trend involving kinetic energy. The change is due to the difficulty of deflecting a more energetic projectile, not due to loss of alpha-particle charge.

Explanation: The nuclear charge is \(+Ze\). A larger value of \(Z\) means a larger positive charge in the nucleus. The repulsive Coulomb force on an alpha particle contains the product of the alpha-particle charge and the nuclear charge. Stronger repulsion increases the chance of larger deflections for comparable incoming particles. This is why the target atomic number affects scattering intensity at larger angles.

Explanation: Rutherford observations showed that most alpha particles pass with little or no deflection. Only a small fraction scatter through moderate angles, and very few scatter through large angles. A rapidly falling \(N\)-versus-\(\theta\) graph represents exactly this pattern. Large angles require close approach to the tiny nucleus, so they are rare. The graph supports a small nuclear region rather than a nucleus filling the whole atom.

Explanation: A thicker foil contains more atomic layers along the path of the alpha beam. More target atoms mean more opportunities for alpha particles to pass near nuclei and scatter. For a qualitative comparison, increasing foil thickness increases the number of scattering events. This does not mean that thick foil is preferred for clean interpretation, because too much thickness can introduce multiple scattering. The trend with thickness concerns event count, while thin foil is chosen to keep the geometry simpler.

Explanation: Increasing target \(Z\) strengthens nuclear repulsion, so large-angle scattering becomes more likely. Increasing alpha-particle kinetic energy makes the projectile harder to deflect, so scattering into a fixed large angle becomes less likely. Increasing foil thickness moderately gives more target atoms and can increase the number of scattering events. However, the count decreases sharply as scattering angle increases; it does not increase sharply. Row S reverses the observed angular trend of Rutherford scattering.