If the Thomson model of the atom was correct, then the result of Rutherford's gold foil experiment would have been:

In the Thomson “plum-pudding’’ model the atom is imagined as a sphere of positive charge whose magnitude is spread out uniformly over the whole atomic volume, while the negatively charged electrons are embedded like small “plums’’ inside this continuous positive “pudding”.

Because the positive charge is not concentrated at a point but is smeared out uniformly, an incoming $$\alpha$$-particle (which itself carries a positive charge $$+2e$$) would experience two effects:

1. All along its path inside the atom it would feel the repulsive electrostatic force due to this diffuse positive charge. Since the force acts continuously, the $$\alpha$$-particle would lose a little kinetic energy throughout the traversal, therefore its speed would become slightly less on emerging from the foil.

2. The direction of the electrostatic force would keep changing gradually as the $$\alpha$$-particle moves, because at each instant the net electric field inside the uniform positive sphere points toward the centre (this follows from Gauss’s law for a uniformly charged sphere). Consequently, instead of one sharp “kick’’ the particle would suffer a series of tiny deflections that, when summed, amount to only a small overall angular deviation from its original straight-line path.

Mathematically, inside a uniformly charged sphere of radius $$R$$ and total positive charge $$+Ze$$, Gauss’s law gives the magnitude of the electric field at a distance $$r<R$$ from the centre as

$$E(r)=\frac{1}{4\pi\varepsilon_0}\,\frac{Z e\, r}{R^3},$$

which is directly proportional to $$r$$. Hence the force on an $$\alpha$$-particle of charge $$+2e$$ is

$$F(r)=2e\,E(r)=\frac{1}{4\pi\varepsilon_0}\,\frac{2 Z e^{2}\, r}{R^{3}}.$$

This force always points radially outward, so while the $$\alpha$$-particle traverses the atom it is slowed down (its kinetic energy decreases) and its trajectory bends slightly outward, but because the field grows only linearly with $$r$$ the maximum force it ever feels is modest. Therefore the cumulative deflection angle is small, and there is no possibility of a large-angle ( >$$90^\circ$$ ) scattering or a full $$180^\circ$$ rebound.

Putting the two conclusions together, the Thomson model predicts that:

• every $$\alpha$$-particle will pass through the gold foil;
• each particle will be deflected only by a small angle;
• each particle will emerge with a speed a little less than the incident speed because of the continual but weak electrostatic repulsion inside the diffuse positive charge.

This description is precisely what is stated in Option D: “$$\alpha$$-particles pass through the gold foil deflected by small angles and with reduced speed.” The other options conflict with one or more of the above inferences. In particular, Option A ignores the expected slight energy loss, Option B speaks of large-angle scattering, and Option C predicts total back-scattering—none of which align with the consequences of a diffuse positive charge distribution.

Hence, the correct answer is Option D.