The graph which depicts the results of Rutherford gold foil experiment with α-particales is :
θ : Scattering angle
Y : Number of scattered α-particles detected (Plots are schematic and not to scale)

Solution

Physics sir · 3 steps

Most alphas fly straight through and a rare few bounce back, so Y falls steeply but never reaches zero.

1. What Rutherford saw. Almost all α-particles passed through the gold foil nearly undeflected, so the count is largest at small θ. A very small number were turned through large angles, and a few came almost straight back (θ close to π).

   Y ∝ 1/sin⁴(θ/2)

2. Read the formula. As θ grows from near 0, sin(θ/2) grows, and its fourth power makes Y drop very sharply. Y at θ = 10° is about 1.7 × 10⁴ times Y at θ = π. At θ = π, sin⁴(π/2) = 1, so Y reaches its smallest value, and that value is not zero.
3. So. The graph must start very high, fall steeply, then flatten out at a small non-zero value up to π. Only graph (c) has this shape.

The idea to keep. Rutherford scattering follows Y ∝ 1/sin⁴(θ/2): a huge count at small angles, a steep drop, and a small but non-zero count at backscattering, which is the evidence for a tiny dense nucleus.

Answer from the NTA final answer key. Solution by PrepWiser.