Question:

Why do only a few \( \alpha \)-particles rebound from the gold nucleus in Rutherford's experiment?
S1: Size of gold nucleus is very small compared to the gold atom.
S2: Impact parameter of \( \alpha \)-particles is very small.
S3: Nuclear charge of \( \mathrm{He^{2+}} \) particles is very small compared to gold.
S4: Very few \( \alpha \)-particles undergo head-on collision.
Then correct statements are:

Updated On: Apr 6, 2026
  • S1 and S2
  • S1, S2 and S3
  • S1, S2 and S4
  • S1 and S3
Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is C

Solution and Explanation

Step 1: Understanding the Concept:
Rutherford's alpha-scattering experiment showed that most of the atom is empty space. Rebounding (scattering at large angles or \( 180^\circ \)) only occurs when the alpha particle comes extremely close to the dense, positively charged nucleus.

Step 2:
Key Formula or Approach:
The number of scattered particles is related to the cross-sectional area of the nucleus and the impact parameter (\( b \)). Large-angle scattering implies a very small impact parameter.

Step 3:
Detailed Explanation:
- S1: True. The nucleus is about \( 10^{-5} \) times the size of the atom, making the "target" for a head-on collision extremely small. - S2: True. Large-angle scattering occurs only for very small impact parameters (the perpendicular distance from the center of the nucleus to the initial velocity vector). - S4: True. A "rebound" is the result of a head-on or near head-on collision. Since the nucleus is tiny, such collisions are rare. - S3: Irrelevant. While true that the gold nucleus has a much higher charge (Z=79) than the alpha particle (Z=2), this explains the force of repulsion, not the frequency or probability of rebounding.

Step 4:
Final Answer:
The correct statements are S1, S2, and S4.
Was this answer helpful?
0
0