Question

In the head-on collision of two alpha particles \( \alpha_1 \) and \( \alpha_2 \) with the gold nucleus, the closest approaches are 31.4 fermi and 94.2 fermi respectively. Then the ratio of the energy possessed by the alpha particles \( \alpha_2 / \alpha_1 \) is:

• A. 1 : 3
• B. 9 : 1
• C. 3 : 1
• D. 1 : 9

Hint:

In head-on collisions, the closest approach \( r \) is inversely proportional to the energy \( E \). Hence, the ratio of the energy is the inverse ratio of the closest approaches.

Answer 1

The Correct Option is A

In a head-on collision of charged particles with a nucleus, the closest approach (\( r \)) is inversely proportional to the energy (\( E \)) of the incoming particle. This is based on the concept of the electrostatic potential energy, where: \[ E \propto \frac{1}{r} \] Thus, if the closest approach for the two alpha particles are \( r_1 = 31.4 \, \text{fm} \) and \( r_2 = 94.2 \, \text{fm} \), the ratio of the energy possessed by the particles \( \alpha_2 \) and \( \alpha_1 \) will be: \[ \frac{E_2}{E_1} = \frac{r_1}{r_2} \] Substitute the given values: \[ \frac{E_2}{E_1} = \frac{31.4}{94.2} = \frac{1}{3} \] Thus, the ratio of the energies possessed by the alpha particles is: \[ \frac{E_2}{E_1} = 1 : 3 \]