Question

A proton and an alpha particle having equal velocities approach a target nucleus. They come momentarily to rest and then reverse their directions. The ratio of the distance of closest approach of the proton to that of the alpha particle will be:

• A. \( \frac{1}{2} \)
• B. 2
• C. \( \frac{1}{4} \)
• D. 4

Hint:

For electrostatic potential energy problems involving charged particles, remember that the charge affects the potential energy directly and the mass indirectly through kinetic energy.

Answer 1

The Correct Option is B

Step 1: Analyzing the forces and energies involved. The distance of closest approach can be determined by equating the kinetic energy to the electrostatic potential energy at the point of closest approach: \[ \frac{1}{2} mv^2 = \frac{Zke^2}{r} \] where \( m \) is the mass, \( v \) is the velocity, \( Z \) is the atomic number of the target nucleus, \( k \) is Coulomb's constant, \( e \) is the charge, and \( r \) is the distance of closest approach. Step 2: Comparing proton and alpha particle. The alpha particle has twice the charge of a proton (since it contains two protons) and four times the mass. Since both particles have equal velocities, their kinetic energies differ but the potential energy of the alpha particle is higher due to its greater charge. Step 3: Calculating the ratio of distances. The formula rearranges to \( r = \frac{Zke^2}{mv^2} \). For the alpha particle, both the charge and mass affect the distance: \[ r_{proton}} = \frac{Zke^2}{m_pv^2}, \quad r_{alpha}} = \frac{Zke^2}{4m_pv^2} \] Thus, \( r_{proton}} = 2r_{alpha}} \), giving the ratio of 2.