Question

The distance of closest approach of an alpha particle to a nucleus when the alpha particle moves towards the nucleus with linear momentum P is d. The distance of closest approach of alpha particle to nucleus, if the linear momentum of the alpha particle is 1.5 P

• A. \( \frac{2d}{3} \)
• B. \( \frac{3d}{2} \)
• C. \( \frac{4d}{9} \)
• D. \( \frac{9d}{4} \)

Hint:

The distance of closest approach is determined by the conservation of energy. The initial kinetic energy is converted into potential energy at the closest approach. Remember that kinetic energy is related to linear momentum by \( K = P^2 / (2m) \), and the electrostatic potential energy is inversely proportional to the distance. The distance of closest approach is inversely proportional to the square of the initial linear momentum.

Answer 1

The Correct Option is C

The distance of closest approach occurs when the initial kinetic energy of the alpha particle is completely converted into electrostatic potential energy at the closest distance from the nucleus.
The initial kinetic energy \( K \) of the alpha particle with linear momentum \( P \) and mass \( m \) is given by: $$ K = \frac{P^2}{2m} $$ The electrostatic potential energy \( U \) between the alpha particle (charge \( 2e \)) and the nucleus (charge \( Ze \)) at a distance \( r \) is given by: $$ U = \frac{k (2e) (Ze)}{r} = \frac{2kZe^2}{r} $$ where \( k \) is Coulomb's constant.
At the distance of closest approach \( d \), the kinetic energy is equal to the potential energy: $$ \frac{P^2}{2m} = \frac{2kZe^2}{d} $$ From this, we can express \( d \) in terms of \( P \): $$ d = \frac{4mkZe^2}{P^2} $$ So, the distance of closest approach \( d \) is inversely proportional to the square of the linear momentum \( P \): $$ d \propto \frac{1}{P^2} $$ Now, let the initial linear momentum be \( P_1 = P \) and the distance of closest approach be \( d_1 = d \).
Let the new linear momentum be \( P_2 = 1.
5 P \) and the new distance of closest approach be \( d_2 \).
Using the proportionality: $$ \frac{d_2}{d_1} = \left( \frac{P_1}{P_2} \right)^2 $$ $$ \frac{d_2}{d} = \left( \frac{P}{1.
5 P} \right)^2 $$ $$ \frac{d_2}{d} = \left( \frac{1}{1.
5} \right)^2 = \left( \frac{1}{3/2} \right)^2 = \left( \frac{2}{3} \right)^2 $$ $$ \frac{d_2}{d} = \frac{4}{9} $$ $$ d_2 = \frac{4}{9} d $$ The distance of closest approach when the linear momentum is \( 1.
5 P \) is \( \frac{4d}{9} \).