Question

When an $\alpha$-particle of mass 'm' moving with velocity 'v' bombards on a heavy nucleus of charge 'Ze', its distance of closest approach from the nucleus depends on m as :

• A. $\frac{1}{\sqrt{m}}$
• B. $\frac{1}{m^2}$
• C. m
• D. $\frac{1}{m}$

Answer 1

The Correct Option is D

Kinetic Energy and Electrostatic Potential Energy 

At the closest distance of approach, the kinetic energy of the particle gets completely converted into electrostatic potential energy. This can be represented by the following equation:

\(\frac{1}{2} mv^2 = \frac{KQq}{d}\)

Where:

• \( m \) is the mass of the particle,
• \( v \) is the velocity of the particle,
• \( K \) is Coulomb's constant,
• \( Q \) and \( q \) are the charges involved,
• \( d \) is the closest distance of approach.

 

Rearranging the equation to solve for \( d \), we get:

\(d \propto \frac{1}{m}\)

Conclusion:

This means that the closest distance of approach \( d \) is inversely proportional to the mass \( m \) of the particle. As the mass increases, the closest distance of approach decreases.