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Transcribed Image Text:**Problem Statement:** How much kinetic energy must an alpha particle have before its distance of closest approach to a gold nucleus is equal to the nuclear radius (7.0 fm)? **Explanation:** To solve this problem, one must understand the concept of the distance of closest approach, which relates to how close an alpha particle can get to a nucleus before kinetic energy is completely converted to electrical potential energy due to repulsion. The gold nucleus has a given nuclear radius of 7.0 femtometers (fm). The kinetic energy can be calculated using formulas from classical physics involving the conservation of energy and electrostatic forces.

Expert Solution

Concept and Principle:

When an alpha particle that has a positive charge gets close to the gold nucleus which is also positive, the kinetic energy of the alpha particle will reduce, and its potential energy will increase.

The distance at which the kinetic energy of the particle becomes zero is called the distance of the closest approach. It is given by,

$d = \frac{1}{4 π ε_{0}} \frac{z Z e^{2}}{K}$

Here ε₀ is the permittivity of free space, z is the atomic number of the projectile, Z is the atomic number of the target, e is the charge of the electron, and K is the kinetic energy of the projectile.

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