What is the distance of closest approach when a 5.0 MeV proton approaches a gold nucleus?

The distance of closest approach refers to the point at which the proton's kinetic energy is entirely converted into electrostatic potential energy due to the Coulomb interaction with the gold nucleus. At this point, the proton momentarily comes to rest before being repelled by the gold nucleus. We can calculate this distance using energy conservation principles, where the initial kinetic energy of the proton is equal to the electrostatic potential energy at the closest approach.

Step 1: Use the Energy Conservation Equation

The total mechanical energy of the proton at the closest approach is conserved. The proton starts with kinetic energy, which is converted into electrostatic potential energy as it approaches the nucleus. Thus, the equation for energy conservation can be written as:

Where:* K.E. is the kinetic energy of the proton.* U is the electrostatic potential energy between the proton and the nucleus.

Step 2: Write the Expression for Kinetic Energy

The kinetic energy of the proton is given by:

Step 3: Write the Expression for Electrostatic Potential Energy

The electrostatic potential energy between two charged particles is given by Coulomb’s law:Where:* k\_e is Coulomb’s constant (8.99 × 10⁹ N·m²/C²).* Z is the atomic number of gold (Z = 79 for gold).* e is the elementary charge (1.6 × 10⁻¹⁹ C).* r is the distance of closest approach, which we are solving for.

Step 4: Set Kinetic Energy Equal to Potential Energy

At the point of closest approach, all the proton’s kinetic energy is converted into potential energy:Simplify:Convert the units of energy into joules for consistency:1 eV = 1.6 × 10⁻¹⁹ J, so:Now, solving for the distance of closest approach $r$:Simplify:

Conclusion

The distance of closest approach of the 5.0 MeV proton to the gold nucleus is approximately 2.27 femtometers (fm).