Question

Rutherford's scattering experiments gave the first indications that an atom consists of a small, dense, positively charged nucleus surrounded by negatively charged electrons. His experiments also allowed for a rough determination of the size of the nucleus. In this problem, you will use the uncertainty principle to get a rough idea of the kinetic energy of a particle inside the nucleus. Consider a nucleus with a diameter of roughly 5.0×10−15 meters.

Part A: The uncertainty Δp sets a lower bound on the average momentum of a particle in the nucleus. If a particle's average momentum were to fall below that point, then the uncertainty principle would be violated. Since the uncertainty principle is a fundamental law of physics, this cannot happen. Using Δp=1.06×10−20 kilogram-meters per second as the minimum momentum of a particle in the nucleus, find the minimum kinetic energy Kmin of the particle. Use m=1.67×10−27 kilograms as the mass of the particle. Note that since our calculations are so rough, this serves as the mass of a neutron or a proton.

Answer 1

The uncertainty in particle momentum is, h Δκ.Ap > 2л _ The change in momentum is, Ap mv Ap v = 1.06x 10 20 1.67x10 kg.m/s -27 kg 6.35x 106 m/s The minimum kinetic energy is 1 K -m 2 167x10 ke)(6.35-10' ms (1.67x10 kg) (6.35x10 m/s) 3.37x1014 J -14 J Hence, the minimum kinetic energy is 3.37 x10