Thursday, February 24, 2011

Mastering Physics: ± Uncertainty in the Atomic Nucleus

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 \times 10^{-15} meters.

Part A
Consider a particle inside the nucleus. The uncertainty Deltax in its position is equal to the diameter of the nucleus. What is the uncertainty Deltap of its momentum? To find this, use \Delta x \Delta p \ge \hbar.
Express your answer in kilogram-meters per second to two significant figures.

  Deltap  = 2.10×10−20
  \rm kg \cdot m/s

Part B
The uncertainty Deltap 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 \Delta p = 2.1 \times 10^{-20} kilogram-meters per second as the minimum momentum of a particle in the nucleus, find the minimum kinetic energy K_min of the particle. Use m=1.7 \times
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.
Express your answer in millions of electron volts to two significant figures.

  K_min  = 0.81
  \rm MeV
Compare this to the normal energy scale for electrons in an atom, which is on the order of single electron volts. The characteristic energy scale for the nucleus seems to be roughly one million times that for electrons in an atom. This difference can be seen in calculations of the energy output per unit mass from coal-burning power plants, which utilize chemical energy (energy associated with the electrons of an atom) compared to the energy output from nuclear reactors (power plants that harness the much higher energies of nuclei).


  1. There is a mistake
    The right equation is ΔxΔp≥ℏ/2
    So the right answer in part a) is two times smaller

  2. Part A is actually 6.63*10^-20