PAC without hyperfine field

homepage Forums PAC curves without hyperfine interaction PAC without hyperfine field

Viewing 1 post (of 1 total)
  • Author
    Posts
  • #3202 Score: 0
    mmvdwiel
    Participant
    1 pt

    1a)
    As indicated by the Heisenberg uncertainty principle, the chance that a particle with a certain energy level will decay at a given time is constant and given by the energy difference between the excited and final state.
    As such, the total amount dN/dt of nuclei in a sample decaying will be proportional to the excited amount of nuclei N left and the constant chance L of decaying:

    -dN/dt = LN

    This means the amount of nuclei that send out gamma rays are:
    N_0exp(-Lt)

    Of course only a fraction of the emitting gamma’s will be detected by detector 1 and 2, but because in the absence of a hyperfine field, the nuclear moments will not change directions, no oscillations will occur, and no deviation from the exponential decay can be noticed.

    1b)
    Due to the anisotropy of the angular distribution of the radiated gamma rays, most of the gamma’s will be detected in the same direction (parallel or antiparallel) as the nuclear moment. As the nuclear spin doesn’t change direction in the absence of a hyperfine field, most of the gamma’s that are radiated in the second decay will radiate in the same direction (parallel or antiparallel) as the first decay a few ns before. As detector 1 and detector 3 are in the same direction, and detector 2 is perpendicular to both, most of the gamma2’s will be observed by the third detector, what can be seen on the graphs.

    Of course, this is only so if only gamma’s are detected, which have a dipole distribution. The same cannot be said for beta-radiation, as this has an asymmetrical distribution.

Viewing 1 post (of 1 total)
  • You must be logged in to reply to this topic.