A nucleus of helium with mass 5 amu breaks up from rest into a nucleus of ordinary helium (mass = 4 amu) plus a neutron (mass = 1 amu). The energy liberated in the break-up is 0.89 MeV, which is shared (not equally) by the products in the form of kinetic energy.
a) Using the conservation of momentum find the kinetic energy of the neutron.
b) The lifetime of the original nucleus is 1.0E-21 s. Because we are not sure when during this lifetime this reaction will occur (i.e., decay), the uncertainty of the time is equal to the lifetime. What range of values of the neutron kinetic energy might we measure in the laboratory as a result of the uncertainty relationship?
a) Since the He-5 nucleus is initially at rest, the initial momentum is 0. And since even in quantum mechanics, momentum is conserved, that means that the final momentum will also be zero:
1. p(He-5) = 0 = p(n0) + p(He-4),
where p(n0) is the momentum of the free neutron and p(He-4) is the momentum of the product nucleus of He-4. From here on in, I will just refer to the He-4 nucleus as "He."
Rearranging (1), we get
2. p(n0) = -p(He),
and from the definition of momentum, this means
3. m(n0)*v(n0) = -m(He)*v(He).
Plugging in masses in amu, and calling v(n0) = u, and v(He) = U,
4. u = -4U.
We also know the total kinetic energy of the products,
5. K(tot) = K(n0) + ...
How a nucleus of helium breaks up from rest into a nucleus of ordinary helium plus a neutron.