Alpha Decay

Heavy nuclei often decay by emitting an a particle. The Z of the emitting nucleus is decreased by 2, and a helium nucleus is observed with a definite kinetic energy. We can write the alpha decay in terms of this general decay equation:

X(Z,N) -> X'(Z-2,N-2) + 4He

where X is a nucleus with Z protons and N neutrons. X' is another nucleons that has two less protons and neutrons each.

Let's take the case of americium, Z = 95, (243Am) which is a man-made isotope often used in commercial smoke detectors. The decay can be written as

243Am -> 239Np + 4He or 243Am -> 239Np + a

The masses of the nuclei in the reaction are:

 243Am: 243.061374 u 239Np: 239.052932 u 4He: 4.002603 u

The energy released in the reaction can be calculated as

(243.061374 u - 239.052932 u - 4.002603 u) = 0.005839 u

You see that these mass differences are quite small when expressed in terms of the mass unit u. (Tip:) Therefore you need to carry as many decimal places as possible in your calculations. We can convert this mass difference to the energy unit MeV via:

0.005839 u·931.5 MeV/u = 5.439 MeV.

This energy takes the form of kinetic energy of the a particle and the Np nucleus.

Because the a particle is much lighter than the Np nucleus, the a particle gets most of the energy. Momentum is conserved in the decay, and one can easily show that the kinetic energy of the Np nucleus is approximately

E(Np) = 5.439 MeV·(4/243) = 0.895 MeV,

and the kinetic energy of the a particle is

E(a) = 5.439 MeV·(239/243) = 5.35 MeV.

Derivation:

This division of the total excitation energy into the kinetic energies of the two emitted particles can be shown by using momentum and energy conservation laws, as follows:

1. Conservation of energy:
E* = E1 + E2 = p12 / 2m1 + p22 / 2m2
2. Conservation of momentum: The momenta of the two emitted particles have to be equal and opposite:
p1 = -p2
3. We therefore have for the ratio
E1 / E2 = (p12 / 2m1) / (p22 / 2m2) = (p12 / 2m1) / (p12 / 2m2) = m2 / m1
-> E1 = E2 m2 / m1
4. We can now use this result and plug it into the first equation:
E* = E1 + E2 = E2 m2 / m1 + E2 = E2 (1 + m2 / m1) = E2 ( m1 + m2) / m1
-> E2 = E * · m1 / ( m1 + m2)
5. In the same way, we get
E1 = E * · m2 / ( m1 + m2)
6. Now all we have to do is plug in the numbers for the excitation energy and the masses to get the above results.