To illustrate this point, we can compute, from the known masses of each particle
and the parent and daughter nuclei, the energy Qreleased when various particles are
emitted by a heavy nucleus. This is given byQ(mimfmx)c^2 (12.10)where mimass of initial nucleus
mfmass of final nucleus
mxparticle massWe find that the emission of an alpha particle in some cases is energetically possible,
but other decay modes would need energy supplied from outside the nucleus. Thus
alpha decay in^23292 U is accompanied by the release of 5.4 MeV, while 6.1 MeV would
be needed for a proton to be emitted and 9.6 MeV for a^32 He nucleus to be emitted.
The observed disintegration energies in alpha decay agree with the predicted values
based upon the nuclear masses involved.
The kinetic energy KEof the emitted alpha particle is never quite equal to the
disintegration energy Qbecause, since momentum must be conserved, the nucleus
recoils with a small amount of kinetic energy when the alpha particle emerges. It is
easy to show (see Exercise 23) from momentum and energy conservation that KEis
related to Qand the mass number Aof the original nucleus byKE Q (12.11)The mass numbers of nearly all alpha emitters exceed 210, and so most of the disin-
tegration energy appears as the kinetic energy of the alpha particle.Example 12.7
The polonium isotope^21084 Po is unstable and emits a 5.30-MeV alpha particle. The atomic mass
of^21084 Po is 209.9829 u and that of^42 He is 4.0026 u. Identify the daughter nuclide and find its
atomic mass.
Solution
(a) The daughter nuclide has an atomic number of Z 84 2 82 and a mass number of
A 210 4 206. Since Z82 is the atomic number of lead, the symbol of the daughter
nuclide is^20682 Pb.
(b) The disintegration energy that follows from an alpha-particle energy of 5.30 MeV isQ KE(5.30 MeV)5.40 MeV
The mass equivalent of this Qvalue ismQ0.0058 uHence
mfmimmQ209.9829 u4.0026 u0.0058 u205.9745 u5.40 MeV
931 MeV/u210
210 4A
A 4A 4
AAlpha-particle
energyDisintegration
energyNuclear Transformations 433
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