14
2 Radioactivity
2.2 Emission of Nuclear Particles
Radioactive atoms possess an excess of either protons or neutrons than required to
be a stable isotope. The easiest way to remove excess neutrons or protons would
seem to be by ejecting them directly from the nucleus. But in reality, such processes
are very rare. Why? Since radioactive decay is a natural process, the unstable nuclei
must be able to do this without receiving energy from outside. Neutron or proton
emission is not possible unless energy equivalent to the binding energy of the ejected
proton or neutron is put into the nucleus. For example, the average binding energy
per nucleon of the last few nucleons for elements of mass number between 209 (i.e.,
209 Bi) and 238 (i.e.,
238 U) is about 5.5 MeV and for lower mass number elements, it
is about 7–8 MeV. Hence, the acquisition of energy equivalent to this (i.e., 7–8 MeV)
would be necessary for one nucleon to be expelled out of the nucleus. However,
on the other hand, if two protons along with two neutrons are to be removed from
the nucleus of a heavy element, then an energy equivalent to about 22 MeV would
be required. This energy is less than the binding energy of an α-particle (i.e., about
28.28 MeV as calculated earlier). Thus, the ejection of an α-particle from the nucleus
of a heavy element is exoergic and spontaneous, and α-decay is able to take place
rather than decay by the emission of either protons or neutrons. Radioactive decay
for nuclides of lower mass number such as, α-decay is usually not feasible, owing
to higher binding energy of nucleons. In order to explain these two aspects of the
decay process, let’s take two examples: decay of
238 U and
39 K isotopes.
238 U 92
→
234 Th 90
+
4 He 2
(238.05076 a.m.u.)
(234.043570 a.m.u.) (4.002603 a.m.u.)
The difference in mass between the parent (
238 U 92 ) nuclei and daughter nuclei
(
234 Th 90 ) is 0.0045 a.m.u. This excess mass is equivalent to 4.26 MeV. Therefore,
it is possible for this isotope to decay by an α-decay process energetically. On the
other hand,
39 K
39 K 19
→
35 Cl 17
+
4 He 2
(38.971458 a.m.u.)
(34.9688545 a.m.u.) (4.002603 a.m.u.)
cannot spontaneously undergo α-decay because the total mass of the two daughter nuclei (
35 Cl 17 +
4 He 2 ) is heavier (i.e., 38.97l458 a.m.u.) than the parent nuclei
(
39 K 19 ) (i.e., 34.9688545 a.m.u.), and so the available energy for transition is negative. These two examples suggest that on the basis of mass calculation, it is possible
to theoretically predict the possibility of α-decay of any isotope.
The process of α-decay can be symbolically represented as
A K z →
A−4 Y z−2 +
4 He 2
2 Radioactivity
2.2 Emission of Nuclear Particles
Radioactive atoms possess an excess of either protons or neutrons than required to
be a stable isotope. The easiest way to remove excess neutrons or protons would
seem to be by ejecting them directly from the nucleus. But in reality, such processes
are very rare. Why? Since radioactive decay is a natural process, the unstable nuclei
must be able to do this without receiving energy from outside. Neutron or proton
emission is not possible unless energy equivalent to the binding energy of the ejected
proton or neutron is put into the nucleus. For example, the average binding energy
per nucleon of the last few nucleons for elements of mass number between 209 (i.e.,
209 Bi) and 238 (i.e.,
238 U) is about 5.5 MeV and for lower mass number elements, it
is about 7–8 MeV. Hence, the acquisition of energy equivalent to this (i.e., 7–8 MeV)
would be necessary for one nucleon to be expelled out of the nucleus. However,
on the other hand, if two protons along with two neutrons are to be removed from
the nucleus of a heavy element, then an energy equivalent to about 22 MeV would
be required. This energy is less than the binding energy of an α-particle (i.e., about
28.28 MeV as calculated earlier). Thus, the ejection of an α-particle from the nucleus
of a heavy element is exoergic and spontaneous, and α-decay is able to take place
rather than decay by the emission of either protons or neutrons. Radioactive decay
for nuclides of lower mass number such as, α-decay is usually not feasible, owing
to higher binding energy of nucleons. In order to explain these two aspects of the
decay process, let’s take two examples: decay of
238 U and
39 K isotopes.
238 U 92
→
234 Th 90
+
4 He 2
(238.05076 a.m.u.)
(234.043570 a.m.u.) (4.002603 a.m.u.)
The difference in mass between the parent (
238 U 92 ) nuclei and daughter nuclei
(
234 Th 90 ) is 0.0045 a.m.u. This excess mass is equivalent to 4.26 MeV. Therefore,
it is possible for this isotope to decay by an α-decay process energetically. On the
other hand,
39 K
39 K 19
→
35 Cl 17
+
4 He 2
(38.971458 a.m.u.)
(34.9688545 a.m.u.) (4.002603 a.m.u.)
cannot spontaneously undergo α-decay because the total mass of the two daughter nuclei (
35 Cl 17 +
4 He 2 ) is heavier (i.e., 38.97l458 a.m.u.) than the parent nuclei
(
39 K 19 ) (i.e., 34.9688545 a.m.u.), and so the available energy for transition is negative. These two examples suggest that on the basis of mass calculation, it is possible
to theoretically predict the possibility of α-decay of any isotope.
The process of α-decay can be symbolically represented as
A K z →
A−4 Y z−2 +
4 He 2