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Predict a likely mode of decay for each of the following unstable nuclides. EXPLAIN why this might be so.
a) Mo-109
b) Fr- 202
c) Rn- 196
d) Sb – 132
e) P – 27
f) Ru – 90
There are a variety of radioactive decay of particles in unstable isotopes of elements such as proton neutron, electron/beta and positron particle decay (gamma radiation, which is not a particle type that emits radiation). There are many isotopes that exist for elements, there’s an area on a graph of the number of neutrons in relation to the amount of protons found in the nucleus of an isotope, which is called the stability zone. This is where the ratio of protons to neutrons creates an isotope that is stable. The stability zone is primarily over the 1:1 neutron to proton ratio line. The kind of decay an unstable isotope undergoes will depend on whether it’s either above or below the stabilization zone of neutron to proton ratio for the element.
The general rule of thumb is that if more the 1.5 is the ratio, beta is the preferred mode of decay. If less than 1.5 is the ratio, heavy elements will decay by alpha mode while lighter elements will decay by postiron. If the ratio is more the 1.5, the preferred mode of decay is beta. If the ratio is less than 1.5, then heavy elements will begin to decay by alpha mode and lighter elements by postiron.
a) Mo-109
Mass no. 109
Number of Atopmics: 42
Number of neutrons (n), (109-42) =67; No protons (p), 42
n/p = 67/42 = 1.59
When the / p ratio is high then beta decay will be preferred mode of decay.
b) Fr- 202
Mass no. 202
Atomic number: 87
Number of neutrons (n), (202-87) =115; Number of protons (p), 87
n/p = 115/87 = 1.32
When the / p-ratio is more, then positron decomposition is in mode. This element is heavy and will prefer alpha decay.
c) Rn- 196
Mass no. 196
Atomic number: 86
No of neutrons, (n) = (196-86), 110; No protons (p), 86
n/p = 110/86 = 1.27
When the / ratio is more, and it’s a heavy element, it will prefer alpha decay mode. Some of them may also undergo positron decay.
d) SB – 132
Mass no. 132
Atomic number 51
No of neutrons, (n) = (132-51) = 881; No protons (p),: 51
n/p = 81/51 = 1.58
The / ratio is high. The preferred mode of decay for this is beta decay.
e) P – 27
Mass no. 27
Atomic number: 15
Number of neutrons (n), (27-15) = 12, No of protons, (p): 15.
n/p = 12/15 = 0.8
When the / ratio is more, the decay mode is positron decay.
f) Ru 90
Mass no. 90
Atomic number 44
Number of neutrons (n), (90-44) = 46. No protons (p), 44
n/p = 46/44 = 1.04
When the / ratio is more, the decay mode is positron decay.