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
General rule of thumb: if n/p ratio is more the 1.5, the preffered mode of decay is beta and if the n/p ratio is less than 1.5 the heavy elements decay by alpha mode and the lighter elements by postiron decay mode. If the
a) Mo-109
Mass no: 109
Atopmic number : 42
No of neutrons (n): (109-42) = 67; No of protons (p): 42
n/p = 67/42 = 1.59
when the n/p ratio is high, beta decay is preffered mode of decay.
b) Fr- 202
Mass no: 202
Atomic number : 87
No of neutrons (n): (202-87) = 115; No of protons (p): 87
n/p = 115/87 = 1.32
when the n/p ratio is less, positron decay is mode of decay. It is a heavy element, it will prefer alpha decay mode
c) Rn- 196
Mass no: 196
Atomic number : 86
No of neutrons (n): (196-86) = 110; No of protons (p): 86
n/p = 110/86 = 1.27
when the n/p ratio is less and its a heavy element and it will prefer alpha decay mode. some of them will also decay by positron decay.
d) Sb - 132
Mass no: 132
Atomic number : 51
No of neutrons (n): (132-51) = 81; No of protons (p): 51
n/p = 81/51 = 1.58
The n/p ratio is high , the preferred mode of decay is by beta decay.
e) P - 27
Mass no: 27
Atomic number : 15
No of neutrons (n): (27-15) = 12; No of protons (p): 15
n/p = 12/15 = 0.8
when the n/p ratio is less the decay mode is by positron decay.
f) Ru - 90
Mass no: 90
Atomic number : 44
No of neutrons (n): (90-44) = 46; No of protons (p): 44
n/p = 46/44 = 1.04
when the n/p ratio is less the decay mode is by positron decay.
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