Question

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

Answer 1

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.