Question

Spontaneous radioactive decay occurs only when the decay products have less mass than the parent, and it tends to produce a daughter that is more stable than the parent. Explain how this is related to the fact that more tightly bound nuclei are more stable. (Consider the binding energy per nucleon.)

a) Higher binding energy per nucleon leads to more stable nuclei.

b) Lower binding energy per nucleon leads to more stable nuclei.

c) Tightly bound nuclei are less stable due to higher energy.

d) Loosely bound nuclei are more stable due to lower energy.

Answer 1

Nuclei with higher binding energy per nucleon are more stable and less likely to undergo spontaneous radioactive decay. This is because the decay products have less mass than the parent nucleus, and the daughter nucleus produced in the decay tends to be more stable due to its higher binding energy per nucleon. Option a is the correct answer.

Step-by-step explanation:

Nuclei with higher binding energy per nucleon are more stable. Higher binding energy means that the nucleus is held together more tightly, making it less likely to undergo spontaneous radioactive decay. This is related to the fact that nuclei with more tightly bound nucleons have greater stability.

When a nucleus undergoes radioactive decay, the decay products have less mass than the parent nucleus. This is because during the decay process, some of the mass is converted into energy. The daughter nucleus that is produced in the decay tends to be more stable because it has a higher binding energy per nucleon than the parent nucleus.

Therefore, the correct option is: a) Higher binding energy per nucleon leads to more stable nuclei.