Question about Nuclear Binding energy and stability

[maverick280857]

Hello all

Here's a question I need some help with:

There are two nuclei X and Y [Binding Energy of X = a and Binding Energy of Y = 2a]. Also Binding Energy per nucleon for X = 2b and Binding Energy per nucleon for Y is b. Then which one of the following is true:

(A) X is always more stable than Y
(B) X is more stable than Y if mass number of X is greater than that of Y
(C) Y is always more stable than X
(D) None of the above

I want to know which one (and why) of the two, i.e. Binding Energy or Binding Energy per nucleon decides the criterion for stability. [I know that Binding Energy is the energy released when a nucleus is formed from its constituents, so the lower the energy released the more stable the nucleus should be. From this line of reasoning, the answer is (A). ]

Thanks for your help..

Cheers
Vivek

 

[Andrew Mason]

Hello all

Here's a question I need some help with:

There are two nuclei X and Y [Binding Energy of X = a and Binding Energy of Y = 2a]. Also Binding Energy per nucleon for X = 2b and Binding Energy per nucleon for Y is b. Then which one of the following is true:

(A) X is always more stable than Y
(B) X is more stable than Y if mass number of X is greater than that of Y
(C) Y is always more stable than X
(D) None of the above

I want to know which one (and why) of the two, i.e. Binding Energy or Binding Energy per nucleon decides the criterion for stability. [I know that Binding Energy is the energy released when a nucleus is formed from its constituents, so the lower the energy released the more stable the nucleus should be. From this line of reasoning, the answer is (A).

Stability is a measure of the probability of a nucleon being released from the nucleus. So it depends on the binding energy per nucleon. If that binding energy is lower, there is a greater probability that a nucleon will acquire sufficient energy to overcome the nuclear force and escape the nucleus.

AM

 

[maverick280857]

Hi Andrew

Thanks for your reply.

The way I see it, the stability of the whole nucleus is being talked of here that is, whether the nucleus will remain as such or will it spontaneously undergo an emission (alpha, positron or beta). The answer given is (A) but I can't figure out a very sound explanation for it. So I think I need to understand why per nucleon energies are important rather than the total energies (or the other way round...whichever is correct!)

I think what your reply implies is that X with a higher binding energy per nucleon is more stable. Is this correct? Is this a valid statement always? Why?

-Vivek

 

[Andrew Mason]

Hi Andrew

Thanks for your reply.

The way I see it, the stability of the whole nucleus is being talked of here that is, whether the nucleus will remain as such or will it spontaneously undergo an emission (alpha, positron or beta). The answer given is (A) but I can't figure out a very sound explanation for it. So I think I need to understand why per nucleon energies are important rather than the total energies (or the other way round...whichever is correct!)

I think what your reply implies is that X with a higher binding energy per nucleon is more stable. Is this correct? Is this a valid statement always? Why?

The higher the (average) nuclear binding energy per nucleon, the more energy is required to remove a nucleon from the nucleus and, hence, the more stable the nucleus.

It gets complicated with large nuclei whose radii approach the limit of the strong nuclear force. For large nuclei, the average nuclear binding energies per nucleon may not be a reliable basis for determining stability as Coulomb forces start competing with nuclear forces.

AM