Why does decay increase Binding energy ?

[kenshi64]

Can someone explain this excerpt from my textbook please?
"IF nucleus A was made from its constituents (Binding energy) and it released 400MeV, but making nucleus B(product) releases 405MeV. Therefore if A changes to B, 5 MeV of energy must be released. This excess energy is given to the particle emitted"

Thanks guys! :D

 

[sambristol]

Binding energy is a negative number, its just a (confusing) convention to leave out the minus sign. Everything in nature tries to decrease its energy so A decays to B and goes from -400MeV to -405MeV which is lower. The sum reads -400 - (-405) = 5 MeV released

 

[Bill_K]

Everything in nature tries to decrease its energy

I'm sure you don't mean that, sambristol. There is no such law! Probably what you're thinking of is the tendency for energy to redistribute itself so as to increase the total entropy.

 

[Lobezno]

That's the layman's term he used...

When an atom is over excited, it will attempt to jump down to a lower energy state (think Uranium decaying by gamma radiation).

 

[kenshi64]

Yeah I think I got it guys! sambristol's reply hold true for me, since I'm not studying too hi-fi physics, but it is right, right?
"to go to a higher BE(binding energy) an atom has to lose energy'' now it makes sense since, as higher BE is actually more negative! :D
PLEASE tell me if I'm right, doubts in Physics just kill me!

 

[sambristol]

Agreed Bill_K but I wanted to leave entropy out of it but in practice the effect is the same.

Kenshei64 you have got it

Regards
Sam

 

[Dickfore]

I'm sure you don't mean that, sambristol. There is no such law! Probably what you're thinking of is the tendency for energy to redistribute itself so as to increase the total entropy.

How can a nucleus increase its entropy?

 

[Bill_K]

A nuclear decay increases total entropy because there are many states the decay products can occupy.

 

[Dickfore]

A nuclear decay increases total entropy because there are many states the decay products can occupy.

So, even if the reaction was endothermic, you are suggesting that nuclear reactions can happen as long as the entropy of a chunk of matter increases, right?

 

[Bill_K]

Dickfore, You know as well as I do that the kinetic energy of the decay products must be positive, but here's my point. Consider nucleus X with decay products Y and Z. What drives X to decay? Is it because nuclei have a mysterious urge to decrease their energy?

No, in fact energy is conserved. So why does X → Y + Z happen and not the inverse, Y + Z → X? If there were just a single state in the decay, the inverse process would happen just as often, and an equilibrium would be reached. But in fact Y + Z represents a large number of states, and the odds of X → Y + Z happening outweighs the chance of Y + Z → X. Although the inverse reaction is possible, it would be unlikely for Y and Z to come together in just the right way to reform X.

This is exactly the same reason why a collection of air molecules will expand to fill a room rather than congregate in one corner. Both configurations are energetically the same, but one is statistically more likely.

So it is increase in entropy that drives an isolated nucleus to decay, not an urge to decrease its energy. In an environment where Y and Z are plentiful, the inverse reaction would be more likely, resulting in a tendency to form the nucleus X, which has more binding energy, not less.

 

[Dickfore]

No, in fact energy is conserved.

This seems to disagree with the previous statement:

Dickfore, You know as well as I do that the kinetic energy of the decay products must be positive, but here's my point.

It is true that total relativistic energy is conserved, but kinetic energy is NOT. This is because part of the rest energy of the original nucleus is converted into kinetic energy of the products.

We can look at it in another way: Binding energy for a nucleus is defined as:
$$B = \sum{\mathrm{rest \; energies \; of \; protons \; and \; neutrons}} - \mathrm{rest \; energy \; of \; nucleus}$$
This is a positive quantity. The bigger it is, the more stable an isotope is. Why? Because it will take that much energy to disintegrate the nucleus into its free constituents. Conversly, we would gain that much energy if we assembled the nucleus from its constituents.

Now, you can think of a reaction in the following way:
1) Disintegrate nucleus X. You need to spend B(X) amount of energy to do so;

2) Use some of the protons and neutrons to make Y. You gain B(Y) amount of energy;

3) Use the rest of the protons and neutrons to make Z. You gain B(Z) amount of energy.

All in all, you had gained:
$$Q = B(Y) + B(Z) - B(X)$$

amount of energy. If $Q > 0$, the reaction is exothermic and it can spontaneously happen. The energy difference is distributed along the constituents, making sure linear momentum is conserved. If $Q < 0$, the reaction is endothermic and the particle X is stable from decaying through the channel described by:
$$X \rightarrow Y + Z$$

So why does X → Y + Z happen and not the inverse, Y + Z → X? If there were just a single state in the decay, the inverse process would happen just as often, and an equilibrium would be reached. But in fact Y + Z represents a large number of states, and the odds of X → Y + Z happening outweighs the chance of Y + Z → X. Although the inverse reaction is possible, it would be unlikely for Y and Z to come together in just the right way to reform X.

I think you need to specify what exactly you mean by 'state' because:

This is exactly the same reason why a collection of air molecules will expand to fill a room rather than congregate in one corner. Both configurations are energetically the same, but one is statistically more likely.

the notion of state in this case means a point (or a small volume) in the 6N - dimensional phase space of the gas of N molecules. According to the laws of statistical physics, in the limit $N \rightarrow \infty$, there is a uniform probability density of occupying a particular volume in phase space as long as that part of the phase space is consistent with conservation laws (energy, momentum, linear momentum). Entropy is defined as the logarithm of this volume (measured in units of $(2\pi \hbar)^{3 N}$) (measured in units of $k_{B}$ - Boltzmann constant).

Then, the law of increase of entropy means that the equilibrium macroscopic state of a macrocopic system is the one that can be realized by the biggest volume in phase space.

So it is increase in entropy that drives an isolated nucleus to decay, not an urge to decrease its energy. In an environment where Y and Z are plentiful, the inverse reaction would be more likely, resulting in a tendency to form the nucleus X, which has more binding energy, not less.

As I said, entropy (unlike energy) is a concept that has a meaning only for macroscopic systems. A nucleus is not such a system. It is an experimental fact that the half-life of a nucleus is an intrinsic property independent of the distribution of other nuclei around it. Therefore, your analogy is meaningless.

 

[Dickfore]

It is true that total relativistic energy is conserved, but kinetic energy is NOT. This is because part of the rest energy of the original nucleus is converted into kinetic energy of the products.

We can look at it in another way: Binding energy for a nucleus is defined as:
$$B = \sum{\mathrm{rest \; energies \; of \; protons \; and \; neutrons}} - \mathrm{rest \; energy \; of \; nucleus}$$
This is a positive quantity. The bigger it is, the more stable an isotope is. Why? Because it will take that much energy to disintegrate the nucleus into its free constituents. Conversly, we would gain that much energy if we assembled the nucleus from its constituents.

Now, you can think of a reaction in the following way:
1) Disintegrate nucleus X. You need to spend B(X) amount of energy to do so;

2) Use some of the protons and neutrons to make Y. You gain B(Y) amount of energy;

3) Use the rest of the protons and neutrons to make Z. You gain B(Z) amount of energy.

All in all, you had gained:
$$Q = B(Y) + B(Z) - B(X)$$

amount of energy. If $Q > 0$, the reaction is exothermic and it can spontaneously happen. The energy difference is distributed along the constituents, making sure linear momentum is conserved. If $Q < 0$, the reaction is endothermic and the particle X is stable from decaying through the channel described by:
$$X \rightarrow Y + Z$$

BTW, this is what was meant by the excerpt from your book, OP.

 

[sambristol]

Gentlemen,

you are both arguing to the same conclusion from different viewpoints.

The essential physics is the phase space for the decay has larger Hilbert measure ('volume' in lay terms) than the original nucleus.

Therefore the entropy (which is well defined for quantum systems contrary to previous posts) must increase, hence the energy of the decay products will redistribute the energy among themselves.

Cf 'The Gibbs paradox'

I hope this ends this thread.

Regards

Sam