Why Is Energy Released in Nuclear Decay Equal to Change in Binding Energies?

[sdfsfasdfasf]

Homework Statement

x

Relevant Equations

E = mc^2

If we consider a nuclear decay in which element X decays into Y and an alpha particle, we can see that the total energy released is $c^2(-m(\alpha) -m(y) + m(x))$.

If we replace $m(alpha)$ with $s(alpha) - be(\alpha)/c^2$, and similarly for the other particles, X and Y, we can see that the energy released is equal to$c^2( s(x) - be(x)/c^2 - s(\alpha) + be(alpha)/c^2 - s(y) + be(y)/c^2)$

Tidying up gives $c^2(s(\alpha) +s(y) -s(x)) +be(y) + be(alpha) - be(x)$, we can see that $s(\alpha) +s(y) -s(x)$ is equal to $0$ (imagine breaking up X, Y and Z, Y and Z are clearly made of the same stuff as X, hence same mass), therefore the energy released is equal to the change in binding energies, $be(y) + be(\alpha) – be(x)$.

Now imagine having X in front of you, you put $be(x)J$ of energy in, and form Y and alpha, which releases $be(y) + be(\alpha) J$, leaving a "net" energy release of $be(\alpha) + be(y) - be(x)$. Is this identical to what happens in decay? I don't think so, as the entire nucleus has no need to break down for alpha decay.
So where does the energy required to break off the alpha particle come from?

 

[Steve4Physics]

if we consider a nuclear decay in which element X decays into Y and an alpha particle, we can see that the total energy released is c^2(-m(alpha) -m(y) + m(x)). If we replace m(alpha) with c(alpha) - be(alpha)/c^2, and similarly for the other particles, X and Y, we can see that the energy released is equal to c^2( c(x) - be(x)/c^2 - c(alpha) + be(alpha)/c^2 - c(y) + be(y)/c^2), tidying up gives c^2(c(alpha) +c(y) - c(x)) +be(y) + be(alpha) - be(x), we can see that c(alpha) +c(y) - c(x) is equal to 0 (imagine breaking up X, Y and Z, Y and Z are clearly made of the same stuff as X, hence same mass), therefore the energy released is equal to the change in binding energies, be(y) + be(alpha) - be(x).
Now imagine having X in front of you, you put be(x)J of energy in, and form Y and alpha, which releases be(y) + be(alpha) J, leaving a "net" energy release of be(alpha) + be(y) - be(x). Is this identical to what happens in decay? I don't think so, as the entire nucleus has no need to break down for alpha decay. So where does the energy required to break off the alpha particle come from?

That hurts my eyes!

Please break large blocks of text into paragraphs. And use LateX to make formulae readable

 

[phinds]

What is the difference between a paragraph and a wall of text?

Walls of text are long paragraphs that aren't written, edited, or formatted to be readable on the web. Mostly a mistake of formatting, like bricks packed closely together, the text forms an almost opaque wall.

 

[Steve4Physics]

Walls of text are long paragraphs that aren't written, edited, or formatted to be readable on the web. Mostly a mistake of formatting, like bricks packed closely together, the text forms an almost opaque wall.

Much more poetic than what I wrote!

 

[sdfsfasdfasf]

That hurts my eyes!

Please break large blocks of text into paragraphs. And use LateX to make formulae readable

Is there a beginners guide to Latex anywhere?

 

[Frabjous]

Is there a beginners guide to Latex anywhere?

https://www.physicsforums.com/help/latexhelp/
Look below lower left corner of reply box.

 

[sdfsfasdfasf]

Made some changes, should be readable now!

 

[Steve4Physics]

Made some changes, should be readable now!

If I were splitting it into paragraphs, I'd put a blank line between paragraphs and try to have smaller paragraphs. Something like this:

​

If we consider a nuclear decay in which element X decays into Y and an alpha particle, we can see that the total energy released is c^2(-m(alpha) -m(y) + m(x)).​

​

If we replace m(alpha) with c(alpha) - be(alpha)/c^2, and similarly for the other particles, X and Y, we can see that the energy released is equal to c^2( c(x) - be(x)/c^2 - c(alpha) + be(alpha)/c^2 - c(y) + be(y)/c^2)/​

​

Tidying up gives c^2(c(alpha) +c(y) - c(x)) +be(y) + be(alpha) - be(x), we can see that c(alpha) +c(y) - c(x) is equal to 0 (imagine breaking up X, Y and Z, Y and Z are clearly made of the same stuff as X, hence same mass), therefore the energy released is equal to the change in binding energies, be(y) + be(alpha) – be(x).​

​

Now imagine having X in front of you, you put be(x)J of energy in, and form Y and alpha, which releases be(y) + be(alpha) J, leaving a "net" energy release of be(alpha) + be(y) - be(x). Is this identical to what happens in decay? I don't think so, as the entire nucleus has no need to break down for alpha decay.​

​

So where does the energy required to break off the alpha particle come from?​

Still needs Latexing though!

What does, for example, 'c(alpha)' mean? I assume you intend it to be $c_{\alpha}$. But since 'c' is already used to mean the speed of light, 'c(alpha)' makes no sense and (for me anyway) contributes to making the question impossible to understand.

 

[sdfsfasdfasf]

At the top of my post it says $c()$ means the mass of the constituents. Therefore $c(alpha)$ means the mass of two neutrons + two protons.

 

[Steve4Physics]

At the top of my post it says $c()$ means the mass of the constituents. Therefore $c(alpha)$ means the mass of two neutrons + two protons.

Your post #1 is getting better but here are a few hints for future use.

Using the letter 'c' for mass is not the best idea! I presume you mean the sum of the masses of the unbound constituent particles. Maybe use 'S' for this.

To get a single character subscript, use an underscore.
E.g. x_1 becomes $x_1$.

To get a multiple character subscript, use an underscore and enclose the multiple characters in curly brackets.
E.g. x_{abc} becomes $x_{abc}$.

To get a Greek letter use a forward slash in front of the letter's name.
E.g. \alpha becomes $\alpha$ (lower case)
E.g. \Delta becomes $\Delta$ (upper case)

Combining these operations for example, S_{\alpha} becomes $S_{\alpha}$.

Another example: S_{\alpha} =2m_n + 2m_p becomes $S_{\alpha} =2m_n + 2m_p$.

 

[sdfsfasdfasf]

Ok. Thanks for the tips I will try make some changes now.
Are you able to help answer the question?

 

[Steve4Physics]

Now imagine having X in front of you, you put $be(x)J$ of energy in, and form Y and alpha,

If you supplied $be_x$ to X, you wouldn't form Y and $\alpha$. You would end up with the completely separated neutrons and protons from X. Nothing else.

So where does the energy required to break off the alpha particle come from?

Not sure if this will answer your question, but here goes…

Remember, binding energy is defined as the energy which must be supplied to pull things apart. The bigger a system’s binding energy, the harder it is to pull it apart – i.e. the more stable the system is.

A system naturally tries to change from it’s current state to a more stable state. If this happens, the system releases energy. The amount of energy released equals the increase in binding energy. A bit like a stone falling from a wall (initial state); there is a release of heat and sound when the stone reaches the ground (final state).

In $\alpha$ (and other radioactive) decay, the total binding energy increases for the new arrangement of protons and neutrons. The final total binding energy is bigger than the initial total binding energy. That’s the change which drives the decay – the tendency for the nucleons to evolve to a more stable state. (In $\alpha$ decay, this means some of the nucleons combine to become separate $\alpha$ particle.)

The difference in initial and final total binding energies gives the energy ‘lost’ – mainly in the form of the kinetic energy acquired by the $\alpha$ particle.

For a fuller understanding, it helps to become familiar with the concept of ‘binding energy per nucleon’. You can research this yourself. Radioactive decay happens because a system tends to evolve in a way that increases its stability – binding energy per nucleon is increased.

 

[sdfsfasdfasf]

If you supplied $be_x$ to X, you wouldn't form Y and $\alpha$. You would end up with the completely separated neutrons and protons from X. Nothing else.


Not sure if this will answer your question, but here goes…

Remember, binding energy is defined as the energy which must be supplied to pull things apart. The bigger a system’s binding energy, the harder it is to pull it apart – i.e. the more stable the system is.

A system naturally tries to change from it’s current state to a more stable state. If this happens, the system releases energy. The amount of energy released equals the increase in binding energy. A bit like a stone falling from a wall (initial state); there is a release of heat and sound when the stone reaches the ground (final state).

In $\alpha$ (and other radioactive) decay, the total binding energy increases for the new arrangement of protons and neutrons. The final total binding energy is bigger than the initial total binding energy. That’s the change which drives the decay – the tendency for the nucleons to evolve to a more stable state. (In $\alpha$ decay, this means some of the nucleons combine to become separate $\alpha$ particle.)

The difference in initial and final total binding energies gives the energy ‘lost’ – mainly in the form of the kinetic energy acquired by the $\alpha$ particle.

For a fuller understanding, it helps to become familiar with the concept of ‘binding energy per nucleon’. You can research this yourself. Radioactive decay happens because a system tends to evolve in a way that increases its stability – binding energy per nucleon is increased.

Is the "proof" in my post good enough reasoning for why the change in binding energy is equal to the energy released?

 

[Steve4Physics]

Is the "proof" in my post good enough reasoning for why the change in binding energy is equal to the energy released?

I assume you mean your derivation that the net energy released is $be(\alpha) + be(y) - be(x)$.
Yes, the reasoning/algebra look correct to me.

I would have expressed it using slightly different symbols/notation as:
$E = B_{\alpha} + B_Y - B_X$
[edited]

Don't underestimate the importance of making your work easy for others to follow (and easy for you to be able to check). A good choice of symbols and formats can make a big difference. Also, use upper and lower case consistently, e.g. don't use x and X for the same thing.