Calculate the mass deficit in this nuclear reaction

[amazingphysics2255]

Homework Statement

Calculate the mass deficit in this nuclear reaction
$12{\rm C}+\frac{12}{6}{\rm C}\rightarrow \frac{24}{12}{\rm Mg}$
Information given:

- rest mass of a carbon-12 nucleus: $$1.9921157\times 10^{-26}\,{\rm kg}$$

- rest mass of a magnesium nucleus: $$3.9817469\times 10^{-26}\,{\rm kg}$$

- rest mass of a proton: $$1.67353\times 10^{-27}\,{\rm kg}$$

- rest mass of a neutron: $$1.67492\times 10^{-27}\,{\rm kg}$$

- speed of light: $$ 3.00\times 10^{8}\,{\rm m}\,{\rm s}^{-1}$$

Relevant Equations

$$12{\rm C}+\frac{12}{6}{\rm C}\rightarrow \frac{24}{12}{\rm Mg}$$

I have tried using the formula $$Δm=[Zmp+(A−Z)mn−M]$$ But either I plug the wrong thing into the wrong part or it just isn't working. Essentially I want to know the method I need to use. I will then try to solve it and hopefully get it correct.

Thanks

 

[collinsmark]

Hello @amazingphysics2255,

Welcome to PF!

Your $\LaTeX$ isn't working. You need to use an extra dollar sign. Here is a guide to using $\LaTeX$ on PF.

https://www.physicsforums.com/threads/guide-to-using-latex.617570/

Here is your original formula,

$12{\rm C}+\frac{12}{6}{\rm C}\rightarrow \frac{24}{12}{\rm Mg}$

Please show your steps and indicate what excactly is giving you the impression that something is wrong.

Edit: Oh, and by the way, when you make a new post, or edit an existing post, use the "Preview" button to preview your post. You can verify that your $\LaTeX$ is valid that way before you post it.

 

[collinsmark]

Problem Statement: Calculate the mass deficit in this nuclear reaction
$12{\rm C}+\frac{12}{6}{\rm C}\rightarrow \frac{24}{12}{\rm Mg}$
Information given:

- rest mass of a carbon-12 nucleus: $$1.9921157\times 10^{-26}\,{\rm kg}$$

- rest mass of a magnesium nucleus: $$3.9817469\times 10^{-26}\,{\rm kg}$$

- rest mass of a proton: $$1.67353\times 10^{-27}\,{\rm kg}$$

- rest mass of a neutron: $$1.67492\times 10^{-27}\,{\rm kg}$$

- speed of light: $$ 3.00\times 10^{8}\,{\rm m}\,{\rm s}^{-1}$$
Relevant Equations: $$12{\rm C}+\frac{12}{6}{\rm C}\rightarrow \frac{24}{12}{\rm Mg}$$

I have tried using the formula $$Δm=[Zmp+(A−Z)mn−M]$$ But either I plug the wrong thing into the wrong part or it just isn't working. Essentially I want to know the method I need to use. I will then try to solve it and hopefully get it correct.

Thanks

Ok, I see you've edited your original post. Your $\LaTeX$ is working now.

But your original equation doesn't make sense to me. Are you sure it isn't something more like:

$^{12}_{\ 6} \rm{C} + \ ^{12}_{\ 6} \rm{C} \rightarrow \ ^{24}_{12} \rm{Mg} \ \ ?$

In your original equation, you have 13 carbon atoms changing into a single magnesium atom. (Edit: Combining two $^{12}_{\ \ 6} \rm{C}$ into a single $^{24}_{12} \rm {Mg}$ seems like a much more likely reaction.)

Please show your steps and let us know where the trouble is.

 

[amazingphysics2255]

[i[/QUOTE]

Ok, I see you've edited your original post. Your $\LaTeX$ is working now.

But your original equation doesn't make sense to me. Are you sure it isn't something more like:

$^{12}_{\ 6} \rm{C} + \ ^{12}_{\ 6} \rm{C} \rightarrow \ ^{24}_{12} \rm{Mg} \ \ ?$

In your original equation, you have 13 carbon atoms changing into a single magnesium atom. (Edit: Combining two $^{12}_{\ \ 6} \rm{C}$ into a single $^{24}_{12} \rm {Mg}$ seems like a much more likely reaction.)

Please show your steps and let us know where the trouble is.

Yes your right it's carbon 12/6

My steps are non of use at the moment. Do I even need to use a formula to solve it?

 

[collinsmark]

Yes your right it's carbon 12/6

Okay. Don't bother using the long formula for this one. Just find the difference in mass between two $^{12} \rm{C}$ nuclei and one $^{24} \rm{Mg}$ nucleus, since you already know the masses of these nuclei. That should do it.

 

[amazingphysics2255]

Okay. Don't bother using the long formula for this one. Just find the difference in mass between two $^{12} \rm{C}$ nuclei and one $^{24} \rm{Mg}$ nucleus, since you already know the masses of these nuclei. That should do it.

Okay. Don't bother using the long formula for this one. Just find the difference in mass between two $^{12} \rm{C}$ nuclei and one $^{24} \rm{Mg}$ nucleus, since you already know the masses of these nuclei. That should do it.

So i need to add two C12 nuclei together then take that off the mg24 nucleus ? If so -12.157686-13.817469=1.659783 is this the answer or have I done something wrong?

 

[collinsmark]

So i need to add two C12 nuclei together then take that off the mg24 nucleus ? If so -12.157686-13.817469=1.659783 is this the answer or have I done something wrong?

Okay, yes... maybe? But I don't quite see where you came up with those numbers. The mass of a $^{12}\rm{C}$ nucleus is $1.9921157 \times 10^{−26} \ \rm{kg},$ right? That's what you had in your original post. Is $3.9817469 \times 10^{-26}$ [kg] not the mass of a magnesium nucleus?

 

[amazingphysics2255]

Okay, yes... maybe? But I don't quite see where you came up with those numbers. The mass of a $^{12}\rm{C}$ nucleus is $1.9921157 \times 10^{−26} \ \rm{kg},$ right? That's what you had in your original post. Is $3.9817469 \times 10^{-26}$ [kg] not the mass of a magnesium nucleus?

I added up two $1.9921157 \times 10^{−26} \ \rm{kg},$ C-12 nuclei and took it a away from one MG-24 nucleus

but then if I do $1.9921157 \times 10^{−26} \ \rm{kg},$ x2 = 3.98423143- take that away from MG-24, 3.9817469=-0.00248453 Is either of them right? and why am I given the rest mass of a proton and a neutron?

 

[collinsmark]

I added up two $1.9921157 \times 10^{−26} \ \rm{kg},$ C-12 nuclei and took it a away from one MG-24 nucleus

but then if I do $1.9921157 \times 10^{−26} \ \rm{kg},$ x2 = 3.98423143- take that away from MG-24, 3.9817469=-0.00248453 Is either of them right? and why am I given the rest mass of a proton and a neutron?

Well, yeah, but don't forget about the power of $10^{-26}$ exponent.

$1.9921157 \times 10^{-26} \rm{kg} + 1.9921157 \times 10^{-26} \rm{kg} - 3.9817469 \times 10^{-26} \rm{kg} = 0.0024845 \times 10^{-26} \rm{kg}$

Let's take a deeper look at the original reaction.

$^{12}_{\ \ 6}\rm{C} + \ ^{12}_{\ \ 6}\rm{C} \rightarrow \ ^{24}_{12}\rm{Mg}$

How many total protons and neutrons are in the reactants? How many are in the products?

If the total number of protons and neutrons are the same in both the products and reactants, but the mass of products are a bit less than the mass of the reactants, it means that difference in mass must be given off as energy (whether or not the reaction is a direct or indirect). Does that make sense?

That's what nuclear fusion is all about. Sometimes the total number of protons and neutrons do not change between the products and reactants. And yet, when arranged differently --and thus producing different elements -- the mass is slightly different. Some of that mass was converted to energy (or vice versa in the case of heavier elements) when rearranging the nucleons. That's why nuclear fusion reactions release energy (or absorb energy in the case of combining heavier elements).

So why were you given the mass of the proton and mass of the neutron? I don't know, maybe for future problems. Some nuclear reactions involve free neutrons. You should be prepared to deal with those. This time though, for this particular reaction, free neutrons don't seem to apply.

[Edit: Sometimes neutrinos and antineutrinos are also involved in the products and/or reactants, but here I'm just lumping those into the mass deficit.]

 

[amazingphysics2255]

Thanks, So the total amount of protons and neutrons in this reaction are 6 protons and 6 neutrons for one carbon atom so double that 12 protons and 12 neutrons, and then Mg-24 has 12 protons and 12 neutrons.

@collinsmark Is there a way to solve by using the protons and neutrons as we know their masses?

And if I did want to see how much energy was released in the reaction what mass would I use for $$E=mc^2$$

And yes it does make sense, thanks.

I don't know if this results in a different answer to the question but in the question it says: inside large stars, two carbon nuclei fuse together to form magnesium. Calculate the mass deficit in the above nuclear reaction.

 

[collinsmark]

Thanks, So the total amount of protons and neutrons in this reaction are 6 protons and 6 neutrons for one carbon atom so double that 12 protons and 12 neutrons, and then Mg-24 has 12 protons and 12 neutrons.

Yes, that's right!

@collinsmark Is there a way to solve by using the protons and neutrons as we know their masses?

If you're asking "Can we accurately predict the mass of a given element rather than having to measure it? (And without already knowing the mass deficit or the energy differences.)" Then the answer is that there is no easy way.

I'm not saying it's impossible, but just that it's complicated enough to be far from easy. I'm guessing that to make such a prediction, one would require a solid knowledge of quantum chromodynamics (QCD) and a lot of time on the supercomputer. And even then, I'm not sure how accurate the prediction would be.

That's why the masses of each element and isotope are measured in the laboratory. Then if one wants to estimate the energy involved in a nuclear fission or fusion reaction, one looks up the masses and does what you did here in this problem.

And if I did want to see how much energy was released in the reaction what mass would I use for $$E=mc^2$$

Precisely. Yes!

 

[collinsmark]

I don't know if this results in a different answer to the question but in the question it says: inside large stars, two carbon nuclei fuse together to form magnesium. Calculate the mass deficit in the above nuclear reaction.

Yes, what you did here in this problem is the correct answer for that.

 

[amazingphysics2255]

Thanks, E=2.23605x10^-12