# Calculate the difference between initial and the final masses

#### amazingphysics2255

Problem Statement
Polonium-212 decays into Lead-208, emitting an alpha particle. The mass of polonium is $$3.51986x10^-25kg$$ The mass of lead is $$3.45323x10^-25kg$$ and the mass of an alpha particle is $$6.646x10^-27kg$$

Calculate the difference between the initial and final masses of the reaction
Relevant Equations
None given
Don't really know how to go about this exercise as I've been given no information for it I tried googling similar problems but couldn't find any
If someone could give me a process to try I will then try to solve it here in this section, thanks.

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#### RPinPA

This seems very simple. Maybe I'm missing something. You want to know the difference between initial and final mass.
You're told what you have initially (Po-212).
You're told its mass.
You're told what you have finally (Pb-208 + alpha).
You're told their mass.

That's the initial mass and the final mass. So what else do you need to know to find the difference between those two masses?

#### amazingphysics2255

This seems very simple. Maybe I'm missing something. You want to know the difference between initial and final mass.
You're told what you have initially (Po-212).
You're told its mass.
You're told what you have finally (Pb-208 + alpha).
You're told their mass.

That's the initial mass and the final mass. So what else do you need to know to find the difference between those two masses?
Do I need to subtract one from another?

#### RPinPA

Yes, that's what "difference" means.

#### amazingphysics2255

Yes, that's what "difference" means.
So I'm meant to take 3.51986x10−25kg off 3.45323x10−25kg ? giving me 0.06663? why am I given the alpha particles mass? am I supposed to use it?

or was I mean to add 3.45323+6.646 together then take 10.09923-3.51986=6.57937

#### collinsmark

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So I'm meant to take 3.51986x10−25kg off 3.45323x10−25kg ? giving me 0.06663? why am I given the alpha particles mass? am I supposed to use it?

or was I mean to add 3.45323+6.646 together then take 10.09923-3.51986=6.57937
It might make more sense if you write out the nuclear reaction equation first. Keep in mind the alpha particle is the same thing as a helium nucleus, $^4_2\rm{He}.$

So the reactants and products involve $^{212}_{\ 84}\rm{Po}, \ ^{208}_{\ 82}\rm{Pb}$ and $^4_2\rm{He}$

Which are the reactants and which are the products?

#### amazingphysics2255

Well
$$^{212}_{\ 84}\rm{Po}, \ ^{208}_{\ 82}\rm{Pb}, ^4_2\rm{He}$$

Po must be a reactant and Pb is a product

$$^{212}_{\ 84}\rm{Po}\rightarrow \ ^{208}_{\ 82}\rm{Pb}+ ^4_2\rm{He}$$
Is that the reaction written down?

#### collinsmark

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$$^{212}_{\ 84}\rm{Po}\rightarrow \ ^{208}_{\ 82}\rm{Pb}+ ^4_2\rm{He}$$
Is that the reaction written down?
Correct! So now just add up the masses on each side and find the difference. That's the mass deficit.

If the mass of the products is less than the mass of the reactants, it means a net amount of energy was released (as opposed to absorbed) during the reaction.

#### amazingphysics2255

Correct! So now just add up the masses on each side and find the difference. That's the mass deficit.

If the mass of the products is less than the mass of the reactants, it means a net amount of energy was released (as opposed to absorbed) during the reaction.
So 3.51986x10^−25kg-3.45323x10^−25kg= 0.06663x10^-25. Was I meant to add the alpha particles mass onto the lead than deduct?

#### collinsmark

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Gold Member
So 3.51986x10^−25kg-3.45323x10^−25kg= 0.06663x10^-25. Was I meant to add the alpha particles mass onto the lead than deduct?
Do that calculation over. You are correct that you first need to add the mass of the alpha particle to the mass of the lead nucleus first.

The mass deficit is the difference in mass between the total mass of all the reactants and the total mass of all the products.

Assuming the reaction releases a net amount of energy (instead of absorbing energy), the total mass of the products will be less than the total mass of the reactants and the mass deficit will be positive.*

*[Edit: This is also assuming that none of the particles are moving at relativistic speeds. Particles moving at relativistic speeds will be important soon when you study beta decays. But don't worry about that for this problem.]

#### amazingphysics2255

Do that calculation over. You are correct that you first need to add the mass of the alpha particle to the mass of the lead nucleus first.

The mass deficit is the difference in mass between the total mass of all the reactants and the total mass of all the products.

Assuming the reaction releases a net amount of energy (instead of absorbing energy), the total mass of the products will be less than the total mass of the reactants and the mass deficit will be positive.
3.45323x10−25kg+6.646x10−27kg=10.09923^-27kg

3.51986x10−25kg-10.09923^-27kg=6.657937^-27kg

I have a feeling this is still wrong as the Polonium-212 is emitting an alpha particle so should the alpha particle also be added to the Polonium-212 then that amount is taken off the 10.09923?

#### collinsmark

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3.45323x10−25kg+6.646x10−27kg=10.09923^-27kg
Try that calculation again. Keep in mind that $6.646 \times 10^{-27} \ \rm{kg} = 0.06646 \times 10^{-25} \ \rm{kg}$

I have a feeling this is still wrong as the Polonium-212 is emitting an alpha particle so should the alpha particle also be added to the Polonium-212 then that amount is taken off the 10.09923?
There's no need to add the mass of the alpha particle to the mass of the polonium nucleus.

In a sense, the alpha particle is already "inside" the polonium nucleus to begin with. It's already in there. No need to add it again. The mass of polonium nucleus already "contains" (so to speak) the mass of the alpha particle.

#### amazingphysics2255

3.45323x10^−25kg+0.06646×10^−25 kg=3.51969x10^-25kg

3.51986x10−25kg-3.151969x10^-25kg=0.367891x10^-25kg

#### collinsmark

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3.45323x10^−25kg+0.06646×10^−25 kg=3.51969x10^-25kg

3.51986x10−25kg-3.151969x10^-25kg=0.367891x10^-25kg

The first calculation looks correct.

But I think you misstyped an extra "1" digit into one of your numbers that doesn't belong. I've highlighted the digit in red.

#### amazingphysics2255

yeah, I did good catch, so it supposedly is 0.00017x10^-25kg

#### collinsmark

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yeah, I did good catch, so it supposedly is 0.00017x10^-25kg
That looks correct to me.

#### amazingphysics2255

$$E=mc^2$$ so
$$E=0.00017x10^-25(3.00x10^8)^2$$

which = 15278838mj

is this also correct?

#### collinsmark

Homework Helper
Gold Member
$$E=mc^2$$ so
$$E=0.00017x10^-25(3.00x10^8)^2$$

which = 15278838mj

is this also correct?
I think it might be off by a factor of 10, according to my calculations.

[Edit: On the other hand, if you are using units of milli-Joules, then you are off by many orders of magnitude.]

#### amazingphysics2255

So what would I need to do to fix this?

#### collinsmark

Homework Helper
Gold Member
So what would I need to do to fix this?
just try redoing the calculation. Maybe you missed or added a digit or something.

[Edit: Now that I notice your units, your answer seems to be off by many orders of magnitude. Maybe something went wrong with your exponents.]

#### amazingphysics2255

ok I've redone it now getting $$1.53x10^-12kg$$ also what's a good way to convert this to joules, mega joules?

#### collinsmark

Homework Helper
Gold Member
ok I've redone it now getting $$1.53x10^-12kg$$ also what's a good way to convert this to joules, mega joules?
The numerical value is correct, but it should have units of Joules. $1.53 \times 10^{-12} \rm{J}$
(not kg.)

If you want to get rid of the $10^{-12}$ the prefix for that is pico-.

#### amazingphysics2255

Ok, so Let's me practice one last one M(mass)=0.0024845×10^−26kg $$E=mc^2$$ $$E=0.0024845×10^−26kg(3.00x10^8)^2$$ $$=2.23605x10^-12J$$

Last edited:

#### collinsmark

Homework Helper
Gold Member
Ok, so Let's me practice one last one M(mass)=0.0024845×10^−26kg $$E=mc^2$$ $$E=0.0024845×10^−26kg(3.00x10^8)^2$$ $$=2.23605x10^-12J$$
Assuming that one is just for practice (I don't know how that mass applies to this problem specifically), then yes, the calculation looks correct to me.

#### amazingphysics2255

Assuming that one is just for practice (I don't know how that mass applies to this problem specifically), then yes, the calculation looks correct to me.
Yeah that mass has nothing to do with this problem, just making sure I'm proficient

"Calculate the difference between initial and the final masses"

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