Calculating Ion Pairs, Total Charge and Exposure.

[Jason Goddard]

Hello,

I am unsure whether this is in the right place, I do think this is nuclear physics however, I was hoping I could get some idea of whether I am answering this question correctly as unfortunately I won't be able to attend the class that explaining this concept.

34eV of energy, are lost, to produce 1 ion pair, in air. An alpha particle, has 276 302eV of available energy. Three alpha particles, lost all their energy completely, in 4 cm3 of air,

The charge on one electron is 1.6 x 10^-19 C.

1 cm3 of air has a mass of 0.0013 g, under standard conditions.

Calculate, in order, the number of ion pairs created and the total charge. Show your workings and report the exposure, using standard index notation, to three significant figures.

Workings:

Mass of 4cm3 of air = 0.0013g x 4 = 5.2 x 10^-3g

Total energy of alpha particles = 3x 276,302eV = 828,906eV

Number of ion pairs created = 828,906 / 34 = 24, 379.58823529412.
Number of ion pairs created = 24,379.

Total charge = 1.6 x 10^-19 C x 24,379 = 3.901x10^-15C, x 2 = total charge is 7.802 x10^-15

Any advice as to exposure calculations and whether I have even done it correctly?

Thanks very much.

 

[PeterDonis]

Moderator's note: moved to homework forum.

 

[PeterDonis]

Any advice as to exposure calculations and whether I have even done it correctly?

Your calculations of the number of ion pairs and the total charge look OK to me. You might want to specify that half of the total charge is positive and half is negative.

For exposure calculations, what equations do you think would be relevant?

 

[Jason Goddard]

Your calculations of the number of ion pairs and the total charge look OK to me. You might want to specify that half of the total charge is positive and half is negative.

For exposure calculations, what equations do you think would be relevant?

Based on an assumption that the total mass of air comes into play I was thinking it would be something like:

7.802 x10^-15 / 5.2 x 10^-3g = 1.500x10^-12

 

[PeterDonis]

I was thinking it would be something like:

7.802 x10^-15 / 5.2 x 10^-3g = 1.500x10^-12

What units would that be in? What are the units of exposure?

 

[PeterDonis]

1 cm3 of air has a mass of 0.0013 g, under standard conditions.

You might want to check the order of magnitude of this number. It looks off by a factor of 1000 to me (i.e., it should be 0.0013 kg, not 0.0013 g).

 

[Jason Goddard]

What units would that be in? What are the units of exposure?

Well it doesn't specifically state the unit in the question but I am assuming in Ckg-1 with that number?

 

[PeterDonis]

it doesn't specifically state the unit in the question

Perhaps not, but surely there is something somewhere in your textbook that explains what exposure is and gives its units.

I am assuming in Ckg-1 with that number?

My point was that if the units don't look right, you should doubt whether the equation you guessed is correct. What would units of Coulombs per kilogram of air mean? How would that relate to exposure?

 

[PeterDonis]

if the units don't look right, you should doubt whether the equation you guessed is correct

Or, if you think those units are right, you ought to be able to find some corroboration of that somewhere in your textbook. Or you could try looking it up elsewhere.

 

[Jason Goddard]

You might want to check the order of magnitude of this number. It looks off by a factor of 1000 to me (i.e., it should be 0.0013 kg, not 0.0013 g).

Unfortunately my tutor for this particular one has been known to make mistakes a number of times, however he always says that we need to answer the questions it asks.

 

[PeterDonis]

my tutor for this particular one has been known to make mistakes a number of times

So how would you check the information he gives you?

 

[Jason Goddard]

So how would you check the information he gives you?

I wouldn't as that's part of the question he has put in the paper, I can't change it.

Or, if you think those units are right, you ought to be able to find some corroboration of that somewhere in your textbook. Or you could try looking it up elsewhere.

That's what I have spent the last few hours doing, I'm honest enough to say that I don't have enough of a good understanding to even know where to begin looking or what to search for, my textbook which was recommended does not actually seem to have anything on this equation, or I am missing it because I don't even know the real name of this type of equation.

I know Coulomb per Kilogram is meant to refer to the ionisation of air from photons, but I can't understand specifically how to calculate it based on what I can find.

 

[PeterDonis]

That's what I have spent the last few hours doing

What have you tried?

my textbook which was recommended does not actually seem to have anything on this equation

It doesn't discuss exposure at all?

Also, your textbook is not the only possible source of information.

I know Coulomb per Kilogram is meant to refer to the ionisation of air from photons

From gamma rays and X-rays, yes. This question seems to be about extending the concept to ionization by alpha particles.

 

[Jason Goddard]

What have you tried?.

I've been looking online now, for exposure calculations, but none of them seem to actually relate to the same sort of numbers, like the lost Ev and total eV etc which is what is confusing me I think.

From gamma rays and X-rays, yes. This question seems to be about extending the concept to ionization by alpha particles.

Ah right, now I'm getting there, maybe a bit slowly but I'm trying! In terms of alpha radiation then it would be micro-Sv then?

 

[PeterDonis]

I've been looking online now, for exposure calculations, but none of them seem to actually relate to the same sort of numbers, like the lost Ev and total eV etc

But you've already done that part of the calculation--you've converted the energy lost into the amount of charge produced. All you need is to relate the charge produced to exposure.

In terms of alpha radiation then it would be micro-Sv then?

The Sievert is a unit of energy deposited by radiation in an object, such as a human body. I'm not sure that's what the question is looking for; "exposure", as you have noted, usually refers to charge per unit mass.

What I meant by "extending the concept" is that, if the question had been talking about ionization produced by gamma rays or X-rays, you would already have the answer: you know the charge produced, and you know the mass of the air it was produced in. (You already wrote down the answer that this method would give you in post #4, in fact.) So the only real question is whether it makes sense to treat ionization produced by alpha particles the same way. I would have expected your textbook to say something about this, but textbooks don't always cover everything.

 

[Jason Goddard]

So the only real question is whether it makes sense to treat ionization produced by alpha particles the same way. I would have expected your textbook to say something about this, but textbooks don't always cover everything.

The only thing I can find that might be relative would be to convert it to Grays, however that is usually used i absorbed dose, all my research so far has indicated that exposure is in coulomb per kilogram or roentgen but both mainly reference X and Gamma rays. So I haven't found anything to indicate to me specifically what the exposure unit for alpha particles would be as such!

 

[PeterDonis]

I haven't found anything to indicate to me specifically what the exposure unit for alpha particles would be as such!

Then I would say the best you can do is to use the same unit as for the only things for which you've found an explicit definition of exposure units (X and gamma rays).

 

[Jason Goddard]

Well thank you for the assistance, it definitely helped me to think through it and actively make sure that I was exploring all the options!