Nuc. Eng. problem


by gump
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gump
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#1
Jan6-05, 02:14 PM
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okay... I have a problem that I've been asked in my nuclear waste class, and since I have a materials science background, I have no clue where to begin. If someone could let me know where to start with this problem I would greatly appreciate it (don't tell me the answer). I'm supposed to estimate the energy in Mev of an alpha particle from a source of activity 1.0*10(-6) Ci which creates a saturation current of 1.0*10^(-9) A in an ionisation chamber. And I'm supposed to assume e = 1.6*10^(-19) C, 1 Ci = 3.7*10^(10) disintegrations per sec. and 30 eV is needed to produce on ion pair. I know the activity = # of disintegrations/sec = decay constant (lambda) * concentration (n(t)); however, I've been looking at various Nuclear Engineering books and haven't been able to figure out what equations I need for this problem.
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Astronuc
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Jan6-05, 02:30 PM
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The activity rate is not part of the solution.

An alpha particle losses energy by collisions with atoms and the ionization process, and you have mentioned "30 eV is needed to produce on ion pair."

Somewhere in the problem, it must state how far the alpha particle travels from birth to rest, and then one must know the concentration of atoms to get the collisions/(unit length of trajectory).

Are you familiar with Linear Energy Transfer rate?
gump
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#3
Jan6-05, 02:48 PM
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Quote Quote by Astronuc
The activity rate is not part of the solution.

An alpha particle losses energy by collisions with atoms and the ionization process, and you have mentioned "30 eV is needed to produce on ion pair."

Somewhere in the problem, it must state how far the alpha particle travels from birth to rest, and then one must know the concentration of atoms to get the collisions/(unit length of trajectory).

Are you familiar with Linear Energy Transfer rate?
Would the Linear Energy Transfer rate be
E(decay) = [M(x) -[M(y) + n (M(electron)]](931.5 MeV/amu) ??? I'm not exactly sure what the Linear Engergy Transfer rate is since my background is in Glass Science and Engineering. I just got that equation from some of the resources I've been looking at.

and also, the problem doesn't state anything about how far the particle travels. My professor said that we need to back calculate, but I don't see how that can be done with the information given.

Astronuc
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Jan6-05, 03:47 PM
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Nuc. Eng. problem


Sorry, I mis-led you there.

You have a current, which is caused by the number of ion pairs generated per unit time, and the current is just charge (C = coulomb) per unit time (C/s).

The total number of ion pairs is equal to the [E(alpha) x Number of alphas]/(30 eV per ion pair).

Now you also have a source strength - the number of alphas = number of decays/sec.

See if that helps.

Forget about linear energy transfer rate - I was incorrect intially when I mentioned it does not apply in this problem, and the rate of decay does.
gump
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#5
Jan6-05, 05:24 PM
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This is where I'm at so far. I'm getting a negative number, so I'm not sure what's going on with that.

Since we know that the current is caused by the number of ion pairs generated per unit time, and I = Q/t, where Q = amount of charge (e = 1.6x10-19 C) and t = time, the time is 1.6x10-10 sec.

Using the equation (n(ion) /t) = (E(a)n(a) / 30 eV) where n(ion) = number of electron pairs, t = time, E(a) = Energy of an alpha particle, and n(a) = number of alpha decays per second we can solve for E(a).

E(a) = 30 eV (1 ion pair) / n(a)(1.6x10-10 sec.)

Since we know that n(a) = a n(ion), where a = source strength, we get n(a) = (1.0x10-6 Ci 3.7x1010 dis./sec.) = -3.7x1010 dis./sec.

This is where I get confused since I don't think there can be a negative disintegrations per second which would lead to a negative energy.
Astronuc
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Jan6-05, 08:34 PM
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Herein lies the problem and perhaps some confusion,
Since we know that n(a) = a - n(ion), where a = source strength, we get n(a) = (1.0x10-6 Ci - 3.7x1010 dis./sec.) = -3.7x1010 dis./sec.
This not correct. 1 Ci = 3.7 x 1010 dps (dps = 1 disintegration per s) .

The production rate of alphas Ra = Activity * dps/Ci = 1 x 10-6 Ci * 3.7 x 1010 dps/Ci = 37000 alphas/s.

Now the energy produced from the decays E is just Ra * Ea = 37000 E eV/s.

Now each ionization produces an ion pair by which the alpha particle loses approximately Ei = 30 eV.

So the ion production rate is then 37000 E (eV/s)/30 ev = 1233.3E ion pairs/sec.

Now the current I in the ionization chamber is related to the collection of the electrons,

I = q/t = 1.0 x 10-9 A, but the charge comes from the ions, so

I = (Ra * Ea)/ Ei * e, where e = 1.6*10-19 C.

So rearranging this,


Ea = (I * Ei)/(Ra * e) all the terms on the RHS are known.


BTW - please do not double post. This problem belongs in the homework section alone.
gump
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#7
Jan7-05, 06:20 AM
P: 11
Quote Quote by Astronuc
BTW - please do not double post. This problem belongs in the homework section alone.
Sorry... been working on this problem for over two days now and I really wanted to figure it out. I really appreciate your help. It makes sense now. Thanks again


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