Nuc. Eng. problemby gump Tags: None 

#1
Jan605, 02:14 PM

P: 11

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.




#2
Jan605, 02:30 PM

Admin
P: 21,637

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? 



#3
Jan605, 02:48 PM

P: 11

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. 



#4
Jan605, 03:47 PM

Admin
P: 21,637

Nuc. Eng. problem
Sorry, I misled 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. 



#5
Jan605, 05:24 PM

P: 11

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 = 1.6x1019 C) and t = time, the time is 1.6x1010 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)(1.6x1010 sec.) Since we know that n(a) = a – n(ion), where a = source strength, we get n(a) = (1.0x106 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. 



#6
Jan605, 08:34 PM

Admin
P: 21,637

Herein lies the problem and perhaps some confusion,
The production rate of alphas R_{a} = Activity * dps/Ci = 1 x 10^{6} Ci * 3.7 x 10^{10} dps/Ci = 37000 alphas/s. Now the energy produced from the decays E is just R_{a} * E_{a} = 37000 E eV/s. Now each ionization produces an ion pair by which the alpha particle loses approximately E_{i} = 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 = (R_{a} * E_{a})/ E_{i} * e, where e = 1.6*10^{19} C. So rearranging this, E_{a} = (I * E_{i})/(R_{a} * e) all the terms on the RHS are known. BTW  please do not double post. This problem belongs in the homework section alone. 



#7
Jan705, 06:20 AM

P: 11




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