How Do You Calculate the Energy of an Alpha Particle in an Ionization Chamber?

[gump]

I'm having a problem on a homework problem that I have for one of my nuclear waste engineering courses. I'm not exactly sure where to start on it. I've been looking at different resources and I haven't found anything that I can use. The question is as follows:

Estimate the energy in MeV of an alpha particle from a source of activity 1.0 microcuries which creates a saturation current of 1.0*10^(-9) A in an inonisation chamber. 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'm not sure where to start on this problem, so if there are a few equations out there that someone could let me know about, I would be greatly appreciative.

 

[Andrew Mason]

Estimate the energy in MeV of an alpha particle from a source of activity 1.0 microcuries which creates a saturation current of 1.0*10^(-9) A in an inonisation chamber. 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'm not sure where to start on this problem, so if there are a few equations out there that someone could let me know about, I would be greatly appreciative.

I'll take a stab at it. The alpha particle shoots through a cloud of atoms and knocks electrons off a gas molecule. The issue is: how many does it knock off? If you knew that, you would know its energy (ie. n=E/30 eV).

The energy needed to ionise one molecule (creating an ion pair) is 30 eV. When this occurs you get a 'current' because the electrons move away from the +ions. The current, in this case, is the result of all the alpha particles being released from the source/second knocking off n electrons each. The current is the rate of charge flow. In this case, it is the number of ions (in coulombs - ie ne) produced per second.

If all of this is correct (and I am not guaranteeing it is, but it makes sense to me) what is the equation for n? where N is the number of alpha particles released from the source/second, n is the number of ion pairs created per alpha particle, e is the charge of one electron or ion in Coulombs, I =1.0E-9 Amperes. Once you find n then you can find E.

AM

 

[gump]

okay... so this is where I'm at so far.

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 nion = 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 – nion where a = source strength, we get n(a) = (1.0x10-6 Ci – 3.7x1010 dis./sec.) = -3.7x1010 dis./sec.

This negative number is really confusing me, and I'm not sure where I'm going wrong. Do I have one of my equations messed up? or am I going wrong somewhere else?

 

[Andrew Mason]

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.

I don't follow you there. Why not use seconds?

Using the equation (n(ion) /t) = (E(a)n(a) / 30 eV) where nion = 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).

Why not just state the equation for I and worry about E later? I get:
$I = dQ/dt = nNe$ where n is number of ion pairs per alpha, N is the number of alpha particles per second and e is the current created per collision.

When you get n, you can find E = 30n.

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

I think you meant $1.0E-6 * 3.7E10 = 3.7E4$ dis./sec. That is N (no. of alpha particles/sec).

With that you should have no difficulty finding n using n = I/Ne Then find E.

AM

 

[Astronuc]

This problem is also addressed here:

https://www.physicsforums.com/showthread.php?p=421097#post421097

 

[gump]

Thanks for your help guys. It now makes sense. I really appreciate it.