Ionization chamber question

[walkdady]

1. Homework Statement

a) Estimate the number of ion pairs produced by a 2.4 MeV beta particle in an ionization chamber. Assume that 34 eV of the beta particle's energy is absorbed in the creation of each ion pair, and that the beta particle energy is completely absorbed within the chamber.
b) Hence estimate the maximum possible value of the output pulse voltage if the capacitance of the chamber is 25 pF.
c) Clearly, the beta energy is not always completely absorbed in the chamber. Suggest one other reason why the output pulse voltage is likely to be less than the value you have calculated.

Homework Equations

q=f/v?

The Attempt at a Solution

i attemped a) by saying 2400000/34 = 70588 pairs

then b) i said 34*70588 = 2399992 hence 8eV not used?

i have no idea if I am even on the right track, what is the question asking?

 

[anathema]

[/b]

Hello, thank you for your question. I would like to provide you with a more detailed explanation and approach to solving this problem.

a) The first part of the question is asking you to estimate the number of ion pairs produced by a 2.4 MeV beta particle in an ionization chamber. This means that the beta particle is traveling through the chamber and causing ionization, or the creation of positively and negatively charged particles. To estimate the number of ion pairs, you can use the given information that 34 eV of the beta particle's energy is absorbed in the creation of each ion pair. This means that for every 34 eV of energy, one ion pair is created. To find the number of ion pairs, we can divide the total energy of the beta particle (2.4 MeV or 2.4 x 10^6 eV) by 34 eV. This gives us an estimate of approximately 70,588 ion pairs produced by the beta particle.

b) The second part of the question asks you to estimate the maximum possible value of the output pulse voltage if the capacitance of the chamber is 25 pF. To do this, we can use the formula q=CV, where q is the charge, C is the capacitance, and V is the voltage. We know the capacitance (25 pF) and we just calculated the charge (70,588 eV). Plugging these values into the formula, we get V = q/C = (70,588 eV)/(25 pF) = 2,823,520 volts. This is the maximum possible output pulse voltage if all of the charge created by the beta particle is collected by the chamber.

c) Finally, the third part of the question asks you to suggest one other reason why the output pulse voltage is likely to be less than the value you have calculated. One possible reason for this is that the beta particle's energy is not always completely absorbed in the chamber. Some of the beta particle's energy may be lost due to scattering or interactions with other particles in the chamber. This means that not all of the charge created by the beta particle will be collected by the chamber, resulting in a lower output pulse voltage than the maximum calculated in part b). Other factors that could contribute to a lower output pulse voltage include imperfect efficiency of the chamber or losses due to resistance.

I hope this explanation helps you better understand the