Answer to Question #206429 in Atomic and Nuclear Physics for Nick

Question

An ionization chamber is connected to an electrometer of capacitance 0.5 pF and voltage sensitivity of 4 divisions. Calculate the number of ion pairs required and the energy of the alpha particle.

Answer 1

The voltage sensitivity is 4 divisions per volt.

So 0.8 divisions correspond to 0.2 V.

The charge deposited, $Q=CV=0.5\cdot 10^{-12}\cdot 0.2=10^{-13}~C.$ If n ion pairs are released then $ne=Q.$  

Therefore $n=\frac Qe=\frac{10^{-13}}{1.6\cdot 10^{-19}}=6.25\cdot 10^5$ ion pairs. 

$E_{\alpha}=nE_I=6.25\cdot 10^5\cdot 35=21.87~MeV.$