9.3 Solutions 513
or
n 1 +n 2 =n 12 −B (1)
Now, for each particle that is counted, on an average there will be a dead
timeτduring which particles are not counted. Then ifNis the counting rate
(number of counts/second) then the time lost will beNτ,sothatthetrue
counting rate
n=N/(1−Nτ)
Thusn 1 =N 1 /(1−Nτ)etc.
We can then write (1) as
N 1 /(1−N 1 τ)+N 2 /(1−N 2 τ)=N 12 /(1−N 12 τ)−B/(1−Bτ)(2)
Now, in practiceN 1 andN 2 will be of the order of 100 per second,N 12 is of
200 per second,B≈1 per second andτ≈ 10 −^4 second, so thatN 1 τ<<1,
etc. We can then expand the denominators binomially and write to a good
approximation
N 1 (1+N 1 τ)+N 2 (1+N 2 τ)=N 12 (1+N 12 τ)−B(1+Bτ)
N 1 +N 2 −N 12 +B=τ[
N 122 −N 12 −N 22
]
≈τ[
(N 1 +N 2 )^2 −N 12 −N 22
]
Orτ=(N 1 +N 2 −N 12 +B)/ 2 N 1 N 29.37 The voltage sensitivity is 4 divisions per volt.
So 0.8 divisions correspond to 0.2 V.
The charge deposited,Q=CV= 0. 5 × 10 −^12 × 0. 2 = 10 −^13 Coulomb
Ifnion pairs are released thenne=Q
Thereforen=Q/e= 10 −^13 / 1. 6 × 10 −^19 = 6. 25 × 105 ion pairs.
Energy of alpha particles=(number of ion pairs)×(Ionization energy)
=(6. 25 × 105 )(35)= 21. 87 × 106 eV= 21 .87 MeV.
9.38 Letnbeta particles enter the ionization chamber per second and stopped.
Number of ion pairs released when ionization energy isIis given by
n=N/I= 0. 49 × 106 / 35 = 1. 4 × 104
The ionization current, i=Nne
Therefore,N=i/ne= 7 × 10 −^11 / 1. 4 × 104 × 1. 6 × 10 −^19 = 3. 125 × 104
beta particles/second
9.39 IfMis the gas multiplication factor andNis the number of ion pairs released,
Vthe voltage developed,Cthe capacitance andethe electronic charge then
the chargeQdeposited will be
Q=CV=MNe
OrM=CV/Ne= 1. 5 × 10 −^9 × 10 −^3 /(2× 106 /35)× 1. 6 × 10 −^19 = 164
9.40 Number of primary ions produced by a 14 keV electron,N= 14 , 000 / 35 =
400
Charge collectedQ=MNe
whereMis the gas multiplication ande= 1. 6 × 10 −^19 Coulomb the elemen-
tary charge. IfCis the capacitance of the circuit the pulse height will be
V=Q/C=MNe/C= 600 × 400 × 1. 6 × 10 −^19 / 10 × 10 −^12 = 3. 84 ×
10 −^3 V= 3 .84 mV.