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
time τ during which particles are not counted. Then if N is the counting rate
(number of counts/second) then the time lost will be N τ , so that the true
counting rate
n = N /(1 − N τ )
Thus n 1 = N 1 /(1 − N τ ) 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 practice N 1 and N 2 will be of the order of 100 per second, N 12 is of
200 per second, B ≈ 1 per second and τ ≈ 10
−4 second, so that N 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 12
2
− N 1
2
− N 2
2
≈ τ
(N 1 + N 2 )
2
− N 1
2
− N 2
2
Or τ = (N 1 + N 2 − N 12 + B)/2N 1 N 2
9.37 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
If n ion pairs are released then ne = Q
Therefore n = Q/e = 10
−13
/1.6 × 10
−19
= 6.25 × 10
5 ion pairs.
Energy of alpha particles = (number of ion pairs) × (Ionization energy)
= (6.25 × 10
5 )(35) = 21.87 × 10
6 eV = 21.87 MeV.
9.38 Let n beta particles enter the ionization chamber per second and stopped.
Number of ion pairs released when ionization energy is I is given by
n = N /I = 0.49 × 10
6
/35 = 1.4 × 10
4
The ionization current, i = Nne
Therefore, N = i/ne = 7 × 10
−11
/1.4 × 10
4
× 1.6 × 10
−19
= 3.125 × 10
4
beta particles/second
9.39 If M is the gas multiplication factor and N is the number of ion pairs released,
V the voltage developed, C the capacitance and e the electronic charge then
the charge Q deposited will be
Q = CV = MNe
Or M = CV /Ne = 1.5 × 10
−9
× 10
−3
/(2 × 10
6
/35) × 1.6 × 10
−19
= 164
9.40 Number of primary ions produced by a 14 keV electron, N = 14,000/35 =
400
Charge collected Q = MNe
where M is the gas multiplication and e = 1.6 × 10
−19 Coulomb the elementary charge. If C is 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.
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
time τ during which particles are not counted. Then if N is the counting rate
(number of counts/second) then the time lost will be N τ , so that the true
counting rate
n = N /(1 − N τ )
Thus n 1 = N 1 /(1 − N τ ) 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 practice N 1 and N 2 will be of the order of 100 per second, N 12 is of
200 per second, B ≈ 1 per second and τ ≈ 10
−4 second, so that N 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 12
2
− N 1
2
− N 2
2
≈ τ
(N 1 + N 2 )
2
− N 1
2
− N 2
2
Or τ = (N 1 + N 2 − N 12 + B)/2N 1 N 2
9.37 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
If n ion pairs are released then ne = Q
Therefore n = Q/e = 10
−13
/1.6 × 10
−19
= 6.25 × 10
5 ion pairs.
Energy of alpha particles = (number of ion pairs) × (Ionization energy)
= (6.25 × 10
5 )(35) = 21.87 × 10
6 eV = 21.87 MeV.
9.38 Let n beta particles enter the ionization chamber per second and stopped.
Number of ion pairs released when ionization energy is I is given by
n = N /I = 0.49 × 10
6
/35 = 1.4 × 10
4
The ionization current, i = Nne
Therefore, N = i/ne = 7 × 10
−11
/1.4 × 10
4
× 1.6 × 10
−19
= 3.125 × 10
4
beta particles/second
9.39 If M is the gas multiplication factor and N is the number of ion pairs released,
V the voltage developed, C the capacitance and e the electronic charge then
the charge Q deposited will be
Q = CV = MNe
Or M = CV /Ne = 1.5 × 10
−9
× 10
−3
/(2 × 10
6
/35) × 1.6 × 10
−19
= 164
9.40 Number of primary ions produced by a 14 keV electron, N = 14,000/35 =
400
Charge collected Q = MNe
where M is the gas multiplication and e = 1.6 × 10
−19 Coulomb the elementary charge. If C is 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.