#cal of potential signal
#intiation of all variables
# Chapter 8
print"Example 8.1, Page:120 \n \n"
#Given:
e=1.6*10**-19;# electron charge
C=6*10**-12;# in F
N=10**5;# # electron multiplication
#Solution:
e1=N*e;
v=e1/(2*C);
v1=1000*v;
print"The potential signal recorded will be (mV)=",round(v1,2)
#cal of capacitance that would be required
#intiation of all variables
# Chapter 8
print"Example 8.2, Page:120 \n \n"
# Given:
N=10**5;# electron multiplication
v=10**-6;# in V
e=1.6*10**-19;# electron charge
# Solution:
e1=N*e;
C=e1/(2*v);
C1=C*10**9;
print"The capacitance that would be required in (nF)=",C1
#cal of alpha coefficient
#intiation of all variables
# Chapter 8
import math
print"Example 8.3, Page:120 \n \n"
# Given:
n0=1;# initial primary electrons
n=1.6*10**4;
x=2.2; #distance in cm
# Solution:
a=math.log(n/n0)/(x);
print"The alpha coefficient in electrons electron^-1 cm^-1=",round(a,2)
#cal of potential,field
#intiation of all variables
# Chapter 8
print"Example 8.4, Page:121 \n \n"
# Given:
import math
V=1600.;# potential across the electrodes
di=3.;# inner diameter
do=40.;# outer diameter
a=1.5;#in mm
A=20.;#in mm
# Solution:
# Part(a)At the inner surface
r1=1.5;# in mm
V1=V*(math.log(A/r1)/math.log(A/a));
X1=V/(r1*(math.log(A/a)));
print"\n The potential at the inner surface is = (V)",V1
print"\n The field at the inner surface is =(V/cm)",round(X1,2)
# Part(b)At the outer surface
r2=20.;# in mm
V2=V*(math.log(A/r2)/math.log(A/a));
X2=V/(r2*(math.log(A/a)));
print"\n \n The potential at the outer surface is =(V)",V2
print"\n The field at the outer surface in (V/cm)=",round(X2,2)
# Part(c)In mid-way between the cylinder
r3=(A+a)/2.;# in mm
V3=V*(math.log(A/r3)/math.log(A/a));
X3=V/(r3*(math.log(A/a)));
print"\n \n The potential in mid-way between the cylinder is =(V)",V3
print"\n The field in mid-way between the cylinderis =(V/cm)",round(X3,3)
#cal of resolving time
#intiation of all variables
# Chapter 8
print"Example 8.5, Page:122 \n \n"
# Given:
ma1=3600.;# counts in 3 min
mb1=2400.;# counts in 5 min
mab1=9900.;# counts in 6 min
# Solution:
ma=ma1/3;
mb=mb1/5;
mab=mab1/6;
t1=(ma+mb-mab)/(mab**2-ma**2-mb**2);
t2=t1*60;# in seconds
t=t2*1000000;# in microseconds
print"The resolving time of the given system in microseconds is =",round(t)
#cal of The resolving time,true count rate
#intiation of all variables
# Chapter 8
print"Example 8.6, Page:122 \n \n"
# Given:
ma1=3321.;# counts in 3 min
mb1=2862.;# counts in 2 min
mab1=4798.;# counts in 2 min
m=1080.;# counts in 30 min
muk1=5126.;# counts in 2 min
# Solution:
ma=ma1/3.;
mb=mb1/2.;
mab=mab1/2.;
mbc=m/30.;
muk=muk1/2.;
t1=(ma+mb-mab-mbc)/(mab**2.-ma**2.-mb**2.);# in min
t2=t1*60.;# in seconds
t=t2*1000000.;# in microseconds
print"The resolving time of the given system in microseconds is =",round(t)
n=muk/(1-muk*t1);# true count rate
print"\n The true count rate of unknown sample in (cpm)=",round(n)
#cal of counting loss of sample A,B,C
#intiation of all variables
# Chapter 8
print"Example 8.7, Page:123 \n \n"
# Given:
ma=9728.;# cpm
mb=11008.;# cpm
mab=20032.;# cpm
# Solution:
t1=(ma+mb-mab)/(mab**2-ma**2-mb**2);# in min
t2=t1*60;# in seconds
t=t2*1000000;# in microseconds
print"\n The resolving time of the given system in microseconds is =",t
#From true count rate equation we have, n=muk/(1-muk*t).
# This implies, n-m=m^2*t where n-m corresponds to counting loss
na=ma**2*t1;# For sample A
nb=mb**2*t1;# For sample B
nab=mab**2*t1;# For sample AB
print"\n The counting loss of sample A in (cpm)=",round(na)
print"\n The counting loss of sample B in (cpm)=",round(nb)
print"\n The counting loss of sample AB in (cpm)=",round(nab)
# NOTE: The resolving time of the given system in microseconds is give 222.7. This is a calculation error in the textbook.