Chapter6-Binary Nuclear Reactions¶

Ex1-pg142¶

In [1]:
##Chapter 6, Example 6.1, Page 142
import math
import numpy
## Minimum Kinetic energy
Q = numpy.array([1.311, -0.6259, -0.1582]) ##Q in MeV of all the reactions
Ex = ([1.994 ,2.11 ,0.1695])
KE = numpy.array(Q+Ex)
print'%s %.2f %s'%("Kinetic Energy for 13C(d,t)12C =  \n",KE[0],"")
print'%s %.2f %s'%(" Kinetic Energy for 14C(p,n)14N = \n",KE[1],"")
print'%s %.2f %s'%(" Kinetic Energy for 14C(n,a)11B = %f",KE[2],"")

Kinetic Energy for 13C(d,t)12C =
3.30
Kinetic Energy for 14C(p,n)14N =
1.48
Kinetic Energy for 14C(n,a)11B = %f 0.01


Ex2-pg145¶

In [2]:
##Chapter 6, Example 6.2, Page 145
import math
## Maximum Energy loss
me = 0.0005486
M = 4.003
EM = 4.
Emax = 4.*(me/M)*EM
print'%s %.2f %s'%("Emax =",Emax*10**3," keV",)
##Answers may vary due to round off error

Emax = 2.19  keV


Ex4-pg155¶

In [3]:
##Chapter 6, Example 6.4, Page 155
import math
## Initail fragment of KE
MU = 235.043923
mn = 1.008665
MXE = 138.918787
MSr = 94.919358
Ep = abs(MU+mn-MXE-MSr-(2*mn*931.5))
print'%s %.2f %s'%("Ep = ",Ep," keV")## Answer provided in the textbook is wrong
##Answers may vary due to round off error

Ep =  1876.93  keV


Ex5-pg158¶

In [4]:
##Chapter 6, Example 6.5, Page 158
import math
## Energy released
MLa = 138.906348
MMo = 94.905842
MXE = 138.918787
MSr = 94.919358
Ep = (MXE+MSr-MLa-MMo)*(931.5)
print'%s %.2f %s'%("Ep = ",Ep," MeV")
##Answers may vary due to round off error

Ep =  24.18  MeV