Chapter 23:Nuclear Chemistry

Example no:23.2,Page no:994

In [5]:
#Variable declaration
NA=6.022*10**23 #avogadro number
c=3*10**8 #speed of light, m/s
p=1.007825 #mass of proton, amu
n=1.008665 #mass of neutron, amu
mI=126.9004 #atomic mass of I, amu

#Calculation
pI=53*p+74*n #estimated mass of I, amu
deltam=mI-pI #mass defect, amu
deltaE=-deltam*c**2 #energy released, amu m**2/s**2
deltaE=deltaE/(NA*1000) #energy released in J
print"Nuclear binding energy is %.2e"%deltaE,"J"
deltaE=deltaE/127 #binding energy per nucleon, J

#Result
print"The nuclear binding energy per nucleon is :%.2e"%deltaE,"J/nucleon"
Nuclear binding energy is 1.73e-10 J
The nuclear binding energy per nucleon is :1.36e-12 J/nucleon