# Chapter13-Nuclear Fission and Fusion¶

## Ex1-pg600¶

In [9]:
## Exa13.1 : : Page-600 (2011)
#calculate rate of fission and energy released
import math
E = 200.*1.6023e-13;     ## Energy released per fission, joule
E_t = 2.;                ## Total power produced, watt
R_fiss = E_t/E;        ## Fission rate, fissions per sec
m = 0.5;                ## Mass of uranium, Kg
M = 235.;                ## Mass number of uranium
N_0 = 6.023e+26;        ## Avogadro's number, per mole
N = m/M*N_0            ## Number of uranium nuclei
E_rel = N*E/4.08*10**-3;        ## Energy released, kilocalories
print'%s %.2e %s %.2e %s'%("\nThe rate of fission of U-235 = ",R_fiss," fissions per sec"and" \nEnergy released =",E_rel," kcal");

## Result
## The rate of fission of U-235 = 6.24e+010 fissions per sec
## Energy released = 1.006535e+010 kcal

The rate of fission of U-235 =  6.24e+10
Energy released = 1.01e+10  kcal


## Ex2-pg600¶

In [8]:
## Exa13.2 : : Page-600 (2011)
#calculate number of free neutrons in the reactor
import math
E = 200*1.6e-13;        ## Energy released per fission, joules per neutron
t = 10**-3;            ## Time, sec
P = E/t;              ## Power produced by one free neutron, watt per neutron
P_l = 10**9;            ## Power level, watt
N = P_l/P;            ## Number of free neutrons in the reactor, neutrons
print'%s %.2e %s'%("\nThe number of free neutrons in the reactor = ",N," neutrons");

## Result
## The number of free neutrons in the reactor = 3.125e+016 neutrons

The number of free neutrons in the reactor =  3.12e+16  neutrons


## Ex3-pg600¶

In [7]:
##  Exa13.3 : : Page-600 (2011)
#calculate average number of neutrons
import math
N_0_235 = 1.;        ## Number of uranium 235 per 238
N_0_238 = 20.;        ## Number of uranium 238 for one uranium 235
sigma_a_235 = 683.;   ## Absorption cross section for uranium 235, barn
sigma_a_238 = 2.73;  ## Absorption cross section for uranium 238, barn
sigma_f_235 = 583.;   ## Fission cross section, barn
sigma_a = (N_0_235*sigma_a_235+N_0_238*sigma_a_238)/(N_0_235+N_0_238); ##Asorption cross sec, barn
sigma_f = N_0_235*sigma_f_235/(N_0_235+N_0_238);        ## Fisssion cross section
v = 2.43;
eta = v*sigma_f/sigma_a;    ##    Average number of neutron released per absorption
print'%s %.2f %s'%("\nThe average number of neutrons released per absorption = ", eta,"");

## Result
## The average number of neutrons released per absorption = 1.921

The average number of neutrons released per absorption =  1.92


## Ex4-pg600¶

In [6]:
## Exa13.4 : : Page-600(2011)
#find excitaion energy for uranum 236 and 239 and 235
import math
a_v = 14.0;        ## Volume binding energy constant, mega electron volts
a_s = 13.0;        ## Surface binding energy constant, mega electron volts
a_c = 0.583;        ## Coulomb constant, mega electron volts
a_a = 19.3;        ## Asymmetric constant, mega electron volts
a_p = 33.5;        ## Pairing energy constant, mega electron volts
Z = 92.;            ## Atomic number
## For U-236
A = 235.;           ## Mass number
E_exc_236 = a_v*(A+1-A)-a_s*((A+1)**(2./3.)-A**(2./3.))-a_c*(Z**2/(A+1.)**(1./3.)-Z**2/A**(1/3.))-a_a*((A+1-2*Z)**2/(A+1)-(A-2*Z)**2/A)+a_p*(A+1)**(-3./4.);        ## Excitation energy for uranium 236, mega electron volts
## For U-239
A = 238.;            ## Mass number
E_exc_239 = a_v*(A+1-A)-a_s*((A+1)**(2./3.)-A**(2./3.))-a_c*(Z**2/(A+1)**(1/3.)-Z**2/A**(1./3.))-a_a*((A+1.-2.*Z)**2/(A+1)-(A-2*Z)**2/A)+a_p*((A+1)**(-3/4.)-A**(-3/4.));    ## Excitation energy for uranium 239
## Now calculate the rate of spontaneous fissioning for U-235
N_0 = 6.02214e+23;        ## Avogadro's constant, per mole
M = 235.;                ## Mass number
t_half = 3e+17*3.15e+7;        ## Half life, years
D = 0.693/t_half;        ## Decay constant, per year
N = N_0/M;                    ## Mass of uranium 235, Kg
dN_dt = N*D*3600;            ## Rate of spontaneous fissioning of uranium 235, per hour
print'%s %.2f %s %.2f %s %.2f %s '%("\nThe excitation energy for uranium 236 =",E_exc_236," MeV"and "\nThe excitation energy for uranium 239 = ",E_exc_239," MeV"and"\nThe rate of spontaneous fissioning of uranium 235 = ",dN_dt," per hour");

## Result
## The excitation energy for uranium 236 = 6.8 MeV
## The excitation energy for uranium 239 = 5.9 MeV
## The rate of spontaneous fissioning of uranium 235 = 0.68 per hour


The excitation energy for uranium 236 = 6.77
The excitation energy for uranium 239 =  5.90
The rate of spontaneous fissioning of uranium 235 =  0.68  per hour


## Ex5-pg601¶

In [5]:
## Exa13.5 : : Page-601 (2011)
#calculate total energy released during fusion
import math
a = 10**5;        ## Area of the lake, square mile
d = 1/20.;        ## Depth of the lake, mile
V = a*d*(1.6e+03)**3;   ## Volume of the lake, cubic metre
rho = 10**3;            ## Density of water, kg per cubic metre
M_water = V*rho;        ## Total mass of water in the lake, Kg
N_0 = 6.02214e+26;      ## Avogadro's constant, per mole
A = 18.;                 ## Milecular mass of water
N = M_water*N_0/A;      ## Number of molecules of water, molecules
abund_det = 0.0156e-02; ## Abundance of deterium
N_d = N*2*abund_det;    ## Number of deterium atoms
E_per_det = 43/6.;       ## Energy released per deterium atom, mega electron volts
E_t = N_d*E_per_det;    ## Total energy released during fusion, mega electron volt
print'%s %.2e %s'%("\nThe total energy released during fusion = ",E_t," MeV");

## Result
## Total energy released during fusion = 1.53e+039 MeV


The total energy released during fusion =  1.53e+39  MeV


## Ex6-pg601¶

In [4]:
## Exa13.6 : : Page-601 (2011)
#calculate temperature attained by thermonuclear device
import math
r = 1/2.;            ## Radius of the tube, metre
a = math.pi*r**2;        ## Area of the torus, square metre
V = 3*math.pi*a;        ## Volume of the torus, cubic metre
P = 10**-5*13.6e+3*9.81;   ## Pressure of the gas, newton per square metre
C = 1200e-6;        ## Capacitance, farad
v = 4e+4;            ##  potential, volts
T_room = 293;        ## Room temperature, kelvin
N_k = P*V/T_room;    ## From gas equation
E = 1/2.*C*v**2;       ## Energy stored, joules
T_k = 1/6.*E/(N_k*10.);    ## Temperature attained by thermonuclear device, kelvin
print'%s %.2e %s'%("\nThe temperature attained by thermonuclear device =",T_k," K");

## Result
## The temperature attained by thermonuclear device = 4.75e+005 K


The temperature attained by thermonuclear device = 4.75e+05  K


## Ex7-pg601¶

In [3]:
## Exa13.7 : : Page-601 (2011)
#calculate energy released by the sun temperautre of the sun
import math
G = 6.67e-11;        ## Gravitational constant, newton square m per square kg
r = 7e+08;        ## Radius of the sun, metre
M_0 = 2e+30;        ## Mass of the sun, kg
E_rel = 3/5.*G*M_0**2/r;      ## Energy released by the sun, joule
E_dia_shrink_10 = E_rel/9.;  ## Energy released when sun diameter shrink by 10 percent, joule
R = 8.314;       ## Universal gas constant, joule per kelvin per kelvin per mole
T = E_rel/(M_0*R);  ## Temperature of the sun, kelvin
print'%s %.2e %s %.2e %s %.2e %s '%("\nThe energy released by the sun = ",E_rel," joule"and" \nThe energy released when sun diameter is shrinked by 10 percent = ",E_dia_shrink_10," joule"and" \nThe temperature of the sun = ",T," kelvin ");

## Result
## The energy released by the sun = 2.29e+041 joule
## The energy released when sun diameter is shrinked by 10 percent = 2.54e+040 joule
## The temperature of the sun = 1.38e+010 kelvin


The energy released by the sun =  2.29e+41
The energy released when sun diameter is shrinked by 10 percent =  2.54e+40
The temperature of the sun =  1.38e+10  kelvin


## Ex8-pg602¶

In [2]:
## Exa13.8 : : Page-602 (2011)
# estimated Q-value
import math
A_0 = 240.;        ## Mass number of parent nucleus
A_1 = 120.;        ## Mass number of daughter nucleus
B_120 = 8.5;        ## Binding energy of daughter nucleus
B_240 = 7.6;        ##  Binding energy of parent nucleus
Q = 2*A_1*B_120-A_0*B_240;    ## Estimated Q-value, mega electron volts
print'%s %.2f %s'%("\nThe estimated Q-value is = ",Q," MeV");

## Result
## The estimated Q-value is = 216 MeV


The estimated Q-value is =  216.00  MeV


## Ex9-pg602¶

In [1]:
## Exa13.9 : : Page-602 (2011)
#find The asymmetric binding energy
import math
E = 31.7;        ## Energy, MeV
a_a = 5/9.*2**(-2/3.)*E;        ## Asymmetric binding energy term, mega electron volts
print'%s %.2f %s'%("\nThe asymmetric binding energy term = ",a_a," MeV");

## Result
## The asymmetric binding energy term = 11.1 MeV


The asymmetric binding energy term =  11.09  MeV