Homework 2

NUCE 2100

Dane Sabo

Instructions: Complete the problems below being sure to show your work. If you need to lookup nuclear data from an external source please reference the source in your solutions (once is sufficient).

Consider the so-called DT (deuterium, tritium) fusion reaction

^2H + ^3H \rightarrow ^HHe + ?

a. What is the missing product in the reaction?

A neutron!

b. Calculate the binding energy of ^2H, ^{3}H, and ^{4}He

def binding_energy(mass,z,n):
    c = 299792458 # m/s
    h1_mass = 1007825.03190 #micro AMU
    neutron_mass = 1008664.9159 #mirco AMU

    mass_defect = (z*h1_mass + n*neutron_mass - mass)/1e6 #calculate mass defect, micro amu to u
    energy = (mass_defect*1.6606e-27)*c**2 #convert amu to kg, answer in J

    return energy*6.242e12 #Convert J to MeV, and return value
    
twoH_mass = 2014101.77784 #micro AMU
print("2H Binding Energy:", binding_energy(twoH_mass,1,1), "MeV")

threeH_mass = 3016049.281328 #micro AMU
print("3H Binding Energy:", binding_energy(threeH_mass,1,2), "MeV")

fourHe_mass = 4002603.25413 #micro AMU
print("4He Binding Energy:", binding_energy(fourHe_mass,2,2), "MeV")

2H Binding Energy: 2.22482284320947 MeV
3H Binding Energy: 8.482774677373571 MeV
4He Binding Energy: 28.298926363869242 MeV

c. Calculate the Q value of the reaction

neutron_mass = 1008664.9159 #mirco AMU
c = 299792458 # m/s

#Calculate difference in mass of products - reactants, mulitply by c^2
Q = (fourHe_mass + neutron_mass - twoH_mass - threeH_mass)/1e6*1.6606e-27 * c**2
print("Q is ", Q*6.242e12, "MeV")

Q is  -17.59132884328642 MeV

d. Show that the Q value is equal to the change in binding energy

Q_from_binding = binding_energy(threeH_mass,1,2) + binding_energy(twoH_mass,1,1) - binding_energy(fourHe_mass,2,2)
print("Q from binding is ", Q_from_binding, "MeV")

Q from binding is  -17.5913288432862 MeV

Using atomic mass data, compute the average binding energy per nucleon of the following nuclei:

a. ^{6}Li

b. ^{12}C

c. ^{51}V

d. ^{138}Ba

e. ^{235}U

Atomic Mass Data

#For each part, find the binding energy of the nuclide, then divde by the number of nucleons.

sixLi_mass = 6015122.8874 #micro AMU
print("A: Average 6Li Binding Energy:", binding_energy(sixLi_mass,3,3)/6, "MeV per nucleon")

twelveC_mass = 12000000.0 #micro AMU
print("B: Average 12C Binding Energy:", binding_energy(twelveC_mass,6,6)/12, "MeV per nucleon")

fiftyoneV_mass = 50943957.66 #micro AMU
print("C: Average 51V Binding Energy:", binding_energy(fiftyoneV_mass,23,28)/51, "MeV per nucleon")

onethirtyeightBa_mass = 137905247.06 #micro AMU
print("D: Average 138Ba Binding Energy:", binding_energy(onethirtyeightBa_mass,56,82)/138, "MeV per nucleon")

twothirtyfiveU_mass = 235043928.1  #micro AMU
print("E: Average 235U Binding Energy:", binding_energy(twothirtyfiveU_mass,92,143)/235, "MeV per nucleon")

A: Average 6Li Binding Energy: 5.332946257554306 MeV per nucleon
B: Average 12C Binding Energy: 7.681030488461077 MeV per nucleon
C: Average 51V Binding Energy: 8.743093705441115 MeV per nucleon
D: Average 138Ba Binding Energy: 8.394390417035352 MeV per nucleon
E: Average 235U Binding Energy: 7.5917907582946915 MeV per nucleon

Compute the atom densities of ^{235}U and ^{238}U in UO2 of physical density 10.8 g/cm3 if the uranium is enriched to 3.5 w/o in ^{235}U.

twothirtyeightU_mass = 238050786.9 # micro AMU
UO2_density = 10.8 #g/cm^3

N_235 = (100-3.5)/3/100 * UO2_density*0.6022e24/(twothirtyfiveU_mass*1e-6)
#(abundance of U235 in UO2, remember only 1 U for every 2 O!) * converting to atoms / cm^3
print("There are ", N_235, " 235U atoms per cm^3")

N_238 = (3.5)/3/100 * UO2_density*0.6022e24/(twothirtyeightU_mass*1e-6)
print("There are ", N_238, " 238U atoms per cm^3")

There are  8.900646006519835e+21  235U atoms per cm^3
There are  3.187437478703836e+20  238U atoms per cm^3

Calculate the mean free path of 1-eV neutrons in graphite. The total cross section of carbon at this energy is 4.8 b.

Graphite Density Source

graphite_density = 1.82 #g/cm^3
graphite_nuclei_density = graphite_density/12.000*0.6022e24 #atoms/cm^3
#Take the density of graphite, divide by 12.000 g 12C / mole, then multiply by Avogadro's Number to get to atoms
print("The nuclei density of graphite is ",graphite_nuclei_density," atoms/cm^3")

mean_free_path = 1/(4.8*10e-24)/graphite_nuclei_density
#Follow the formula from the slides, where we are converting Barns to cm.
print("The mean free path is ",mean_free_path,"cm in graphite.")

The nuclei density of graphite is  9.133366666666668e+22  atoms/cm^3
The mean free path is  0.22810135729431646 cm in graphite.

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Homework 2

NUCE 2100

Dane Sabo

4.9 KiB

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