University of Dhaka
Department of Nuclear Engineering
Course Name: Advanced Nuclear Reactors Design and Features
Course Code: NE-4201
Homework Assignment 2
Submitted to:
Farhana Islam Farha
Lecturer
Dept. of Nuclear Engineering
University of Dhaka
Submitted by:
Name: Minhajur Rahman
Batch: 5th
Roll No: 25
Submitted on: October 3, 2022
1. [20] A beam of 0.025eV neutrons, which has a cross-sectional area of 1.0cm2 strikes a small,
thin 6Li target, inducing the exothermic reaction 6 Li(n,α )3H . The beam intensity is 5×1015
neutrons/cm2sec. The target has an area of 0.5cm2 and it is 0.005cm in thickness. The cross
section for this reaction at 0.025 eV is 945 barns.
(1) What is density of neutrons in the beam?
(2) What is the atom density of the target at room temperature? Assume the Li-6 target density
is 0.5 g/cm3. Note that the Li-6 atomic density decreases with time as dN6(t)/dt = −σIN6 (t),
where N6 is the atomic density of lithium-6 target, I is the neutron beam intensity, and σ is
the microscopic (n,α) cross section of Li-6.
(3) What is the rate of production of tritium in the target? Note that the H-3 production rate is
a function of time as the Li-6 atomic density is.
(4) What is the maximum tritium activity, which can be induced in the target by this beam?
Note that H-3 decays with a half-life of 12 years and thus the number of H-3 atoms varies
with time as dn3(t)/dt = P(t) − λn3(t), where n3 is the number of H-3 atoms in the target, P(t)
is the H-3 production rate, and λ is the decay constant of H-3.
Solution:
(1) Given that, the energy of the neutrons beam is 0.025 eV and the beam intensity is 5×1015
neutrons/cm2-sec.
The kinetic energy of neutrons is negligible to its rest energy, so we use the classical kinetic
energy formula to find their velocity.
1
𝐸 = 𝑚𝑣 2
2
⇒𝑣=√
=√
2𝐸
𝑚
2 × 0.025 × 1.60 × 10−19
1.67 × 10−27
= 2188.70 𝑚𝑠 −1
The density of neutrons in the beam, n can be found as follows:
𝐼 = 𝑛𝑣
𝐼
⇒𝑛=
𝑣
5 × 1015
=
2188.70 × 100
= 2.28 × 1010 𝑛𝑒𝑢𝑡𝑟𝑜𝑛𝑠/𝑐𝑚3
(2) Given that, the Li-6 target density is 0.5 g/cm3.
The atom density of the target, N6(0) at the beginning is calculated from below relationship:
𝜌𝑁𝐴
𝑁6 (0) =
𝑀
0.5 × 6.023 × 1023
=
6
= 5.02 × 1022𝑎𝑡𝑜𝑚𝑠/𝑐𝑚3
(3) Given that,
Cross-section, σ = 945b; Intensity, I = 5 × 1015 neutrons/cm2 – sec; Target area, A =
0.5cm2 and thickness, X = 0.005 cm.
The rate of tritium generation, dn3/dt must equal to the rate of consumption of 6Li, dn6/dt as
per the following reaction: 6Li +1n = 3H +4α
Now, the production rate of tritium is𝑑𝑛3
𝑑𝑛6
𝑑𝑁6
=−
= −𝐴𝑋
= 𝜎𝐼𝑁6 (𝑡)𝐴𝑋
𝑑𝑡
𝑑𝑡
𝑑𝑡
Now as per (2),
𝑑𝑁6
= −𝜎𝐼𝑁6 (𝑡)
𝑑𝑡
The equation is solved for N6 as follows:
𝑑𝑁6
= −𝜎𝐼𝑑𝑡
𝑁6
Integrating both sides,
𝑁6 (𝑡)
𝑡
𝑑𝑁6
= −𝜎𝐼 ∫ 𝑑𝑡′
𝑁6 (0) 𝑁6
0
(𝑡)
𝑁6
⇒ 𝑙𝑛
= −𝜎𝐼𝑡
𝑁6 (0)
⇒ 𝑁6 (𝑡) = 𝑁6 (0)𝑒 −𝜎𝐼𝑡
∫
∴
𝑑𝑛3
= 𝜎𝐼𝑁6 (0)𝑒 −𝜎𝐼𝑡 𝐴𝑋
𝑑𝑡
= 945 × 10−24 × 5 × 1015 ×5.02× 1022 × 𝑒 −𝜎𝐼𝑡 × 0.5 × 0.005
−6 𝑡
= 5.93 × 1014 × 𝑒 −4.73×10
𝑡𝑟𝑖𝑡𝑖𝑢𝑚𝑠/𝑠𝑒𝑐
(4) Given in the question:
The half life of H-3 is 12 years, hence decay constant is
𝑙𝑛2
𝜆=
12 × 365 × 24 × 3600
= 1.83 × 10−9 𝑠 −1
The number of H-3 atoms varies with time as
𝑑𝑛3 (𝑡)
= 𝑃(𝑡) − 𝜆𝑛3 (𝑡)
𝑑𝑡
where n3 is the number of H-3 atoms in the target, P(t) is the H-3 production rate, and λ is
the decay constant of H-3.
Rearranging and multiplying both sides by an integration factor of 𝑒 𝜆𝑡 leads to:
𝑑𝑛3 (𝑡)
𝑒 𝜆𝑡
+ 𝜆𝑒 𝜆𝑡 𝑛3 = 𝑒 𝜆𝑡 𝑃(𝑡)
𝑑𝑡
𝑑
⇒
[𝑛 𝑒 𝜆𝑡 ] = 𝑒 𝜆𝑡 𝑃(𝑡)
𝑑𝑥 3
The tritium production rate as per question (c) is,
−6
𝑃(𝑡) = 5.93 × 1014 × 𝑒 −4.73×10 𝑡 𝑡𝑟𝑖𝑡𝑖𝑢𝑚𝑠/𝑠𝑒𝑐
The above equation can be written in a more convenient form by taking the constants as
α = 5.93 × 1014 and 𝛽 = 4.73 × 10−6 which results in 𝑃(𝑡) = 𝛼𝑒 −𝛽𝑡
Replacing P(t) and integrating both sides with respect to t gives:
𝑛3 𝑒 𝜆𝑡 = ∫ 𝛼𝑒 (𝜆−𝛽)𝑡 𝑑𝑡
=
𝛼
𝑒 (𝜆−𝛽)𝑡 + 𝐶
𝜆−𝛽
𝛼
At t = 0, n3 = 0 gives, 𝐶 = − 𝜆−𝛽
𝛼
Substituting C in the equation of 𝑛3 𝑒 𝜆𝑡 yields, 𝑛3 = 𝜆−𝛽 [𝑒 −𝛽𝑡 − 𝑒 −𝜆𝑡 ]
Tritium activity is maximum when its concentration is maximum, i.e.,
𝑑𝑛3 (𝑡)
=0
𝑑𝑡
𝑑𝑛3 (𝑡)
𝑑𝑡
=0
𝛼
[−𝛽𝑒 −𝛽𝑡 + 𝜆𝑒 −𝜆𝑡 ] = 0
𝜆−𝛽
⇒ 𝜆𝑒 −𝜆𝑡 = 𝛽𝑒 −𝛽𝑡
𝜆
⇒ 𝑒 (𝜆−𝛽)𝑡 =
𝛽
𝑙𝑛𝜆 − 𝑙𝑛𝛽
𝑡=
𝜆−𝛽
Substituting in the values for B and 𝜆
ln(1.83 × 10−9 ) − ln (4.73 × 10−6)
𝑡=
1.83 × 10−9 − 4.73 × 10−6
= 1.66 × 106 𝑠
Thus, the maximum activity of tritium is,
𝐴 = 𝜆𝑛3
𝜆𝛼
=
[𝑒 −𝛽𝑡 − 𝑒 −𝜆𝑡 ]
𝜆−𝛽
1.83 × 10−9 × 5.93 × 1014 −4.73×10−6×1.66×106
−9
6
=
[𝑒
− 𝑒 −1.83×10 ×1.66×10 ]
−9
−6
1.83 × 10 − 4.73 × 10
= 2.29 × 1011 𝐵𝑞
= 6.20𝐶𝑖
⇒
2. [20] The control rods in the Ford Nuclear Reactor were solid stainless steel rods with a
rectangular cross section, approximately 1 inch x 2 inch x 24 inches long. Boron-10 is the
principal neutron absorber and is added to the control rods by mixing natural boron in with the
stainless steel at a concentration of 2% by weight.
(1) Using the JANIS data (ENDF/B-VIII), calculate the probability that a thermal (0.0253 eV)
neutron normally incident on the face of the control rod will exit the other side (hence
traveling one inch) without a collision. Assume that you only need to consider Fe-56 in the
stainless steel but consider both B-10 and B-11 in the boron.
(2) What is the relative probability of the thermal neutron having a collision (any collision)
with a B-10 nucleus vs. a B-11 nucleus vs. a Fe-56 nucleus? (These must add to unity.).
Solution:
(1) The probability that a incident neutron passes through a material of thickness T uncollided is
given by:
𝐼
= 𝑒 −Σ𝑡𝑇
𝐼0
Where, the total macroscopic cross section Σt =σiNi for all i components of the material.
The density of 304 Stainless steel, commonly used in control rods, has a mass density, ρ ss =
7.74g/cm3
The atom density of each component in the alloy, Ni can be found as follows:
𝜔𝑖 × 𝜌𝑠𝑠 × 𝑁𝐴
𝑁𝑖 =
𝐴𝑖
In natural boron, 18.41% weight is 10B and the rest is 11B. So,
98 × 7.74 × 6.023 × 1023
𝑁𝐹𝑒−56 =
100 × 56
= 8.16 × 1022 𝑐𝑚−3
18.41 × 2 × 7.74 × 6.023 × 1023
𝑁𝐵−10 =
100 × 100 × 10
= 1.72 × 1021 𝑐𝑚−3
81.59 × 2 × 7.74 × 6.023 × 1023
𝑁𝐵−11 =
100 × 100 × 11
= 6.92 × 1021 𝑐𝑚−3
From JANIS WEB:
𝜎𝑡 (𝑏)
Fe-56
B-10
B-11
14.79
3848.11
5.07
∴ Σ𝑡 = (14.79 × 8.16 × 1022 + 3848.11 × 1.72 × 1021 + 5.07 × 6.92 × 1021 ) ×
10−24
= 7.86𝑐𝑚−1
∴ Probability,
𝐼
= 𝑒 −7.86×1×2.54 = 2.14 × 10−4
𝐼0
(2) Relative probability of collision with:
Fe-56:
14.79 × 8.16 × 1022
7.86
= 0.154
B-10:
3848.11 × 1.72 × 1021
7.86
= 0.842
B-11:
5.07 × 6.92 × 1021
7.86
= 0.004
3. [20] Two thermal neutron beams (0.025 eV) are injected from opposite directions along the x- axis
into a sample of pure U235. The thickness of the sample (in the direction of the beams) is 1 cm
and the cross sectional area of the target is 2 cm2. The beam incident on the left side of the sample
has an intensity of 1×1010 neutrons/cm2-s while the beam incident on the right side has an intensity
of 2×1010 neutrons/cm2-s. Use JANIS data (ENDF/B-VIII) and assume that all reactions in U235
remove neutrons from the parallel beams. That is, a scattering reaction will remove a neutron from
the parallel beam because the neutron will go off in a different direction.
(1) What is the scalar flux at x = 0, 0.5, and 1 cm, where x = 0 on the left surface of target?
(2) What is the rate at which fissions are occurring in the target? (i.e., # fissions/s)
(3) What is the rate at which neutron captures are occurring in the target?
(4) How many neutrons/s are going from left to right at x = 0, .5, and 1 cm?
(5) How many neutrons/s are going from right to left at x = 0, .5, and 1 cm?
(6) What is the net number of neutrons/s going in the +x direction at x = 0, .5, and 1 cm? How is
this related to your solution for (5) and (6) above?
Solution:
(1) The scaler flux ϕ(x) is the sum of the fluxes directed at both left and right hand sides:
ϕ(𝑥) = 𝜙𝑙𝑒𝑓𝑡 (𝑥) + 𝜙𝑟𝑖𝑔ℎ𝑡 (1 − 𝑥)
= 𝜙𝑙𝑒𝑓𝑡 (0). 𝑒 −Σ𝑡𝑥 + 𝜙𝑟𝑖𝑔ℎ𝑡 (0). 𝑒 −Σ𝑡(1−𝑥)
Where Σt = σtN235; N235 = atom density of 235U
If mass density of U-235, ρ235= 19.1 g/cm3 and σt = 700.24
barns (from JANIS WEB). Then,
∴ Σ𝑡 = 𝜎𝑡 𝑁235
= 700.24 × 4.89 × 1022
= 34.28𝑐𝑚−1
𝜌235 𝑁𝐴
235
19.1 × 6.023 × 1023
=
235
= 4.89 × 1022𝑐𝑚−3
𝑁235 =
Now, the scalar flux at x = 0, 0.5, and 1 cm are calculated below:
ϕ(0) = 𝜙𝑙𝑒𝑓𝑡 (0) + 𝜙𝑟𝑖𝑔ℎ𝑡 (0). 𝑒 −Σ𝑡
=1 × 10 10 + 2 × 10 10. 𝑒 −34.28 𝑛𝑒𝑢𝑡𝑟𝑜𝑛𝑠/𝑐𝑚2 . 𝑠
=1 × 10 10 𝑛𝑒𝑢𝑡𝑟𝑜𝑛𝑠/𝑐𝑚2 . 𝑠
Φ(0.5) = 𝜙𝑙𝑒𝑓𝑡 (0)𝑒 −0.5Σ𝑡 + 𝜙𝑟𝑖𝑔ℎ𝑡 (0). 𝑒 −0.5Σ𝑡
=1 × 10 10𝑒 −0.5×34.28 + 2 × 10 10 . 𝑒 −0.5×34.28 𝑛𝑒𝑢𝑡𝑟𝑜𝑛𝑠/𝑐𝑚2 . 𝑠
=1.09 × 10 3 𝑛𝑒𝑢𝑡𝑟𝑜𝑛𝑠/𝑐𝑚2 . 𝑠
ϕ(1) = 𝜙𝑙𝑒𝑓𝑡 (0). 𝑒 −Σ𝑡 + 𝜙𝑟𝑖𝑔ℎ𝑡 (0)
=1 × 10 10. 𝑒 −34.28 + 2 × 10 10 𝑛𝑒𝑢𝑡𝑟𝑜𝑛𝑠/𝑐𝑚2 . 𝑠
=2 × 10 10 𝑛𝑒𝑢𝑡𝑟𝑜𝑛𝑠/𝑐𝑚2 . 𝑠
(2) Fission rate in target = ∑𝑈−235
𝜑𝑉𝑡𝑎𝑟𝑔𝑒𝑡 = 𝜎𝑓𝑈−235𝑁235𝜑𝐴𝑡𝑎𝑟𝑔𝑒𝑡 𝑥
𝑓
From JANIS, 𝜎𝑓𝑈−235 = 586.7371𝑏
∴ 𝐹𝑖𝑠𝑠𝑖𝑜𝑛 𝑟𝑎𝑡𝑒 = 586.7371 × 10−24 × 4.89 × 1022 × (1 × 1010 + 2 × 1010 ) × 2 × 1
= 1.72 × 1012 𝑠 −1
(3) Capture rate in target = ∑𝑈−235
𝜑𝑉𝑡𝑎𝑟𝑔𝑒𝑡 = 𝜎𝑐𝑈−235 𝑁235 𝜑𝐴𝑡𝑎𝑟𝑔𝑒𝑡 𝑥
𝑐
From JANIS, 𝜎𝑐𝑈−235 = 99.39𝑏
∴ 𝐶𝑎𝑝𝑡𝑢𝑟𝑒 𝑟𝑎𝑡𝑒 = 99.39 × 10−24 × 4.89 × 1022 × (1 × 1010 + 2 × 1010 ) × 2 × 1
= 2.92 × 1011 𝑠 −1
(4) The flux directed from left to right, 𝜑𝑙𝑒𝑓𝑡 (𝑥) = 𝜑𝑙𝑒𝑓𝑡 (0)𝑒 −Σ𝑡𝑥
So, the rate of neutrons going toward right, 𝑛̇ 𝑙𝑒𝑓𝑡 = 𝜑𝑙𝑒𝑓𝑡 𝐴𝑡𝑎𝑟𝑔𝑒𝑡
At x = 0cm,
𝑛̇ 𝑙𝑒𝑓𝑡 (0) = 𝜑𝑙𝑒𝑓𝑡 (0)𝐴𝑡𝑎𝑟𝑔𝑒𝑡 𝑒 −Σ𝑡×0
= 1 × 1010 × 2
= 2 × 1010 𝑠 −1
At x = 0.5cm,
𝑛̇ 𝑙𝑒𝑓𝑡 (0.5) = 1 × 1010 × 2 × 𝑒 −Σ𝑡×0.5 𝑠 −1
= 7.23 × 102 𝑠 −1
At x = 1cm,
𝑛̇ 𝑙𝑒𝑓𝑡 (0.5) = 1 × 1010 × 2 × 𝑒 −Σ𝑡×1 𝑠 −1
= 2.62 × 10−5 𝑠 −1
(5) The flux directed from right to left, 𝜑𝑟𝑖𝑔ℎ𝑡 (𝑥) = 𝜑𝑟𝑖𝑔ℎ𝑡 (0)𝑒 −Σ𝑡𝑥
So, the rate of neutrons going toward right, 𝑛̇ 𝑟𝑖𝑔ℎ𝑡 = 𝜑𝑟𝑖𝑔ℎ𝑡 𝐴𝑡𝑎𝑟𝑔𝑒𝑡
At x = 0cm,
𝑛̇ 𝑟𝑖𝑔ℎ𝑡 (0) = 𝜑𝑟𝑖𝑔ℎ𝑡 (0)𝐴𝑡𝑎𝑟𝑔𝑒𝑡 𝑒 −Σ𝑡×(1−0)
= 2 × 1010 × 2 × 𝑒 −Σ𝑡×1
= 5.23 × 10−5 𝑠 −1
At x = 0.5cm,
𝑛̇ 𝑙𝑒𝑓𝑡 (0.5) = 2 × 1010 × 2 × 𝑒 −Σ𝑡×0.5 𝑠 −1
= 1.45 × 10−3 𝑠 −1
At x = 1cm,
𝑛̇ 𝑙𝑒𝑓𝑡 (0.5) = 2 × 1010 × 2 × 𝑒 −Σ𝑡×0 𝑠 −1
= 4 × 1010 𝑠 −1
(6) Net rate of neutron passage in +X direction, 𝑛̇ 𝑛𝑒𝑡 = 𝑛̇ 𝑙𝑒𝑓𝑡 − 𝑛̇ 𝑟𝑖𝑔ℎ𝑡
So, the net number of neutrons/s going in the +x direction at x = 0, .5, and 1 cm:
𝑛̇ 𝑛𝑒𝑡 (0) = 2 × 1010 𝑠 −1
𝑛̇ 𝑛𝑒𝑡 (0.5) = 723 − 1450
= −7.27 × 102 𝑠 −1
𝑛̇ 𝑛𝑒𝑡 (1) = 4 × 1010 𝑠 −1
As almost all the neutrons have collided once by the time each flux has traversed the medium,
the neutron passage rate at each face of the medium equals the incident flux at each face. As the
-X directed flux is doubly intense relative to the +X directed flux, the net flux is directed in the X direction at the midpoint between the faces.
4. [10] A neutron beam of intensity I0 neutrons/cm-s going in the +x direction is incident on
purely absorbing semi-infinite slab [0, ∞) with an exponentially varying absorption cross section
Σa (x) = Σa0 e−x/d for x > 0 and d is a positive constant. (The atmosphere is an example of such a
medium because its density decreases exponentially with height above the sea level.)
(1) Find the neutron intensity I (x) for x > 0
(2) Will all the incident neutrons be absorbed (eventually) in the infinite slab? If not, what fraction
of the incident neutrons will not be absorbed?
Solution:
(1) We know the decrease in intensity is given by
𝑥
−𝑑𝐼 = 𝐼Σ𝑎 𝑑𝑥 = 𝐼Σ𝑎0 𝑒 −𝑑 𝑑𝑥
𝐼(𝑥)
𝑥
𝑥
𝑑𝐼
⇒∫
= −Σ𝑎0 ∫ 𝑒 −𝑑 𝑑𝑥
𝐼
𝐼0
0
𝑥
𝐼(𝑥)
⇒ 𝑙𝑛
= Σ𝑎0 𝑑(𝑒 −𝑑 − 1)
𝐼0
⇒ 𝐼(𝑥) = 𝐼0 𝑒 Σ𝑎0 𝑑(𝑒
This is the neutron intensity I(x) for x > 0.
−
𝑥
𝑑 −1)
(2) Since medium is purely absorbing, there can be no scattering interactions, which are the only
way that allows incident neutrons to escape the medium. So, all incident neutrons will be
absorbed.
5. [20] The configuration to be solved is a semi-infinite homogeneous diffusing medium with
diffusion constant D and absorption cross section Σa . There is a beam of thermal neutrons of
strength I0 neutrons/cm2 -s incident normally on the left side of the medium. This could be looked
at as an example of a medical physics application where a beam of thermal neutrons is incident on
a patient.
(1) Set up the diffusion equation and boundary conditions for this problem. The normal beam of
neutrons on the left boundary may be represented as an incoming partial current at x = 0 or J +
(0) = I0.
(2) Solve this system of equations for the scalar flux φ( x) for x > 0.
Solution:
(1) The general diffusion equation (steady -state form)
𝐷𝛻 2 𝜑 − Σ𝑎 𝜑 + 𝑠 = 0
The medium is non-multiplying, i.e., S = 0.
Neutron motion is considered to be one dimensional, i.e., 𝛻 2 𝜑 =
The diffusion equation becomes:
𝑑2𝜑
− Σ𝑎 𝜑 = 0
𝑑𝑥 2
= 𝐿2, where L = diffusion length
𝐷
𝐷
We know, Σ
𝑎
So, the final diffusion equation becomes:
𝑑2𝜑 1
− 𝜑=0
𝑑𝑥 2 𝐿2
𝑑2𝜑
𝑑𝑥 2
Boundary Conditions:
a) Finite flux condition: lim 𝜑(𝑥) < ∞
𝑥→+∞
b) Vacuum boundary condition: at x = 0, neutron current = 𝐼0 , into the medium (in the +X
direction), i.e.,
lim+ 𝐽(𝑥) = 𝐽+ (0) = 𝐼0
𝑥→0
(2) To solve
𝑟𝑒 𝑟𝑥 and
𝑑2𝜑
𝑑𝑥 2
𝑑2𝜑
𝑑𝑥 2
1
− 𝐿2 𝜑 = 0, we will consider 𝜑(𝑥) = 𝑒 𝑟𝑥 as a form of solution. Then
= 𝑟 2 𝑒 𝑟𝑥
1
The equation becomes: 𝑟 2 𝑒 𝑟𝑥 − 𝐿2 𝑒 𝑟𝑥 = 0
1
Either 𝑒 𝑟𝑥 = 0 or 𝑟 2 − 𝐿2 = 0
But rx is undefined as 𝑒 𝑥 → 0 for all x.
1
So, 𝑟 = ± 𝐿
The general solution 𝜑(𝑥) is of the form
𝜑(𝑥) = 𝐴𝑒
Applying the boundary conditions:
−𝑥
𝐿
𝑥
+ 𝐵𝑒 𝐿 , 𝑥 > 0
𝑥
−𝑥
a) As 𝑥 → ∞, 𝑒 𝐿 → ∞, so B = 0 to fulfill this condition, i.e., 𝜑(𝑥) = 𝐴𝑒 𝐿
b) As 𝑥 → 0, 𝜑(0) = 𝐴 = 𝐼0
Therefore, the solution for this system:
𝜑(𝑥) = 𝐼0 𝑒
−𝑥
𝐿 ,𝑥
> 0 [𝐿 = √
𝐷
Σ𝑎
𝑑𝜑
𝑑𝑥
=