Lecture 14
Problem Set-4
Objectives
In this lecture you will learn the following
To illustrate the application of the elastic scattering process and elementary diffusion theory.
First we shall illustrate several examples on scattering that conveys some physics.
Then we illustrate application of connection between flux and current density.
Finally, we shall look at application of superposition for a multiple source problem.
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Lecture 14
Problem Set-4
Question-1
1. A 2-MeV neutron travelling in water has a head-on collision with an O 16 nucleus. (a) What are the energies of the neutron and nucleus after the collision? (b) Would you
expect the water molecule involved in the collision to remain intact after the event?
Solution
For head-on collision, the final energy of the neutron after collision is given by αE 0,
where E 0 is the original energy of the neutron and For O nucleus, A = 16. Hence Therefore final energy = 0.7785 x 2 = 1.557 MeV.
Hence energy of O nucleus = 2.0-1.557 = 0.443 MeV. Since this shall be much higher
than the bond energy, O nucleus will detach and a free radical will be generated.
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Lecture 14
Problem Set-4
Question-2
A 1-MeV neutron strikes a C12 nucleus that is initially at rest. If the neutron is
elastically scattered through an angle of 90o , (a) what is the energy of the scattered
neutron? (b) what is the energy of the recoiling nucleus? (c) at what angle does the
recoiling nucleus appear?
Solution
We had shown that the ratio of the square root of the energy of a neutron after and before
collision can be written as
For θ = 90o , and A = 12, we can write
For E = 1 MeV, we get
This implies that energy of C12 is = 1 – 0.846 = 0.154 MeV.
The following figures show the collision process and momentum balance
It may be understood from the above figure that 1
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Lecture 14
Problem Set-4
Question-3
Show that the average fractional energy loss expressed during elastic scattering for large
A nucleus is given approximately by,
We had shown that the average energy of a neutron after collision as a fraction of initial
energy is
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Lecture 14
Problem Set-4
Question 4
Calculate the average number of collisions required to reduce the energy of a neutron
from 2 MeV to 0.025 eV in H1 and C12 .
Solution
For a neutron originally having energy E after a collision will on the average have
energy
After two collisions, the energy shall be
Thus, after n collisions, the energy shall be
For Hydrogen nucleus
In this problem, E = 2 MeV, En = 0.025 eV
For Carbon nucleus
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Lecture 14
Problem Set-4
Question-5
The neutron flux in a bare spherical reactor of radius 50 cm is given by 5 x
1013(sin(0.0628r)/r) neutron/cm 2-s, where ‘r’ is measured in cm from the centre of the
reactor. The diffusion coefficient for the system is 0.80 cm. (a) What is the maximum
value of the flux in the reactor? (b) Calculate the neutron current density as a function
of position in the reactor. (c) How many neutrons escape from the reactor per second?
Solution
a) The flux is given by
The maximum value of above function shall be at r = 0
n/cm 2 - s
b) Current density
c) Current density at r = 50 cm is given by
n/cm 2 - s
Number of neutrons leaking would be given by
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Lecture 14
Problem Set-4
Question-6
Isotropic point sources each emitting S neutrons/sec are placed in infinite moderator at
the four corners of a square of side ‘a’. Compute the flux and current at its centre.
Solution
Distance from centre to any source is
We have shown that the flux from a point source at a distance r is given by
Thus the flux from a single source shall be
Flux from all the four sources shall be
The current from each source shall be radially outward as shown below
The vector addition of the current shall be 0. Hence current density in any direction is 0.
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