480
8 Nuclear Physics – II
At the surface the neutron density corresponding to r = 9.6 cm and mean
neutron velocity 2.2 × 10
5 cm s
−1
n(r ) =
φ
v
=
3Q
4πλ tr v
e
−9.6/L
9.6
= 2.93 × 10
−9 Q cm
−3
8.88 Consider the diffusion equation
∂n
∂t
= S +
λ tr
3
∇
2
φ − φΣa
(1)
where n is the neutron density, S is the rate of production of neutrons/
cm
3
/s, φΣa is the absorption rate/ cm
3
/s and
λ tr ∇
2
φ
3
represents the leakage
of neutrons. Σa is the macroscopic cross-section, φ is the neutron flux and λ tr
is the transport mean free path.
Since it is a steady state,
∂n
∂t
= 0. Further S = 0 because neutrons are not
produced in the region of interest. As we are interested only in the x-direction
the Laplacian reduces to
d
2
dx 2 . Thus (1) becomes
λ tr
3
d
2
φ
dx 2 − φΣa = 0
( 2 )
or
d
2
φ
dx 2 − K
2
φ = 0
( 3 )
where
K
2
=
3Σa
λ tr
=
3
λ a λ tr
(4)
λ a being the absorption mean free path.The solution of (3) is
φ = C 1 e
Kx
+ C 2 e
−Kx
(5)
where C 1 and C 2 are constants of integration. The condition that the flux
should be finite at any point including at infinity means that C 1 = 0. Therefore,
(5) becomes
φ = C 2 e
−Kx
(6)
We can now determine C 2 . Consider a unit area located in the YZ plane
at a distance x from the plane source as in Fig. 8.14. On an average half
of the neutrons will be travelling along the positive x-direction. As x → 0,
the net current flowing in the positive x-direction would be equal to
1
2
Q; the
diffusion of neutrons through unit area at x = 0 would have a cancelling
effect because from symmetry equal number of neutrons would diffuse in the
opposite direction.
8 Nuclear Physics – II
At the surface the neutron density corresponding to r = 9.6 cm and mean
neutron velocity 2.2 × 10
5 cm s
−1
n(r ) =
φ
v
=
3Q
4πλ tr v
e
−9.6/L
9.6
= 2.93 × 10
−9 Q cm
−3
8.88 Consider the diffusion equation
∂n
∂t
= S +
λ tr
3
∇
2
φ − φΣa
(1)
where n is the neutron density, S is the rate of production of neutrons/
cm
3
/s, φΣa is the absorption rate/ cm
3
/s and
λ tr ∇
2
φ
3
represents the leakage
of neutrons. Σa is the macroscopic cross-section, φ is the neutron flux and λ tr
is the transport mean free path.
Since it is a steady state,
∂n
∂t
= 0. Further S = 0 because neutrons are not
produced in the region of interest. As we are interested only in the x-direction
the Laplacian reduces to
d
2
dx 2 . Thus (1) becomes
λ tr
3
d
2
φ
dx 2 − φΣa = 0
( 2 )
or
d
2
φ
dx 2 − K
2
φ = 0
( 3 )
where
K
2
=
3Σa
λ tr
=
3
λ a λ tr
(4)
λ a being the absorption mean free path.The solution of (3) is
φ = C 1 e
Kx
+ C 2 e
−Kx
(5)
where C 1 and C 2 are constants of integration. The condition that the flux
should be finite at any point including at infinity means that C 1 = 0. Therefore,
(5) becomes
φ = C 2 e
−Kx
(6)
We can now determine C 2 . Consider a unit area located in the YZ plane
at a distance x from the plane source as in Fig. 8.14. On an average half
of the neutrons will be travelling along the positive x-direction. As x → 0,
the net current flowing in the positive x-direction would be equal to
1
2
Q; the
diffusion of neutrons through unit area at x = 0 would have a cancelling
effect because from symmetry equal number of neutrons would diffuse in the
opposite direction.