54
2 Critical Mass, Efficiency, and Yield
The concept of a reaction rate will be used extensively in Sects. 3.3 and 4.5 for
estimating the rates of production of plutonium and polonium within a reactor. For
the present purpose, our interest is in the probability that neutron will pass through
the layer without striking a nucleus. To approach this, it is easier to begin with the
probability that a neutron will in fact precipitate a reaction:
P react =
reactions
per second
incident neutron f lux
per second
=
(R o Σ)(s n σ )
(R o Σ)
= s n σ.
(2.3)
The probability that a neutron will pass through the slab to escape out the back
side will then be:
P escape = 1 − P react = 1 − s n σ.
(2.4)
Now consider a block of material of macroscopic thickness x. As shown in Fig. 2.2,
we can imagine this comprising a large number of thin slabs each of thickness s
placed back-to-back.
The number of slabs is x/s. If N o neutrons are incident on the left side of the
block, the number that would survive to emerge from the first thin slab would be
N o P, where P is the escape probability in (2.4). These neutrons are then incident
on the second slab, and the number that would emerge unscathed from that passage
would be (N o P)P = N o P
2 . These neutrons would then strike the third slab, and so
on. The number that survive passage through the entire block to escape from the
right side would be N o P
x/s , or
N esc = N o (1 − s n σ )
x/s
.
(2.5)
Define z = –snσ. The number of neutrons that escape can then be written as
Fig. 2.2 Neutrons
penetrating a thick target
x
No
incident
neutrons
Ne
escaping
neutrons
s
2 Critical Mass, Efficiency, and Yield
The concept of a reaction rate will be used extensively in Sects. 3.3 and 4.5 for
estimating the rates of production of plutonium and polonium within a reactor. For
the present purpose, our interest is in the probability that neutron will pass through
the layer without striking a nucleus. To approach this, it is easier to begin with the
probability that a neutron will in fact precipitate a reaction:
P react =
reactions
per second
incident neutron f lux
per second
=
(R o Σ)(s n σ )
(R o Σ)
= s n σ.
(2.3)
The probability that a neutron will pass through the slab to escape out the back
side will then be:
P escape = 1 − P react = 1 − s n σ.
(2.4)
Now consider a block of material of macroscopic thickness x. As shown in Fig. 2.2,
we can imagine this comprising a large number of thin slabs each of thickness s
placed back-to-back.
The number of slabs is x/s. If N o neutrons are incident on the left side of the
block, the number that would survive to emerge from the first thin slab would be
N o P, where P is the escape probability in (2.4). These neutrons are then incident
on the second slab, and the number that would emerge unscathed from that passage
would be (N o P)P = N o P
2 . These neutrons would then strike the third slab, and so
on. The number that survive passage through the entire block to escape from the
right side would be N o P
x/s , or
N esc = N o (1 − s n σ )
x/s
.
(2.5)
Define z = –snσ. The number of neutrons that escape can then be written as
Fig. 2.2 Neutrons
penetrating a thick target
x
No
incident
neutrons
Ne
escaping
neutrons
s