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ConorDMK

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- Homework Statement
- The fission of ##^{235}_{92} U## by thermal neutrons is asymmetric with typical fission fragments being ##^{92}_{37} Rb## and ##^{140}_{55} Cs## . The energy release per fission is 149.7 MeV. An average of one neutron per ##^{235}_{92} U## fission escapes the core of a 60 MW fission reactor. A beam of neutrons is created by a hole of area ##100 mm^{2}## at a distance of ##5 m## from the reactor core. Show that ##8 \times 10^{11}## neutrons per second exit through this hole.

A proportional counter filled with gas ##BF_{3}## at STP is used to count the neutrons coming out of the hole. The neutrons create ##\alpha## particles via the reaction ##^{10}B(n, \alpha)^{7}Li## . The size of the counter is much larger than the hole, and its thickness is ##10 mm##. The approximate value of the neutron absorption cross-section for these energies is 1 barn (##10^{28} m^{2}## ). Natural boron contains 20% of the isotope ##^{10}B##. Assuming that the efficiency of the counter for detecting ##\alpha## particles is 100%, find the detected rate. (One mole of gas at STP occupies ##22.4 \times 10^{3} m^{3}##.)

- Relevant Equations
- First reaction ##^{235}_{92} U + n \rightarrow ^{92}_{37} Rb + ^{140}_{55} Cs + 4n + 149.7 MeV ## with one neutron escaping.

Area of sphere = ##4 \pi R^{2}##

Number of particles ##N=n N_{A}##

Second reaction ## ^{10}_{5}B + n \rightarrow ^{7}_{3}Li + \alpha ##

Finding the neutrons per second.

Uncluttering the question: ##P = 6 \times 10^{7} J/s, E_{1} = 149.7 MeV/event, A = 10^{-4} m^{2}, R = 5 m ##.

Number of events per second = ##\frac{P}{E_{1}}## = escaping neutrons per second

Area of ##5m## sphere around reactor = ## 4 \pi R^{2} ##

Fraction of sphere's area that neutrons escape through = ## \frac {A}{4 \pi R^{2}} ##

Number of neutrons passing through hole per second = ## \frac{P}{E_{1}} \frac {A}{4 \pi R^{2}}=7.96... \times 10^{11} ## neutrons per second

Finding the detected rate. This is the part which I'm not too sure on.

Uncluttering the question: ## T = 10^{-2} m, \sigma _{n} = 10^{-28} m^{2}, V_{n} = 22.4 \times 10^{-3} m^{3}/mole ##, 20% of Boron is ## ^{10}_{5}B##.

The neutron beam sweeps an area = ## A ##

The volume of the gas that interacts with the neutron beam = ## AT ##

Number of moles of gas ## n_{gas} = \frac {AT}{V_{n}} ##

Number of gas particles = ## n_{gas} N_{A} = N_{gas} ## = number of Boron nuclei

Fraction of the neutrons that interact with the gas = ## \frac {(8 \times 10^{11}) \sigma _{n} N_{gas}} {A} ##

Fraction of the neutrons that interact with ## ^{10}_{5} B## = 20% of the fraction of the neutrons that interact with the gas = ## \frac {1}{5} \frac {(8 \times 10^{11}) \sigma _{n} N_{gas}} {A} ##

Count rate = Fraction of the neutrons per second that interact with ## ^{10}_{5} B##

So putting everything together:

Count rate = ## \frac {1}{5} \frac {(8 \times 10^{11}) \sigma _{n} T N_{A}} {V_{n}} = 4.3 \times 10^{6}##

Uncluttering the question: ##P = 6 \times 10^{7} J/s, E_{1} = 149.7 MeV/event, A = 10^{-4} m^{2}, R = 5 m ##.

Number of events per second = ##\frac{P}{E_{1}}## = escaping neutrons per second

Area of ##5m## sphere around reactor = ## 4 \pi R^{2} ##

Fraction of sphere's area that neutrons escape through = ## \frac {A}{4 \pi R^{2}} ##

Number of neutrons passing through hole per second = ## \frac{P}{E_{1}} \frac {A}{4 \pi R^{2}}=7.96... \times 10^{11} ## neutrons per second

Finding the detected rate. This is the part which I'm not too sure on.

Uncluttering the question: ## T = 10^{-2} m, \sigma _{n} = 10^{-28} m^{2}, V_{n} = 22.4 \times 10^{-3} m^{3}/mole ##, 20% of Boron is ## ^{10}_{5}B##.

The neutron beam sweeps an area = ## A ##

The volume of the gas that interacts with the neutron beam = ## AT ##

Number of moles of gas ## n_{gas} = \frac {AT}{V_{n}} ##

Number of gas particles = ## n_{gas} N_{A} = N_{gas} ## = number of Boron nuclei

Fraction of the neutrons that interact with the gas = ## \frac {(8 \times 10^{11}) \sigma _{n} N_{gas}} {A} ##

Fraction of the neutrons that interact with ## ^{10}_{5} B## = 20% of the fraction of the neutrons that interact with the gas = ## \frac {1}{5} \frac {(8 \times 10^{11}) \sigma _{n} N_{gas}} {A} ##

Count rate = Fraction of the neutrons per second that interact with ## ^{10}_{5} B##

So putting everything together:

Count rate = ## \frac {1}{5} \frac {(8 \times 10^{11}) \sigma _{n} T N_{A}} {V_{n}} = 4.3 \times 10^{6}##