# Calculating Count Rate for Neutron Beam Detection

• ConorDMK
In summary, the conversation discusses finding the number of neutrons per second and the detected rate in a reactor. It involves calculations using various parameters such as power, energy, area, and volume. The key steps include determining the number of events per second, the fraction of neutrons escaping through a hole in a sphere, and the interaction of neutrons with boron-10. Careful checking of units and scaling with inputs is recommended to avoid errors in calculations.
ConorDMK
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}##

Looks fine.

Checking the units and scaling with the inputs catches most errors in these types of problems (e.g. would the calculated count rate double if the thickness doubles). It won't find if you miss e.g. the 20% boron-10 fraction, but it finds most other things.

mfb said:
Looks fine.

Checking the units and scaling with the inputs catches most errors in these types of problems (e.g. would the calculated count rate double if the thickness doubles). It won't find if you miss e.g. the 20% boron-10 fraction, but it finds most other things.

Thank you very much! I'll keep the advice noted as well.

These sorts of questions always trip me up, I guess I just need to be more confident with my workings.

## What is a neutron beam detection rate?

A neutron beam detection rate is a measure of the number of neutrons that are detected per unit time by a neutron detector. It is used to characterize the intensity of a neutron beam and can provide information about the properties of the neutrons, such as their energy and direction.

## How is the neutron beam detection rate measured?

The neutron beam detection rate is typically measured using a neutron detector, such as a scintillation detector or a gas-filled detector. These detectors are designed to detect and count individual neutrons, and the rate is determined by the number of detections over a specific period of time.

## What factors can affect the neutron beam detection rate?

The neutron beam detection rate can be affected by a variety of factors, including the energy and direction of the neutron beam, the type of neutron detector used, and the distance between the detector and the neutron source. Other factors, such as shielding and background radiation, can also impact the detection rate.

## Why is the neutron beam detection rate important?

The neutron beam detection rate is important because it provides information about the properties of the neutrons being detected. This can be useful in a variety of scientific fields, such as nuclear physics, materials science, and medical research. It can also be used to monitor and control the intensity of a neutron beam in experiments and applications.

## How can the neutron beam detection rate be improved?

There are several ways to improve the neutron beam detection rate, such as using more sensitive detectors, optimizing the neutron source and beam delivery system, and reducing background noise. Additionally, techniques such as neutron moderation and collimation can be used to increase the number of neutrons that reach the detector.

Replies
1
Views
1K
Replies
1
Views
1K
• Nuclear Engineering
Replies
5
Views
1K
Replies
3
Views
2K
Replies
11
Views
1K
Replies
1
Views
2K
• Classical Physics
Replies
20
Views
997
• Introductory Physics Homework Help
Replies
21
Views
401