Calculating Count Rate for Neutron Beam Detection

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

 

[mfb]

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.

 

[ConorDMK]

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.