# Calculate the temperature of neutrons emerging from a reactor

1. Jan 2, 2018

### Physgeek64

1. The problem statement, all variables and given/known data
A collimated beam of thermal neutrons emerges from a nuclear reactor and passes through a speed selector into a detector. The number of neutrons detected in a second with speeds in the range 4000 to 4010 m s−1 is twice as large as the number per second detected with speeds in the range 2000 to 2010ms−1. What is the temperature of the moderator in the nuclear reactor?

2. Relevant equations

3. The attempt at a solution
So the speed distribution is proportional to $v^3e^{-\frac{v^2}{v_{th}^2}}$ so my instinct was to write $v^3e^{-\frac{v^2}{v_{th}^2}}|_{4000}^{4010}=2v^3e^{-\frac{v^2}{v_{th}^2}}|_{2000}^{2010}$ but then i don't know how to solve this.

Many thanks

Last edited: Jan 2, 2018
2. Jan 2, 2018

### TSny

The speed distribution should have a factor of $v^2$ instead of $v^3$. However, the rate at which neutrons strike the detector would have the factor of $v^3$.

What does $v_{th}$ represent? If it has dimensions of speed, then note that the argument of your exponential function is not dimensionless.

Can you explain the notation $v^3e^{-\frac{v^2}{v_{th}}}|_{4000}^{4010}$ as regards the interpretation of $|_{4000}^{4010}$ ?

3. Jan 2, 2018

### Physgeek64

Sorry that is meant to be $v_{th}^2$ I have missed of the $^2$. That notation is meant to be putting in limits from 4000 to 4010

4. Jan 2, 2018

### TSny

Limits of an integration? Did you perform an integration?

5. Jan 2, 2018

### Physgeek64

no, but i probably should have!!

6. Jan 2, 2018

### TSny

Yes. However, an interval of 10 m/s is quite small compared to 2000 m/s or 4000 m/s. So, you can get a decent answer by considering the integrand as constant over the interval of 10 m/s.

7. Jan 2, 2018

### Physgeek64

So can we approximate this as $(4000)^3e^{-\frac{4000}{v_{th}^2}}\int_{4000}^{4010}{dv}=2(2000)^3e^{-\frac{2000}{v_{th}^2}}\int_{2000}^{2010}{dv}$ ?

$T=\frac{m}{K_B}\Big(\frac{4000^2-2000^2}{\ln{\frac{4000^3}{2(2000^3)}}}\Big)^2$

8. Jan 2, 2018

### TSny

Looks good. But did you drop a factor of 2 in the relation between $v_{th}^2$ and $T$?

9. Jan 2, 2018

### Physgeek64

I did- was a typo though. Thank you for your help!! Very much appreciated