# Collision of thermal neutrons and Cobalt

1. May 26, 2014

### skrat

1. The problem statement, all variables and given/known data
The cross section of $^{59}Co$ for capturing the thermal neutrons is $2000 fm^2$. A $10g$, thin paper of $^{59}Co$ is radiated for $100 hours$ in reactor with neutron flux $2\cdot 10^{18} /m^2s$. Density of $^{59}Co$ is $8.9g/cm^3$. Half life time of $^{60}Co$ is $5.2 years$.

2. Relevant equations

3. The attempt at a solution

So what happens is $n+$^{59}Co$->^{60}Co$ which than decays with $t_{1/2}=5.2 years$.

Well, the number of successful collisions is $dN=\frac{NN_{59}\sigma }{S}$ and dividing that by $dt$ gives me $\frac{dN}{dt}=jN_{59}\sigma$.

Of course, some of the $^{60}Co$ will immediately start to decay, therefore:

$\frac{dN}{dt}=jN_{59}\sigma -\frac{N}{\tau }$

Solving this differential equations using $N=C+Ae^{-t/\tau }$

Leaves me with $N=\tau (jN_{59}\sigma -1)(1-e^{-t\tau })$ where $t/\tau <<1$ so

$\frac{N}{t}=jN_{59}\sigma -1= 2.4\cdot 10^{16}Bq$ instead of $6\cdot 10^{11}Bq$.

Note that $N_{59}=\frac{mN_a}{M}$.

Why is this wrong? :(

2. May 26, 2014

### dauto

You're solution seems correct. I'm not sure why you say it's wrong.

3. May 27, 2014

### skrat

Basically the only reason why I think this method is wrong is because the "official" result is $6\cdot 10^{11}Bq$

I also remembered that I could take into account here $\frac{dN}{dt}=jN_{59}\sigma -\frac{N}{\tau }$ that already excited $^{60}Co$ can't get even more exited. Therefore the more exact solution would follow from inserting that $N_{59}=N_{59}^0-N$

Of course "official" results can be wrong too so if you agree with my solution and if nobody else will complain than I think it is safe to assume that what I did is ok.

4. May 27, 2014

### phyzguy

Your solution is incorrect. If you determine the constants C and A correctly, you will find:
$$N = \tau j N_{59} \sigma (1-e^{\frac{t}{\tau}})$$
Then, as you said, t<<τ, so:
$$N = \tau j N_{59} \sigma {\frac{t}{\tau}} = t j N_{59} \sigma$$
This makes sense, since at early times the activity increases linearly with time. So the activity N/τ is:
$$\frac{N}{\tau} = \frac{t}{\tau} j N_{59} \sigma$$

It is this factor of t/τ that you omitted. When you plug in the numbers, you do in fact get 6E11 Becquerels.

5. May 27, 2014

### Staff: Mentor

Where one wrote N/t, it should be N/τ (where τ is tau = 1/λ), lambda = decay constant, and tau is the mean lifetime.

6. May 27, 2014

### skrat

Ah, of course!

Thanks to all of you!