Collision of thermal neutrons and Cobalt

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  • #1
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Homework Statement


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##.


Homework Equations





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? :(
 

Answers and Replies

  • #2
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You're solution seems correct. I'm not sure why you say it's wrong.
 
  • #3
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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
phyzguy
Science Advisor
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Your solution is incorrect. If you determine the constants C and A correctly, you will find:
[tex] N = \tau j N_{59} \sigma (1-e^{\frac{t}{\tau}})[/tex]
Then, as you said, t<<τ, so:
[tex] N = \tau j N_{59} \sigma {\frac{t}{\tau}} = t j N_{59} \sigma [/tex]
This makes sense, since at early times the activity increases linearly with time. So the activity N/τ is:
[tex] \frac{N}{\tau} = \frac{t}{\tau} j N_{59} \sigma [/tex]

It is this factor of t/τ that you omitted. When you plug in the numbers, you do in fact get 6E11 Becquerels.
 
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  • #5
Astronuc
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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.
 
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  • #6
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Ah, of course!

Thanks to all of you!
 

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