Nuclear reactions, cross section, reaction rate

  • Thread starter Incand
  • Start date
  • #1
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Homework Statement


A beam of ##^{50}##Ti nuclei impinged on a ##0.5 mg/cm^2## thick ##^{208}##Pb target in an experiment lasting 176 hours. During that time a total of ##3260 \; ^{257}##Rf nuclei were detected. The beam intensity was throughout the experiment constant at ##6.6\mu A##. Each ##^{50}##Ti beam ion carried a charge of ##q=+15##. The efficiency for detecting ##^{257}##Rf was ##3\%##. Calculate the reaction cross section.

Homework Equations


Reaction rate
##R=N_0\sigma I##
where ##N_0## is the number of target particles ##\sigma## the cross section and ##I## the current of incoming particles (I assume).

The Attempt at a Solution


As I understand it ##3\%## of the true number of ##^{257}##Rf were detected so we don't have to worry about any decay.

Secondly I assume the cross section I should calculate is actually the cross section of the entire target i.e. ##\sigma_{tot} = N_0 \sigma## so the thickness doesn't matter. I don't see how I could calculate anything otherwise since I don't know how big the total sample is.


In that case the number of ##^{50}##Ti hitting the target in one second is
##N_{Ti} = \frac{6.6 \mu A}{15e} = \frac{6.6 \cdot 10^{-6}}{15 \cdot 1.6 \cdot 10^{-19}} = 2.75\cdot 10^{12}.##

And the number of ##^{257}##Rf created in one second is
##N_{Rf} = \frac{3260}{0.03\cdot 176\cdot 3600} = 0.17##
this is then also the reaction rate.

Then ##\sigma_{tot} = \frac{R}{I} = \frac{N_{Rf}}{N_{Ti}} = 6.24\cdot 10^{-14}##.

Now I have no idea if what I did was correct or what I should be doing. There's also a question about units, my cross sections seems to be unit less but shouldn't the unit of a cross section be barn i.e. an area unit? Or in the case of the macroscopic cross section ##m^{-1}##.

Perhaps what I should calculate is the macroscopic cross section but again I don't see how since I don't know the size of the sample.
 

Answers and Replies

  • #2
268
72

Homework Statement


A beam of ##^{50}##Ti nuclei impinged on a ##0.5 mg/cm^2## thick ##^{208}##Pb target in an experiment lasting 176 hours. During that time a total of ##3260 \; ^{257}##Rf nuclei were detected. The beam intensity was throughout the experiment constant at ##6.6\mu A##. Each ##^{50}##Ti beam ion carried a charge of ##q=+15##. The efficiency for detecting ##^{257}##Rf was ##3\%##. Calculate the reaction cross section.

Homework Equations


Reaction rate
##R=N_0\sigma I##
where ##N_0## is the number of target particles ##\sigma## the cross section and ##I## the current of incoming particles (I assume).

The Attempt at a Solution


As I understand it ##3\%## of the true number of ##^{257}##Rf were detected so we don't have to worry about any decay.

Secondly I assume the cross section I should calculate is actually the cross section of the entire target i.e. ##\sigma_{tot} = N_0 \sigma## so the thickness doesn't matter. I don't see how I could calculate anything otherwise since I don't know how big the total sample is.


In that case the number of ##^{50}##Ti hitting the target in one second is
##N_{Ti} = \frac{6.6 \mu A}{15e} = \frac{6.6 \cdot 10^{-6}}{15 \cdot 1.6 \cdot 10^{-19}} = 2.75\cdot 10^{12}.##

And the number of ##^{257}##Rf created in one second is
##N_{Rf} = \frac{3260}{0.03\cdot 176\cdot 3600} = 0.17##
this is then also the reaction rate.

Then ##\sigma_{tot} = \frac{R}{I} = \frac{N_{Rf}}{N_{Ti}} = 6.24\cdot 10^{-14}##.

Now I have no idea if what I did was correct or what I should be doing. There's also a question about units, my cross sections seems to be unit less but shouldn't the unit of a cross section be barn i.e. an area unit? Or in the case of the macroscopic cross section ##m^{-1}##.

Perhaps what I should calculate is the macroscopic cross section but again I don't see how since I don't know the size of the sample.


1) The reaction cross section is not the total but per atom. A cross section is an area and is given in units such as ##cm^2## or barns (##1b=10^{-24}\mathrm{cm}^2##)
2) You are given that you have ##0.5\mathrm{mg}/\mathrm{cm}^2## for the target. Consider now an area of ##1\mathrm{cm}^2##. How many target lead atoms do you have? This will be your ##N_0##.

I hope this helps.
 
  • #3
334
47
Okey so in the area ##A=1cm^{2}## we have
##N_0/A = \frac{0.5mg}{208u} = 1.45\cdot 10^{18}cm^{-2}##.

Now surely ##N_{Rf}## must be the reaction rate and ##I=N_{Ti}## (everything in one second)
##\sigma = \frac{R}{N_0 I} = \frac{N_{Rf}}{N_0/A N_{Ti}}=4.26\cdot 10^{-32}cm^{2} = 42.6 nb##

But what If I picked an area of ##5cm##? Wouldn't I get a different number then? I know that the width of the beam really shouldn't matter but I don't seem to have any term where the areas cancel?
 
  • #4
268
72
Okey so in the area ##A=1cm^{2}## we have
##N_0/A = \frac{0.5mg}{208u} = 1.45\cdot 10^{18}cm^{-2}##.

Now surely ##N_{Rf}## must be the reaction rate and ##I=N_{Ti}## (everything in one second)
##\sigma = \frac{R}{N_0 I} = \frac{N_{Rf}}{N_0/A N_{Ti}}=4.26\cdot 10^{-32}cm^{2} = 42.6 nb##

But what If I picked an area of ##5cm##? Wouldn't I get a different number then? I know that the width of the beam really shouldn't matter but I don't seem to have any term where the areas cancel?

No, it will not. Sorry, if I was a bit unclear in my last post. A useful tool in this respect is dimensional analysis. Let [##X##] be the unit of the quantity ##X##. Let's now apply this to the equation for the reaction rate:

$$[R]=[N_0] [\sigma] [I_{}]$$,
and with ##[R]=[s^{-1}]##, ##[\sigma]=[\mathrm{cm}^2]## and ##[I_{}]=[s^{-1}]##, we get that ##N_0## must have the unit ##\mathrm{cm}^{-2}##. It is therefore convenient to consider a unit area of 1 cm##^{2}##.
 
  • #5
334
47
Thanks, now I believe I understand the formula!
 

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