# Nuclear reactions, cross section, reaction rate

1. Dec 13, 2016

### Incand

1. The problem statement, all variables and given/known data
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.

2. Relevant 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).

3. 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.

2. Dec 13, 2016

### eys_physics

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. Dec 14, 2016

### Incand

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. Dec 14, 2016

### eys_physics

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. Dec 14, 2016

### Incand

Thanks, now I believe I understand the formula!