Bragg curve -> observing dependence on velocity

[pepediaz]

Homework Statement

None, it is just a paper I am trying to understand.

Relevant Equations

https://www.fuw.edu.pl/IIPRACOWNIA/home/Opisy-cwiczen/J14_2015.09.16_publikacja.pdf

We can read: "The velocity dependence of the stopping power, increasing with decreasing velocity, is obvious from Fig.4".

I know why the stopping power depends on velocity as Bethe equation states, but I do not know how I can observe that dependence on a Bragg curve.

 

[Steve4Physics]

Homework Statement:: None, it is just a paper I am trying to understand.
Relevant Equations:: https://www.fuw.edu.pl/IIPRACOWNIA/home/Opisy-cwiczen/J14_2015.09.16_publikacja.pdf

We can read: "The velocity dependence of the stopping power, increasing with decreasing velocity, is obvious from Fig.4".

I know why the stopping power depends on velocity as Bethe equation states, but I do not know how I can observe that dependence on a Bragg curve.

Possibly a simple mistake. I think 'Fig. 4' should be 'Fig. 3' - then it makes sense.

 

[pepediaz]

ah okay, so it is because at equal displacements (the markers on abscissa), the plotted function starts kinda lineally, but then decays steeply. As a result, energy decreases steeply -> the squared velocity decreases steeply -> the stopping power increases, isn't it?

Thanks!

 

[TSny]

Maybe you can think about it this way. Suppose the stopping power did not depend on the speed of the particle. Then doubling the absorber thickness $\Delta x$ would double the energy loss $\Delta E$. So, the ratio $\Delta E/\Delta x$ would not change if you double the thickness. So, what would Fig. 4 look like if the stopping power were independent of the speed of the particle?

 

[pepediaz]

We would have the Fig.3 as a decreasing line function and Fig.4 as a constant value.

But how do we know from Fig 3 or 4 that it is velocity and not other parameter what creates dependence for dE/dx ?

 

[TSny]

We would have the Fig.3 as a decreasing line function and Fig.4 as a constant value.

Yes.

But how do we know from Fig 3 or 4 that it is velocity and not other parameter what creates dependence for dE/dx ?

For a given medium, I can't think of any parameter other than the speed (or energy) of the alpha particle that would be relevant. As the particle slows down, it might capture one or more electrons and so the charge of the particle could change. But, it seems to me that the probability of electron capture would be determined by the speed of the particle. However, I have not studied this topic in any detail. So, I could be wrong.

 

[Steve4Physics]

Let me have another go...

First, note that stopping power is really $-\frac {dE}{dx}$ but the authors have dodged the minus sign by referring to stopping power as the ‘rate of energy loss’. I’ll stick with their usage.

The (empirical) relationship between stopping power and velocity might best be deduced by considering Fig. 3 and Fig. 4. together. Remember Fig. 4 is directly derived from Fig. 3, being essentially a graph of Fig. 3’s gradient vs. absorber thickness. Note $\frac {ΔE}{Δx}$ has been used in Fig. 4 rather than $\frac {dE}{dx}$.

The kink in Fig. 3 and the peak shown in Fig. 4 are anomalous, as explained at the end of the paper’s 2nd paragraph. So consider absorber thickness (x) up to x=36cm only; i.e. exclude the kink and peak

Consider thickness x=3cm.
From Fig. 3, E ≈ 5.2 MeV so velocity ≈ $\sqrt {5.2} = 2.3$ (funny units).
From Fig. 4, stopping power = $\frac {ΔE}{Δx}$ ≈ 1.0MeV/cm.

Consider thickness x=36cm.
From Fig. 3, E ≈ 1.3MeV so velocity ≈ $\sqrt {1.3} ≈ 1.1$ (funny units).
From Fig. 4, stopping power = $\frac {ΔE}{Δx}$ = 2.0MeV/cm.

Observe that up to x=36cm, the Fig. 3 graph is monotonically decreasing and the Fig.4 graph is monotonically increasing. So we see the general rule: stopping power increases as velocity decreases.

 

[pepediaz]

Ah okay, I think that speed (or kinetic energy) is to look when studying stopping power. We see that matter absorbs and scatters alpha particles that traverse it.

I can also see that Fig.4 is the gradient of Fig.3 , and recognise the shape of the functions.

The key on this is that stopping power increases as velocity (kinetic energy) decreases, and we can check that comparing Figs. 3 and 4.

Thank you!