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- Thread starter sportfreak801
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Simon Bridge

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You identify elements by the patterns of peaks in the spectra - the peaks occur because the atomic electrons and nucleons scattered from live in energy shells which depend on the atomic number of the atom. Basically, we have put known samples in the apparatus and taken note of where peaks occur so we can ID them from an unknown sample.

The peaks will be shifted down in energy though, the amount of the shift depends on the thickness of the material.

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Thank you for your response!

I do, however, have some questions that I would like clarified.

Thanks in advance,

sport

I do, however, have some questions that I would like clarified.

Does this mean that multiple peaks in the spectra correspond to the same element? For example, if in my spectra I have three distinct peaks that occur at different energy levels and have different counts, those three peaks each represent a different energy shell of the same element for which electrons were scattered?You identify elements by the patterns of peaks in the spectra.

When noting where peaks occur, what exactly do you look for? (I have attached the example spectra from class for your benefit.)we have put known samples in the apparatus and taken note of where peaks occur so we can ID them from an unknown sample.

Thanks in advance,

sport

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Simon Bridge

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Notice how the thickness shifts the energy peak - smearing it out.

This should also allow you to work out what you are looking for to locate the peak too.

IRL we would do interpolation to locate the top of the peak but anything near-by will probably be fine.

The width of the "thin" peaks (measured at half the height) is usually solely a characteristic of the detector. It is used to estimate the systematic uncertainty in the measurement.

Looks like your data has one double-peak ... maybe the same element responsible, you'll know when you do the math to find Z. The fat jagged peaks at low energy are often artifacts. Here's another example:

http://www.eaglabs.com/training/tutorials/rbs_theory_tutorial/silicide.php

notice - silicone is thicker than the tantalum, and the heavy tantalum is to the far-right close to the maximum energy. You'd expect the far right peak in your example to be Bismuth - giving you two Si peaks - perhaps your two different layers?

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[itex]\left ( \frac{d\sigma}{d\Omega} \right )=\left [ \frac{Z_{1}Z_{2}e^{2}}{4E} \right ]^{2}\frac{4\left \{ \left [ 1-\left ( \left ( M_{1}/M_{2} \right )sin^{2}\theta \right ) \right ]^{1/2} + cos\theta \right \}^{2}}{sin^{4}\theta \left [ 1-\left ( \left ( M_{1}/M_{2} \right )sin^{2}\theta \right ) \right ]^{1/2}}[/itex]

Where [itex] \frac{d\sigma}{d\Omega}[/itex] is the Rutherford Cross Section

[itex]Z_{1}[/itex] is the atomic number of the incident particles

[itex]Z_{2}[/itex] is the atomic number of the target particles

[itex]M_{1}[/itex] is the mass of the incident particles

[itex]M_{2}[/itex] is the mass of the target particles

[itex]E[/itex] is the incident energy

And to calculate [itex]M_{2}[/itex] we use the equation of the kinematical factor

[itex]K = \left ( \frac{M_{1}cos\theta + \sqrt{M_{2}^{2}-M_{1}^{2}\left ( sin^{2}\theta \right )} }{M_{1}+M_{2}} \right )^{2}[/itex]

Where [itex]E_{1} = KE_{0} [/itex] and [itex]E_{1}[/itex] can be determined from the peaks of the spectra. In calculating would I just take the difference between the max and min energy for each peak to be equal to [itex]E_{1}[/itex]?

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[itex] K = \frac{E_{1}}{E_{0}}=\left ( \frac{M_{1}cos\theta + \sqrt{M_{2}^{2}-M_{1}^{2}\left ( sin^{2}\theta \right )} }{M_{1}+M_{2}} \right )^{2} [/itex]

Looking at the first peak of my example spectra, I took

[itex]M_{1}= 4[/itex] (Helium)

[itex]E_{0}= 1 MeV[/itex]

[itex]E_{1}= 125 KeV[/itex]

[itex]\theta= 165 degrees[/itex]

So the equation becomes:

[itex] \frac{.125 MeV}{1 MeV}=\left ( \frac{4*cos(165) + \sqrt{M_{2}^{2}-4^{2}\left ( sin^{2}(165)\right )} }{4+M_{2}} \right )^{2} [/itex]

And I find that [itex]M_{2}=8.545 Amu[/itex]. So would this correspond with Beryllium?

Now knowing the atomic number of Beryllium, I can calculate the areal density [itex]Nt_{Be}[/itex] using the relation

[itex]\frac{A_{Be}}{A_{Bi}}=\left (\frac{Z_{Be}}{Z_{Bi}} \right )^{2}\frac{Q_{Be}*Nt_{Be}}{Q_{Bi}*Nt_{Bi}}[/itex]

Where [itex]A_{Be} [/itex],[itex]Q_{Be} [/itex],[itex]Nt_{Be} [/itex],and [itex]Z_{Be}[/itex] come from a reference experiment with the quantities:

[itex]A_{Be} = 24000 [/itex](Number of particles detected)

[itex]Nt_{Be} = 5.65*10^{15} Atoms/cm^{2} [/itex] (Given quantity)

[itex]Z_{Be} = 83 [/itex]

We are provided a given [itex]dose = 10\mu c[/itex] and I0 ~ 70 nA:

[itex]Q_{Be} = \frac{Dose}{Charge of Helium} = \frac{10^{-5} C}{2e} [/itex] (Number of incident particles) Is this how one calculates Q?

[itex]Nt_{Be}=\frac{A_{Be}}{A_{Bi}}\left (\frac{Z_{Bi}}{Z_{Be}} \right )^{2}\frac{Q_{Bi}*Nt_{Bi}}{Q_{Be}}=\frac{22000}{24000}\left (\frac{83}{4} \right )^{2}{5.65*10^{15}}=2.23*10^{18} Atoms/cm^{2}[/itex]

This does not seem to be right. Any help that you could provide me would be greatly appreciated!

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Simon Bridge

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What is it about your result that suggests it may not be correct?

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It seems that the Nt for Beryllium is quite large compared to the Nt for Bismuth. There is a factor of [itex]10^{3}[/itex] difference, are such densities plausible?

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Simon Bridge

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I don't know how the sample was prepared so cannot comment on that ... however, just looking at it, how have you used the Q's?

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Aside: LaTeX tip - the multiplication character [itex]\times[/itex] is \times - you don't need it between constants (if you need to be explicit, put a point like: [itex]{\bf u}\cdot {\bf v} = uv\cos(\theta)[/itex] thats: \cdot) but you do need it for scientific notation as in 1.03 \times 10^{-4} which is [itex]1.03 \times 10^{-4}[/itex] :)

---------------------------------

Aside: LaTeX tip - the multiplication character [itex]\times[/itex] is \times - you don't need it between constants (if you need to be explicit, put a point like: [itex]{\bf u}\cdot {\bf v} = uv\cos(\theta)[/itex] thats: \cdot) but you do need it for scientific notation as in 1.03 \times 10^{-4} which is [itex]1.03 \times 10^{-4}[/itex] :)

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Simon Bridge

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That may be a possibility; however, we were provided what I had presumed to be two different samples. The first being the reference sample made up of known layers: a Silicon layer of [itex]115\times10^{15} atoms/cm^{2}[/itex] then a Bismuth layer of [itex]5.65\times10^{15} atoms/cm^{2}[/itex] and then a Silicon substrate of [itex]10^{28} atoms/cm^{2}[/itex].

The second sample being the example sample where we were provided the energy spectra and have to find the element and its concentration for each layer.

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