# Tag Info

4

The formula you cite gives wavelength related to the energy of an individual photon. The perceived brightness of a light source also depends on the number of photons which are present. The lower energy levels were brighter because they were being excited more frequently, because it's easier to drive an atom to less excited states. This means that more yellow ...

2

First of all I would recommend you to plot your spectra as a function of energy rather than wavelength. This is because wavelength is a reciprocal scale which makes it difficult to compare linewidths and, moreover, if you do spectroscopy you are mostly interested in energy differences anyway. Now your linewidths. Typically the dominant broadening effect ...

1

What you say is true. If you are dealing with pure counting statistics and a constant source then the uncertainty in the countrate determined from the average of 10 measurements ought to be the same as the uncertainty derived from a measurement that is 10 times as long. So what could be gained? Firstly you get to test whether the statistics are Poissonian ...

1

I faced exactly this problem in my PhD, though it was silver on glass not aluminium. All the really precise methods are time consuming, so I ended up using them to make a calibration curve for the optical absorption. Then I measured the thickness using a standard lab spectrometer. This wasn't stunningly accurate, but allowed me to process the dozen or so ...

0

I have a project in which thin metal films up to 5000 Angstroms thickness are sputtered onto various substrates, and the tool that is regularly used to determine thickness is just a mechanical contact or stylus profilometer. That's the preferred method because it's fast and fairly accurate. We also have an optical profilometer which uses optical interference ...

1

if your sample is partially coated and partially bare with the subtract, you use the old interferometry method very easily and precisely. See this reference. Otherwise, you may want to check out Ellipsometry and Profilometry for precise measurement.

3

In its simplest form binning of data from something like a CCD camera (as may be used for spectroscopy) means adding together mutiple pixels. On a 2d array detector (like an everyday camera) you could for example use 2x2 binning to add together 4 pixels into one output value. Each pixel contributes exactly once to the output data, i.e. appears in exactly ...

18

Suppose you are analysing the weights of people in the UK to see what the distribution of weights looks like. Suppose also you can measure the weight to arbitrary precision, so that no two people's weights will be exactly the same. When you're finished you plot your data on a histogram, but the trouble is that because everyone has a different weight you get ...

4

"Bin" as a verb means to divide/discretize data into a group of (frequently equal-width) ranges, to facilitate various sorts of analysis and visualization. See: https://en.wikipedia.org/wiki/Data_binning In particular, binning is the basis of histogram plots among other things. (https://en.wikipedia.org/wiki/Histogram) As a noun, a "bin" refers to one of ...

3

You'll need to amplify the light coming in. There are two classes of devices which do this: cameras and telescopes. You can understand them by analogy to your eye. A telescope works by having a very wide "pupil", letting more light in, but then using lenses to focus this light onto a smaller "retina" (which is in fact your pupil). A camera is slightly ...

2

I've built several CD based spectrometers myself. The best resolution is obtained with a CD-RW because the data lines are much closer together than on a ordinary CD. But evaluating the light from stars is seriously difficult because so very little light reaches the spectrometer because stars are so far away. At the very least you'll need a good quality ...

1

I believe that a more important effect of using semicondctor detectors (Ge(Li), HPG, Si(Li) and others) in a neutron-ridden environment is the dislocation damage of the crystalline structure due to neutron irradiation. Dislocations in the crystal increase the charge carrier collection time and decrease the efficiency of charge collection. This results in ...

0

Unless it's been isotopically enriched on purpose, it's safe to assume natural abunances. If you need better information, the best way to measure the ratios is with the gamma spectrum from thermal neutron activation, which it sounds like you're going to get for free. Depending on your lithium doping you may care about $\rm ^7Li(n,2n){}^6Li$.

1

Broadband Emission from 400 to 2200 nm"...which means it will transmit all mixed wavelength of infrared light between 400 to 2200? Yes, this is correct. But the light intensity will likely not be equal at all wavelengths, but will have some dependence, which may be in your spec sheet. Note that 400 - 700 nm is the visible range. I don't know that ...

3

The hydrogen atom has an infinite number or quantum mechanically allowed energy levels, as explained on this web page. Using that same link, scroll up the page a bit to better understand how transitions between these energy levels give rise to absorption or emission of photons of very specific frequencies. Then scroll further down to see how the hydrogen ...

5

The 511-keV photons in the $^{22}Na$ spectrum are annihilation photons. They definitly have Compton interactions as seen in the diagram. The 1250-keV peak is a gamma in the daughter of the sodium positron decay. It also has a Compton edge and a backscatter region.

3

Photons are photons. If photons from one source experiences a certain kind of physics, then photons from other sources do too. So, short answer: yes. And they can produce Compton edges as well.

Top 50 recent answers are included