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🌀Unit 3

4 min read•august 6, 2020

Dalia Savy

Kanya Shah

Dylan Black

In chemistry, many times we want to calculate how light reacts to a certain colored solution with solute in it. Well, spectrophotometry can do just that!

To get a better understanding of what spectrophotometry, let's take a look at a spectrophotometer itself:

Image Courtesy of ResearchGate

There are 3 main parts to this machine: the monochromator, the sample, and the detector.

The Monochromator

The monochromator has three main parts: the entrance slit, the dispersion device, and the exit slit. The entrance and exit slips are simple, they're where a beam of light enters and exits the monochromator. However, the dispersion device is the important part to the monochromator. Essentially, the dispersion device takes a beam of white light and splits it into the full spectrum of colors🌈. The exit slit is placed based on what wavelength of light the experimenter needs for the solution at hand, typically the opposite color (red solution --> green light, etc.)

The Sample

The sample is just that - the sample! It can be any colored solution you wish, from a solution of Copper (II) Sulfate, to red Gatorade, which we'll take a look at in a minute. Light from the monochromator is run through the sample and light is absorbed in the sample (That's why we want to use the opposite color, that's the most absorbed light!).

Image Courtesy of Research Gate

On the right is the Incident Light (I0) and on the left is the transmitted light (I).

The Detector

Finally, after all of this has occurred, the remaining light is detected and the amount absorbed is spit out on a screen.

Let's say a chemist is interested in calculating the molar concentration of red food dye🔴 in Gatorade. Well, this is a perfect time to talk about spectrophotometry in action! A chemist would fill a small test tube🧪 (though spectrophotometer test tubes are rectangular) with red gatorade and run it through a spectrophotometer with green light at around 560–520 nm. Then, he would get results as to the absorption, but how would he find the concentration? Well, luckily scientists have a formula for that:

The formula for the Beer-Lambert Law is actually quite simple. The **Beer-Lambert law **(also known as Beer's law) is a linear relationship between the absorption of light and the concentration of the absorbing species.

Once all written out, it may seem a little confusing, but the two variables a and b have are both quite simple to understand and are most often constant.

Back to our chemist friend. Let's say he looked up the molar absorptivity of his food dye, Red-40 and found it to be 2.13 * 10^4 L/(mol)(cm). He runs his solution through the spectrophotometer and finds an absorption of 0.500. Assuming a path length of 1cm, we can use Beer's law to calculate the molarity!

A = abc

0.5 = (2.13 * 10^4)(1)(c)

c = 0.5/2.13 * 10^4 = 2.3 * 10^-5 mol/L

And boom. We've found the concentration of red-40 in gatorade with an absorption of 0.5.

To enact the lab you would do for this key topic, you can use PhET simulations:

In one of our previous guides, we went over part a of #5 on the 2019 AP Chemistry Exam. Let's go over part b since we now know the formulas🤓.

To look over the in depth explanation of part a, read 1.6!

Part b is asking us for the wavelength of *something*. First things first, there are two equations that involve wavelength of light:

- E=hv, where h = 6.626 x 10^-34
- c=λv, where c = 2.998 x 10^8

That something is the energy needed to remove an electron from the valence shell, or the ionization energy/binding energy⚡. So let's solve for the wavelength of this energy!

The only piece of information they give us is the actual energy itself, but there are 6 different types of energy listed😕.

To recall, where the binding energy is the highest is where the nucleus is located. Thinking about valence electrons, we should be using the lowest binding energy. Now we know that E = 0.980 x 10^-18 and we can solve for frequency!

E = hv --> 0.980 x 10^-18 = (6.626 x 10^-34) (v)

v = 1.48 x 10^15

Now that we have frequency, we can solve for wavelength using c=λv.

2.998 x 10^8 = (λ) (1.48 x 10^15)

λ = 2.03 x 10^-7 m

This is an example of using both equations to solve an FRQ!🧠

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