Learn the toughest concepts covered in your Analytical Chemistry class with step-by-step video tutorials and practice problems.

Fundamentals of Spectrophotometry

1

The Nature of Light

The video is coming soon

Was this helpful?

So here we're going to say that visible light represents a small portion of the continuum of radiant energy. This image that we have here below, known as the electro magnetic spectrum. So our electromagnetic spectrum consists of different forms of electromagnetic radiation. We have gamma rays, X rays UV. Which is ultraviolet light. The small slither here is our visible light spectrum. The portion that we can see without the aid of any instruments. We have infrared, which is IR microwave. And then our radio waves can be separated into FM and am as well as our long radio waves now beyond gamma rays. We also have cosmic rays but we won't go into those. Now realize here that the small portion that we can see is the visible light spectrum that can be expanded out. We have to remember Roy G biv. So Roy G bev, which stands for red, orange, yellow, green, blue indigo and violet. Many more modern electromagnetic spectrum images tend to just combine violet and indigo into just violet. We're gonna say the visible light spectrum ranges from about 380 nm To around 750 nm realize that when we talk about the electromagnetic spectrum, we talk about three variables, We talk about frequency, wavelength and energy, realize as we're proceeding this way that our energy which will just say is E and frequency are both increasing. So as we go from radio waves to gamma rays, the energy is increasing. They are inversely proportional to wavelength. So our wavelength would be dropping same thing with the visible light spectrum it follows the same pattern going this way the energy is increasing as well as the frequency, realize here that ultraviolet is next to violet from the visible light spectrum and that infrared is next to read from the visible light spectrum. Now with each one of these different types of electromagnetic radiation, we have atomic and molecular transitions uh that tend to pop up now with x rays, X rays have the highest energy. If we exempt cosmic rays because they have the highest energy, they're going to cause nuclear changes within a given atom. The energy is so great that it actually can destroy the nucleus of any atom that's exposed to it with X rays, it affects the core electrons within a given atom. So here we can say that this can result in bond breaking and possibly ionization of an element. Next we have is ultraviolet or UV. This has a tendency affecting the valence electrons of a given atom. We're gonna say that this causes electronic excitation. Yeah, visible light also can affect the valence electrons of a given atom. Next infrared causes molecular vibrations. So when Adam or a compound absorbs enough infrared energy, it'll eventually release that excess energy and as it does so it begins to vibrate and these different vibrations cannot help us identify the different types of functional groups within any given unknown compound. Next we have microwaves, microwaves cause molecular rotations within a given element or adam and this can also cause electron spin within a given atom. Finally, radio waves, radio waves can cause nuclear spins within a given element. Now, some of these are connected to the processes involved in determining an unknown compound with radio waves. It's connected to nuclear magnetic resonance, which is NMR. That helps us to determine the different types of carbon and hydrogen structures within a given element. Here, we can say that infrared can help us determine the functional groups of a given element, visible light is used for conjugated systems. That basically means you have a compound that has alternating double and single bonds. These conjugated systems can be identified through the use of what's called UV visible spectroscopy. Now just realize these are different forms of electromagnetic radiation. Each of them has a different effect on a given atom. The higher the energy, the more deeply they can penetrate a given atom and cause much greater effects in terms of that given element or atom. Now that we've seen this, click onto the next video and see how we relate even further the ideas of energy, frequency and wavelength to the electromagnetic spectrum

2

The Nature of Light

The video is coming soon

Was this helpful?

So in terms of the electromagnetic spectrum we said that we look at three variables energy, frequency and wavelength. Now in terms of frequency and wavelength, we're gonna say here that mu represents our frequency. It's the number of waves you have per second and it's using the units of seconds inverse or hurts. Then we're gonna say that lambda represents our wavelength. It's the distance from one crest of a wave to the other and expressed in units of meters here, if we take a look at a typical electromagnetic spectrum wave that's basically done on plane polarized light, we're gonna say here we have our Y axis, our Z axis and our X axis here. Here we're gonna say the top of this wave, the crest to the other wave would represent our wavelength lambda. Then we'd say here that the number of waves we get within a given second would represent my frequency or mu We'd say here that this here represents our electromagnetic wave and then here that's protruding out on the Y axis. This would represent our magnetic field vectors. Now the energy involved is directly proportional to frequency. So the more waves we get per second, the greater the energy would be. We can also say that the height of one wave would be our amplitude. Our amplitude does not directly influence the energy of our particular wave. Again, that is frequencies. Job. So these are the variables that you need to remember. This is how they're related to one another. Remember frequency and wavelength are inversely proportional. If there are several waves within a given second, the frequency will be high, but then the distance between those waves will be very small. So as your frequency is increasing, your wavelength has to be decreasing. Remember, they're connected to the energy of the different types of electromagnetic spectrum radiation that we saw up above.

3

Properties of Light

2m

Play a video:

Was this helpful?

So here we want to say that the speed of a typical wave is equal to the product of mu times lambda. We're going to say in a vacuum. All forms of electromagnetic radiation travel at 2.998 times 10 to the eight m per second. This is called our speed of light. Now our speed of light is equal to the variable C. So speed of light equals the product of mu which is your frequency times lambda, which is your wavelength. We're gonna say that the physicist max Plank and Albert Einstein theorized that light was made up of small packets of electromagnetic energy, which they called quanta. And the energy of a single photon could be calculated by this formula. So here the energy of a single photon equals H times mu H. Here represents plank's constant and it's equal to 6.626 times 10 to the negative 30 for jules times seconds. Now, in addition to this, we can say that energy of a single photon equals Planck's constant times frequency. It is also equal to Planck's constant times your wave number. Now, your wave number is equal to the inverse of your wavelength. Just like frequency is equal to the inverse of your wavelength. Now here, if we were to isolate our wavelength, we'll isolate our frequency here. We could say that divide both sides here by lambda. We can say that frequency is equal to speed of light over lambda through substitution. We can say here that the energy of a single photon equals planks constant times the speed of light divided by the wavelength in meters. So there are two versions of the equation that we can utilize. We can utilize this version here when they give us the frequency and they want us to determine the energy of a single photon. And we can use this version here when they give us the wavelength. And they want us to figure out the energy of a single photon here, this portion and showing the inverse relationship that frequency has with wavelength with the incorporation of your speed of light. So relates back to this portion right here. So just remember and keep in mind the formulas that relate wavelength and frequency to each other and how we can use that information to help us determine the energy of a single photon. Now that we've seen this, move on to the example one and see if you can calculate the wavelength from the given information. Once you do come back and see if your answer matches up with mine.

4

Properties of Light

2m

Play a video:

Was this helpful?

So here it says to calculate the wavelength in nanometers of the red light emitted by neon sign with a frequency of 4.16 times 10 to the eight megahertz. Alright, so they're asking us to determine wavelength which is represented by the variable lambda and frequency which is represented by mu. They're connected by the equation that speed of light equals frequency times wavelength. Here we're looking for wavelength so we need to isolate wavelength. So my wavelength equals speed of light divided by my frequency speed of light. Remembers 2.998 times 10 to the eight m per second, divided by your frequency in hertz, not megahertz. So we have to do some converting. So we have 4.16 times 10 to the eight megahertz. Remember here that one MHz is equal to 10 to the six hurts. That gives me 4.16 times 10 to the 14 Hz. Take that now and plug it in. So we have 4.16 times 10 to the 14 hertz hurts and seconds inverse are the same. So they're going to cancel out at this point. This will give me my wavelength in meters. So I get 7.21 times 10 to the negative seven m. But remember I didn't ask for the wavelength in meters, I asked for it in nanometers. So we need to do one last conversion to get to our final answer. We have 7.21 times 10 to the negative seven m For every one nanometer, it's 10 to the negative nine m. So meters cancel out. And what we get at the end is 721 nm as our final answer. So that number represents the wavelength of the red light that's being emitted by a neon sign. This answer is reasonable because it lies within the visible light spectrum of the electromagnetic spectrum 721 is around the range of red light. Now that we've seen. Example, one take a look at example to see if you can answer that final question here, we're asking for the energy of a mole of photons. So at this point, you should be able to at least calculate the energy of a single photon. If you don't know what to do after that. Don't worry, come back and see how I approach example, too.

5

Properties of Light

2m

Play a video:

Was this helpful?

So here it asks, what is the energy in jewels of a mole of photons associated with visible light of wavelength for 93 nanometers. Alright, what we're gonna do first is we're going to determine what the energy of a single photon is before we figure out the mole of photons. So we're gonna say the energy of a single photon equals planks constant times. Because the question is dealing with wavelength, we're gonna say times speed of light divided by wavelength. So plank's constant is 6.626 times 10 to the negative 30 for jules, times seconds. Speed of light is 2.998 times 10 to the eight m per second. And we need wavelength to be in units of m. So we have to convert the 493 nm into meters. So one nanometer is equal to 10 to the -9 m. So it gives me 4.93 times 10 to the -7 m, plug that in. So seconds cancel out meters, cancel out what we're going to get at this point is 4.29 times 10 to the negative 19 jewels per photon. Now we have to convert jewels per photon into jewels, per mole of photons. So bring down that value. We're gonna have to get rid of those photons. We're gonna say here one mole of photons just like one mole of anything is equal to avocados, number 6.22 times 10 to the 23 photons. So photons here cancel out. And I'll have as my final answer jewels per mole of photons, so that gives me 2.43 times 10 to the five jewels per mole. So realize we utilized the first equation in order to figure out the energy of a single photon, but then to convert it into molds a photon, we have to use avocados number. Remember that step in order to figure out the final answer for any question that appears like this.

6

Properties of Light Calculations 1

3m

Play a video:

Was this helpful?

So here we're told laser pulse produces 1.242 killer jewels of energy. It was experimentally determined that the pulse contains 3.50 times 10 to the 22 photons determine the wavelength of light in meters emitted by one photon. Alright, so we're talking about energy and we're talking about wavelength. Remember the equation that connects them together is energy of a photon equals plank's constant times the speed of light divided by wavelength in meters. Now we're going to plug in the values that we know from the information given. We want to isolate our wavelength. So here if we rearrange this equation, we'll be able to isolate our wavelength. So wavelength equals planks, constant times speed of light, divided by the energy of a single photon, plug in plank's constant plug in speed of light. So when we do that, all we have to do now is determine what the energy of a single photon is. And plug that into. The formula were given what the total energy is. And we're told that total energy is the result of this many photons. So all we're gonna do here is we're gonna convert my total energy into jewels. So killer jewels cancel out. Now I have jewels and we're gonna divide that by the total number of photons so that we can figure out the energy of a single photon. So we have 1.242 times 10 to the three jewels total divided by the number of photons will give me the energy of a single photon. So there goes the energy of a single photon and we can take that and plug it into our formula. So here jules cancel out seconds, cancel out. We'll have a final answer here in meters. So when we plug all that in, we're gonna get 5.60 times 10 to the negative six m. So just realize for a question like this, the equation that connects wavelength and energy together is this one. But here the energy is of a single photon. So we have to take the total energy and divided by the number of photons to find out what that energy is. Once you do plug it in to isolate your wavelength at the end and meters. Now that you've seen this example, move on to example to look to see if you can determine what the final answer will be for this question. Don't overthink it. Read into what the question is asking you to find and then do any types of manipulations needed to get your final answer. Once you do move, move on and take a look at the video in which I explain how to approach. Example too

7

Properties of Light Calculations 1

2m

Play a video:

Was this helpful?

So here it says how much total energy in micro jewels per mole would it take to remove the electrons from a mole of hydrogen atoms here, we're told that the ionization energy for a hydrogen atom is 2.178 times 10 to the negative 18 jewels. Alright realize here that they're talking about the ionization energy for a single atom. Adam. Photon. We can say here that this represents the jewels per adam or jewels per photon based on what we know from before. We know that if we have jewels per photon or jewels per atom, we can utilize avocados number in order to convert it to moles of atoms or moles of photons. So we have 2.178 times 10 to the negative 18 jewels per atom, which is again replaceable with photon. We want micro jewels. So we're gonna say here one micro jewel is equal to 10 to the negative six jewels jules cancel out. Now we have micro jewels per atom. And finally we're gonna say that for every one mole of atoms We have 6.02, 2 times 10 to the 23 atoms. So adam's here cancel out. And what we have left at the end is micro jewels per mole of atoms. So that gives me 1.3, 1 times 10 to the 12 micro jewels per mole. So this question was pretty straightforward if you remembered the steps necessary to cancel out Adams or photons and isolate moles. Now that you've seen example to finish off this page by taking a look at the practice example left on the bottom of the page. Once you do that, come back and see if your answer matches up with mine.

8

Problem

A low-pressure mercury-vapor lamp has a characteristic emission line at 253 nm. Knowing that this lamp is putting out 11.8 watts of light energy, how many mercury atoms are emitted per second during operation?

A

7.86x10^{-19} Hg atoms/s

B

4.73x10^{5} Hg atoms/s

C

1.50x10^{19} Hg atoms/s

D

1.50x10^{28} Hg atoms/s

E

9.27x10^{-18} Hg atoms/s

© 1996–2023 Pearson All rights reserved.