Properties of Light - Online Tutor, Practice Problems & Exam Prep

The electromagnetic spectrum encompasses various types of radiation, from long radio waves to gamma rays. Visible light, a small segment, ranges from approximately 700 nm (red) to 380 nm (violet). Key relationships include energy (E), frequency (ν), and wavelength (λ), where E = hν and c = νλ, with c being the speed of light (2.998 × 10⁸ m/s). As frequency increases, wavelength decreases, and energy rises, affecting atomic and molecular transitions, such as electron excitation in UV-visible spectroscopy and core electron interactions with X-rays.

The Nature of Light

Hey everyone. So in this video, we're going to take a look at the electromagnetic spectrum. Now, here we're going to say that visible light represents a small portion of the continuum of radiant energy known as the electromagnetic spectrum. And if we take a look at our electromagnetic spectrum, we can say that we have, starting on the left side, long radio waves and as we're moving from left to right, we're transitioning into radio waves, ones that we should be familiar with based on our car stereo. We have AM and FM radio. From there, we transition to microwaves, then infrared. This small portion here represents our visible light spectrum. That's the portion that we can see with our own eyes without the aid of any instrumentation.

After that, we have UV, then X-rays, and then gamma rays. We can say here that the trend is as we head from left to right, we say here that we have an increase in our frequency, which is represented by f, and we also have an increase in our energy which is represented by E. These two are directly proportional; they both increase or decrease together. Also, as we're heading towards the left side, we're going to say that our wavelength, which is represented by λ, would be decreasing.

Now here, if we were to zoom in more into the visible light spectrum itself, we'd see this and we can remember the colors based on ROYGBIV. And remember here, ROYGBIV would be red, orange, yellow, green, blue, indigo, and violet. Now, a lot of literature nowadays tends to group together indigo and violet into just violet, but if we wanted to really separate them, we'd have indigo, then followed by violet. Now, here it's a range of approximately 700 to 400 nanometers, but if you want it to be as specific as possible, it's actually 700 nanometers to 380 nanometers.

Now, with these transitions from different forms of electromagnetic radiation, we have atomic and molecular transitions involved. We can say that with radio waves, what we typically have is nuclear spin of an atom or element. With microwaves, we tend to have molecular rotations, and electron spins. And with this, we tend to think of NMR, different types of processes if you take organic chemistry. Next, we have infrared, which deals with molecular vibrations. This helps us to determine different types of functional groups. For those who've taken organic chemistry, we have infrared spectroscopy, so that deals with vibrations of chemical bonds, and from that, we can tell different types of functional groups involved, like alcohols, aldehydes, etc.

Next, we have the visible light spectrum. This deals with valence electrons, as does ultraviolet. Now these two, they're part of a process known as UV-Visible or UV-Vis spectroscopy. This examines what we call conjugated systems. Conjugated systems are basically compounds that have alternating double and single bonds. They go double bond, single bond, double bond, single bond. The study of valence electrons really deals with electronic excitation. We're going, jumping up to a higher energy state. Next, we have X-rays. X-rays affect core electrons. Now, what you should see in terms of as we go from radio waves down, what's happening? Our energy is increasing. Our frequency is increasing. As the energy increases, that energy is better able to penetrate an atom.

So you can see that deeper and deeper into the atom is being affected. So when we get to X-rays, we're talking about core electrons themselves being affected. Now here, this deals with bond breaking and even ionization. Finally, if we're dealing with gamma rays, we're talking about nuclear reactions. So gamma rays are extremely energetic; they're dealing with nuclear reactions. Right. So just remember, we're taking a look at the electromagnetic spectrum, and with them, we have all these different types of electromagnetic radiation. The only portion that we can see without the aid of instruments is the visible light spectrum, but we have different types of electromagnetic radiations above and below it. Right? So keep in mind all that we've gone over here in terms of looking at their order as well as the different types of atomic and molecular transitions involved.

The Nature of Light

So in terms of the electromagnetic spectrum, we said that we look at 3 variables: energy, frequency, and wavelength. Now, in terms of frequency and wavelength, we're going to say here that μ represents our frequency. It's the number of waves you have per second and it's using the units of seconds inverse or Hertz. Then we're going to say that λ represents our wavelength. It's the distance from one crest of a wave to another 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 going to say here we have our y-axis, our z-axis, and our x-axis here. Here we're going to say the top of this wave, the crest of the other wave would represent our wavelength λ. Then we'd say here that the number of waves we get within a given second would represent my frequency or μ. 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 you 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 frequency's job. So these are the variables that you need to remember; this is how they're related to one another. Remember that 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. And remember, they're connected to the energy of the different types of electromagnetic spectrum radiation that we saw up above.

Properties of Light

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 \(2.998 \times 10^8\) meters per second. This is called our speed of light. Now, our speed of light is equal to the variable \(c\). So, the speed of light equals the product of mu, which is your frequency, times lambda, which is your wavelength.

We're going to say that the physicist Max Planck and Albert Einstein theorized that light was made up of small packets of electromagnetic energy, which they called quanta. 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 Planck's constant and it's equal to \(6.626 \times 10^{-34} \text{ joules} \cdot \text{ seconds}\). Now, in addition to this, we can say that the 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 or isolate our frequency, we could say that by dividing 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 Planck's 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 shows the inverse relationship. 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 from the given information. Once you do, come back and see if your answer matches up with mine.

Properties of Light

So here it says to calculate the wavelength in nanometers of the red light emitted by a neon sign with a frequency of 4.16 times 10 to the 8 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 the 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, remember, is 2.998 times 10 to the 8 meters per second divided by your frequency in hertz, not megahertz. Hertz, not megahertz. So we have to do some converting. So we have 4.16 times 10 to the 8 megahertz. Remember here that one megahertz is equal to 10 to the 6 hertz. That gives me 4.16 times 10 to the 14 hertz. Take that now and plug it in, so we have:

c μ

Hertz 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 7 meters. 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:

7.21 × 10 - 7 meters 10 - 9 meters/nanometer

Meters cancel out and what we get at the end is 721 nanometers as our final answer. So that number represents the wavelength of the red light that's being emitted. And 721 is around the range of red light.

Now that we've seen example 1, take a look at example 2 and 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 2.

Properties of Light

So here it asks, what is the energy in joules of a mole of photons associated with visible light of wavelength 493 nanometers? Alright. What we're going to do first is determine the energy of a single photon, before we figure out the mole of photons. So we're going to say the energy of a single photon equals Planck's constant times, because the question is dealing with wavelength, we're going to say times speed of light divided by wavelength. So Planck's constant is 6.626×10-34 joules times seconds. Speed of light is 2.998×108 meters per second. And we need wavelength to be in units of meters. So, we have to convert the 493 nanometers into meters. So, 1 nanometer is equal to 10-9 meters. So, it gives me 4.93×10-7 meters. Plug that in. So seconds cancel out. Meters cancel out. What we are going to get at this point is 4.029×10-19 joules per photon. Now we have to convert joules per photon into joules per mole of photons. So bring down that value. We're going to have to get rid of those photons. We're going to say here 1 mole of photons, just like 1 mole of anything, is equal to Avogadro's number, 6.022×1023 photons. So photons here cancel out, and I'll have as my final answer joules per mole of photons. So that gives me 2.43×105 joules per mole. So, realize we utilize the first equation in order to figure Avogadro's number. Remember that step in order to figure out the final answer for any question that appears like this.

Properties of Light Calculations 1

Here we're told a laser pulse produces 1.242 kilojoules of energy. It was experimentally determined that the pulse contains \(3.50 \times 10^{22}\) photons. Determine the wavelength of light in meters emitted by 1 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 Planck's constant times speed of light divided by wavelength 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 Planck's constant times speed of light divided by the energy of a single photon.

Plug in Planck'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. We're given what the total energy is and we're told that total energy is the result of this many photons. So all we're going to do here is we're going to convert my total energy into joules. So kilojoules cancel out. Now I have joules and we're going to divide that by the total number of photons so that we can figure out the energy of a single photon. So we have \(1.242 \times 10^{3}\) joules total, divided by the number of photons, will give me the energy of a single photon. Photon and we can take that and plug it into our formula. So here, joules cancel out. Seconds cancel out. We'll have a final answer here in meters.

So when we plug all that in, we're going to get \(5.60 \times 10^{-6}\) meters. 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 divide it 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 in meters.

Now that we've seen this example, move on to example 2. 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 on and take a look at the video in which I explain how to approach example 2.

Properties of Light Calculations 1

So here it says, how much total energy in microjoules 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 joules. Alright. Realize here that they're talking about the ionization energy for a single atom. Atom, photon, we can say here that this represents the joules per atom or joules per photon. Based on what we know from before, we know that if we have joules per photon or joules per atom, we can utilize Avogadro's number in order to convert it to moles of atoms or moles of photons. So we have 2.178-18 joules per atom which is, again, replaceable with photon. We want micro joules. So we're gonna say here, 1 microjoule is equal to 10-6 joules. Joules cancel out. Now, we have microjoules per atom. And finally, we're going to say that for every 1 mole of atoms, we have 6.02223 atoms. So atoms here cancel out and what we'll have left at the end is microjoules per mole of atoms. So that gives me 1.3112 microjoules per mole. So this question was pretty straightforward if you remembered the steps necessary to cancel out atoms or photons and isolate moles. Now that you've seen example 2, 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.

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. 4.73x10⁵ Hg atoms/s
• C. 1.50x10¹⁹ Hg atoms/s
• D. 1.50x10²⁸ Hg atoms/s
• E. 9.27x10^-18 Hg atoms/s

Here’s what students ask on this topic:

What is the electromagnetic spectrum and what are its different types of radiation?

The electromagnetic spectrum encompasses all types of electromagnetic radiation, ranging from long radio waves to gamma rays. The spectrum includes, in order of increasing frequency and energy: long radio waves, radio waves (AM and FM), microwaves, infrared, visible light, ultraviolet (UV), X-rays, and gamma rays. Visible light, which humans can see without instruments, ranges from approximately 700 nm (red) to 380 nm (violet). As we move from radio waves to gamma rays, the frequency and energy increase, while the wavelength decreases.

How are energy, frequency, and wavelength related in the electromagnetic spectrum?

Energy (E), frequency (ν), and wavelength (λ) are interrelated in the electromagnetic spectrum. The energy of a photon is given by the equation $E=h\nu$, where $h$ is Planck's constant (6.626 × 10^-34 J·s). The speed of light (c) is related to frequency and wavelength by the equation $c=\nu \lambda$, where $c$ is approximately 2.998 × 10⁸ m/s. As frequency increases, wavelength decreases, and energy increases.

What is the visible light spectrum and its range in nanometers?

The visible light spectrum is the portion of the electromagnetic spectrum that can be seen by the human eye without the aid of instruments. It ranges from approximately 700 nanometers (nm) to 380 nm. The colors in the visible spectrum can be remembered using the acronym ROYGBIV, which stands for Red, Orange, Yellow, Green, Blue, Indigo, and Violet. Red light has the longest wavelength (~700 nm), while violet light has the shortest (~380 nm).

What is the speed of light and how is it related to frequency and wavelength?

The speed of light (c) in a vacuum is approximately 2.998 × 10⁸ meters per second (m/s). It is related to frequency (ν) and wavelength (λ) by the equation $c=\nu \lambda$. This means that the speed of light is the product of the frequency and the wavelength of the electromagnetic wave. As frequency increases, wavelength decreases, and vice versa, to maintain the constant speed of light.

What is Planck's constant and how is it used to calculate the energy of a photon?

Planck's constant (h) is a fundamental constant in physics with a value of 6.626 × 10^-34 joules·seconds (J·s). It is used to calculate the energy (E) of a photon using the equation $E=h\nu$, where $\nu$ is the frequency of the photon. Alternatively, if the wavelength (λ) is known, the energy can be calculated using $E=\frac{hc}{\lambda }$, where $c$ is the speed of light.