In these series of work problems we’re going to study the mechanics of light. To begin, we need to establish some background knowledge on what the electromagnetic spectrum is and how ‘light’ plays into it.
What we often refer to as ‘light’ is a narrow portion of a broad spectrum of electromagnetic radiation. This spectrum stretches from radio waves with a wavelength of 10³ m to gamma rays with a wavelength of 10⁻¹² m. We work mainly with light in the visible spectrum in this lesson, but the fundamentals learned here can be applied to any portion of the EM spectrum.
Before we jump in, I wanted to address some prerequisites about the mechanics of these problems.
For one, it is accepted that the speed of light, or how far a beam of light can travel in a given amount of time, is 3.00 * 10⁸ meters/second. This is notated with the letter ‘C’ for these equations.
We can also assume that the speed of light is equal to the wavelength multiplied by its frequency. The wavelength is defined as the length spanning between either crest to crest or trough to trough (the space between two peaks or two valleys of the wave). Wavelength is usually measured in nanometers, and is represented by the greek letter lambda, λ
The frequency of light is defined as the number of wavelengths that pass a certain point during a given period of time. One unit used for this is Hertz (Hz), but the SI unit is accepted as s⁻¹, sometimes writtten as 1/s⁻¹. Frequency is represented in these equations as ν, nu.
When we write out the relationship using their signifiers, it’s represented as: C=λν.
Because of the relationship between these variables, we’ll also need to note the conversions between nanometers and meters. I always write them down first when addressing a problem like this because it’s too easy to get wrapped up in units. Writing them down first when you have a clear head will help you know you’re working with the right proportions.
Let’s finally look at a work problem.
Given Problem: Europium emits a red glow at 625 nanometers. What is the frequency of this radiation?
Step 1: Establish our givens and what we need.
We know that C=λν, C = 3.00 * 10⁸ m/s, and λ = 625 nanometers.
Rearrange C=λν to find ν. We can do this by dividing both sides by λ, ending up with:
C/λ = ν
Step 2: Plug in what we know and convert any units necessary.
When we plug the values given into the equation, we initially have:
(3.00*10⁸ m/s) / (625 nm) = ν
We can’t go forward with dividing until we convert nanometers to meters. This is done by dimensional analysis, make sure you have your units in the appropriate spots. You want to end up with meters, so that will be in your top half.
You should end up with something between the 9th and 7th power, depending on what the appropriate sig figs are. I stick with my value to the *10⁻⁷ power. We’re now ready to run the full equation.
Step 3: Multiply/divide through and check answers.
Your equation should mirror:
(3.00 * 10⁸ m/s) / (6.25 * 10⁻⁷ meters) = ν
First, divide the coefficients, 3.00 / 6.25 = 0.48. Then move to the powers.
I have always learned that you subtract through when dividing powers 10^x / 10^y. In this case, 8 – (-7) = 15. Make sure to note the nanometers cancel out.
Altogether you should end up with something in the realm of:
This is a realistic value for the frequency of an EM wave. Decimals may shift around but it should be between the 14th and 15th power or so. If you calculate a 10¹ power (or just 10), check your work again. I got confused and then checked my powers rules.
Ok, so we can find the wavelength or frequency of a certain EM emittance. What about the energy of that beam?
We can apply the speed of light, wavelentgh, and frequency to energy using Planck’s constant. This formula relates the energy of a photon with its frequency.
A photon is defined as the smalled amount or quantum of EM radiation. We can calculate the energy of a single photon and expand that to find the energy of a mole of photons at a certain wavelength or frequency.
Photons of light can transfer their energy to photoemissive materials. This means that the energy transferred between the photon and the material is enough to dislodge an electron of that material and release it to the surrounding atmosphere. This is known as the photoelectric effect. This will come up in a later post so it doesn’t apply to what we’re working on today, but I thought it was neat.
The energy of a singular photon can be found by multiplying Planck’s constant with the frequency of the light. Furthermore, we can use algebra to manipulate this equation depending on what is asked in the question. (We’ll get there).
Planck’s constant is calculated as 6.626*10⁻³⁴ Joules/second. It’s defined as the letter ‘h‘ in the overall energy equation.
As you can see, we can use the C/λ = ν relationship to substitute into this equation.
Finally, this equation shows that the overall energy of a photon or group of photos is directly proportional to the frequency of the light.
1. How much energy is in one photon with the wavelength of 540 nanometers?
2. How much eneryg is in one mole of photons at 540 nanometers in wavelength?
Question 1, Step 1: First as always, define your givens. You have given constants for the speed of light, and Plancks constant. Your wavelength is defined. You know which equation to use, E = hv.
Step 2: Convert the wavelength from nanometers to meters. You can work with your wavelength with the 10⁻⁹ power but it’ll be easier later on if you work at the 10⁻⁷ power.
Step 3: Plug in what you know. You’ll end up with:
E = (6.626*10⁻³⁴ Joules/second) * (3.00 * 10⁸ m/s) / (6.50 * 10⁻⁷ m)
The meters and seconds will cancel out. This will leave you with Joules as your unit, which is what we want.
Split up your equation into the coefficients and the powers:
(6.626)(3.00) / (6.50) and (10⁻³⁴ )(10⁸) / (10⁻⁷)
Remember when you multiply exponents you add those multiplied together and subtract those being divided. PEMDAS still applies to add your numerators here first and then subtract your denominator.
((-34) + (8)) – (-7), or -26 + 7 = -19.
You should end up with something around 3.68 * 10⁻¹⁷ Joules/ photon at this wavelength.
As you see, I had a hard time with my exponents this time. I’m a little rusty on my math, it’s been a while. So if I went wrong somehwere, please let me know.
Question 2, Step 1: Take your previously calculated value and multiply it by Avogadro’s number. This will apply the energy of one photon towards a mole of them.
(3.68 * 10⁻¹⁷ Joules/photon)(6.022*10²³ photons/mole) = Joules/mole of photon
Step 2: Again, work through the coefficients first and then the powers. Multiply 3.68*6.022 first. Then add the -17 and 23 together. You should end up with something close to 22.1609 * 10⁶ Joules/mole of photons. Adjust for sig figs and you’ll get something like 2.21*10⁵ Joules/mole of photons.
Final check, we answered both questions and both answers are realistic and have the correct units.
Like I said, if I got anything wrong, let me know. If you have any questions, go ahead and leave a comment or email them to us. Thanks!