So in this problem we have a photon that has 740 nanometers as its wavelength. So that's whats being given to us right here is a wavelength of 740 nanometers. And we want to know the energy of this single photon. Well, what we have is an equation that gives us the energy of a photon. And this equation is E equals h times the frequency. But, that's not the frequency. We have an equation that relates frequency to wavelength, C equals lambda nu. Gets you the frequency from the wavelength. So if I were to take these two qu, equations and combine them. Now I like to memorize as little as I can get by with, that's just my nature and I've already got these two equations in my head. So I will take the second equation and I will sail, solve it for frequency. So C over lambda equals frequency, and I can compute this portion into this equation, and have that the energy of a photon is equal to h times c over lambda. And that's a way to get the energy of a photon if you know the wavelength of a photon. I never memorized this equation. I can derive it kind of quickly in my head, but I don't memorize it. Because it's really obtained from the two pieces that I do know. So, now, I can take the information and plug it into this equation. The Planck's constant is 6.63 times 10 to the minus 34 joules times seconds. The speed of light is three times ten to the eighth meters per second and the wavelength is 740 nanometers. Now where students make mistakes is they don't look at their units. They just blindly plug these numbers in and never write down the units. As we cancel units, the seconds will cancel with the seconds, but this meter will not cancel with a nanometer. So, we know we have to convert the nanometers to meters. So, ten to the minus nine meters equals one nanometer. Now the nanometers are cancelling, the meters are cancelling, and we're left with the unit of joule, which is a good unit for energy. Now let's see here, if we do all the mathematics of this, we end up with 2.69 times 10 to the minus 19 joules. Now, what we need to understand about this is this is not the joules of a bunch of photons. This is the joules of one photon. So, we could say it is a joule per photon, like that. Keeping, always keeping that in mind as we use this equation to obtain either this equation or this equation, to obtain the energy of the photon. So in part B, what we have is trying to figure out if I have ten kilojoules of energy from this wavelength, how many photons would that require? So the total energy that I have is ten kilojoules. Now I'm going to go ahead and convert the kilojoules to joules. So that I can use the converter that I just obtained in part A. So it takes a thousand joules to equal the, a kilojoule and then we're trying to figure out how many photons this would require. Well, I know that 2.69 times 10 to the minus 19th joules are in one photon. That's what that converter that I obtained from up here tells me. So if I were to do the multiplication and the division here I would end up with a value of 3.72 times 10 to the 22 photons. So it takes quite a few photons. I ran out of space here, I'll write photons here. >> Mh-hm. >> It takes quite a few photons to give you ten kilojoules of energy.
