 So this problem says, calculate the wavelength and frequency of light emitted when an electron changes from n equals 3 to n equals 1 in the hydrogen atom. What region of the spectrum is this radiation found? So n equals 3 to n equals 1. So in order to do this, since the hydrogen atom you've got to use is the Rieberg equation. So the Rieberg equation, remember, is 1 divided by a wavelength equals the Rieberg constant, which has to be given to you up here, times 1 divided by n2 squared minus 1 divided by n1, just write it down. So here the number of significant figures is infinite, so you can put it to whatever amount of significant figures you want, because the only numbers this problem gave you are 3 and 1, and both of those have an infinite number of significant figures. I'll usually do it to like 3 or 4 whenever that's the case, but not that it's a big problem. So 1 divided by 1 squared minus 1 divided by n. This is 1 minus 1 ninth, so I am going to calculate it to the 1, 2, 3, and this is 1 meter like that. So wavelengths aren't in per meters, they're in meters or appropriate type units. So what we're going to do is divide, say 1 divided by this number, and that will give us our, or invert this, right, because this is 1 divided by a wavelength. So if we want to get wavelength, we're going to take this number and invert it. So the wavelength equals 1 meter divided by 9.75 times 10 to the 6, like that. Is everybody okay with doing that? It's 6 times 10 to the negative 7 meters, but let's convert this into nanometers because it asks us what part of the electromagnetic spectrum this is in. So 1 meter is 1 times 10 to the ninth nanometer, so that's going to be 102.6 nanometers. So that'll be the wavelength of light. So that's going to be in the ultraviolet region. I know you guys haven't memorized the electromagnetic spectrum, but remember, visible is 400 to 750, so below visible, right below visible, is ultraviolet. That's also what's the frequency of this light. So I'm going to erase most of this stuff. Is that okay if I erase most of this stuff? So the only thing I really want to not erase, as I have to answer, is this number here because that's the number I'm going to use to help me get the frequency. So if you recall, if you don't, this is something you need to remember, speed of light equation C equals lambda nu, right? So frequency is nu, and we've got lambda, and we know the speed of light that's given to us, right? Speed of light is 2.998 times 10 to the eighth meters per second. So if we're looking for frequency, that's C divided by lambda, and 2.998 times 10 to the eighth meters per one second, and we've got meters here, so let's just use that number. So the frequency of this is 2.923 times 10 to the 15th per second, right? And per second, remember, is per second, or if you want to look at it this way, per one second, right? And that means per second. Is that on this one? So I know we kind of did three different problems, because the Rydberg equation, right? And then we, I don't know, maybe student problems, the Rydberg equation, and then the figuring out the frequency of using the speed of light. Any questions on this one? Again, I think the hardest part is to remember these equations. So remember the Rydberg equation.