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The emission spectrum of cesium contains two lines whose frequencies are (a) $3.45 \times 10^{14} {Hz}$ and $({b}) 6.53 \times$ $10^{14} {Hz}$ . What are the wavelengths and energies per photon of the two lines? What color are the lines?

$\begin{array}{cl}{\lambda_{1}=8.69 \cdot 10^{-17}(\text { Red })} & {\lambda_{2}=4.59 \cdot 10^{-7}(\text {Blue})} \\ {E_{1}=2.29 \cdot 10^{-19}} & {E_{2}=4.33 \cdot 10^{-19}}\end{array}$

Chemistry 101

Chapter 6

Electronic Structure and Periodic Properties of Elements

Electronic Structure

Periodic Table properties

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of nine from Chapter six is asking us to calculate the wavelength and energy of two lines from the emission spectrum of sexy. So we're gonna need to your questions first. Ah, we'll use the equation that relates speed, wave length and frequency. And then we'll also use planks equation, which relates energy clanks, constant and frequency. Um, you can either song for energy or wavelength. First, it doesn't really matter in this case because you're not using the answer. One toe. Answer the other one. Um, I chose the soul for wavelength first, but you can solve the energy first, if you like. So to solve for, um, wave like we're just gonna have to rearrange this equation. Move the frequency over on this side of the equation and that will get us wave link equals speed. Constant divided by the frequency on I solved for line A and line be at the same time. But I'll walk through each, um so excuse me, I've done them in tandem here. So if we plug in the values for a line A, we have the speed constant which is 2.998 times 10 to the eighth meters per second, divided by the provided frequency in the question 3.45 times 10 to the 14th. Inverse second wth e question gives us the number and Hertz, which is equal to one over one over second. So that's why there's a second inverse here. Uh, so the same for B. This is line be here. So speed constant 2.998 times 10 to the eighth meters per second, divided by the provided frequency for line B which is 6.53 times 10 to 14 University. So what we do cancel the second inverse unit on both sides and then that leaves us with meters on either side. So for line, eh? We get a wavelength of 8.69 times 10 to minus seven meters. And for line B, we get a wavelength of 4.59 times 10 to the minus seven meters. We'll leave these in leaders for no. And then if we need to convert them later weekend the next, we will compute energy. Um, this one we do not have to rearrange the equation, for we just are multiplying the provided frequency for Line A's and lines and be multiplied by blanks constant. So I'm gonna do the same. Gonna calculate them both in tandem, but I'll explain both. So when we plug in the frequencies for Line A and B here, it's like a it might be when we plug in planks. Constant for line, eh? Multiplied by the frequency. We get 6.66 times 10 to minus there before Jewel second multiplied by 3.45 times 10 of the 14 2nd in verse and for B, same collects constant multiplied by 6.53 times in the 14th 2nd in verse. Um, we're gonna, um, as we were being multiplied against each other, we're going to get a unit of jewels. So the seconds counsel outs a second, an inverse second, we're gonna cancel out cause we're multiplying, and then you get an answer in jewels and same for this side. You get the answer in your jewels. Great. So once you multiply those out, you're gonna get a energy of 2.29 times 10 to the minus 19 jewels for a and for B, you'd get 4.33 times 10 to the minus 19 Jules. So then the next part of the question is asking us to determine which color these lines are hand to do that I have found this diagram that I thought was a little bit more or gave a little bit more resolution. Um, for tthe e wavelengths. Um, if you just google visible light spectrum, you confined, um, images like this. So all we need to do is figure out which colors, um, lines a and line B R on this visible light spectrum. So if we go back to line a the wave length of line A, which is here, it's a 0.69 times 10 minus seven, and the wavelength for line B is 4.59 times 10 to the minus of it. So keep that in mind, we're gonna scroll back down here, and I have included them both here. Um, As you can see, this wavelength spectrum is in nanometers. So we're just gonna convert, um, are meters into Nana meters and we do that. Bye. Multiplying are wavelength in meters by tending the night, um so are meters will cancel out on both sides for lines and be, and that's gonna give us a wave like the 869 animators for A and for B that's going to give us a wavelength of 459 animators. So if we go to the scale, let's look at line a first 869 Nana meters. So here's 700 so it's off in the red, so it doesn't have an 800 here, but best answer to select would be 800. So gonna draw a here. And for B. It's 459 and a meter. So if we go back to our skill four hundreds, like four fifties, hear about that's blue. So did Robbie here her line. All right, and that's the answer.

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