The energies for an electron in the K, L, and M shells of the tun...

Question

The energies for an electron in the K, L, and M shells of the tungsten atom are -69,500 eV, -12,000 eV, and -2200 eV, respectively. Calculate the wavelengths of the Ka and Kb x rays of tungsten.

Accepted Answer

Hey everyone. So this problem is dealing with the atomic structure and x-rays, let's see what it's asking us. We have a high energy beam in a particle accelerator that knocks a K shell electron of an atom which creates a vacancy in the K shell. This vacancy can be filled with an electron in the LM or N shell. Each cell has a distinct energy. The electrons energy in the K shell is given as negative 68,500 electron volts. The L shell is given as negative 11,000 electron volts in the MS shell is a negative 2100 electron volts. We're asked to determine the wavelength of the characteristic x-rays K alpha and K beta. And our multiple choice answers are given in PICO meters where we have a 21.6 and 18. B, 32.5 and 17.7 C 15.2 and 19.1 or D point sorry. Yeah, 28.4 and 13.9. OK. So the first key to solving this problem is understanding what these characteristic x-rays are. So we can recall that the alpha K is when we go from an electron in the L shell to the show. And our beta is when we go from the m shell to the show. And so now we can use the given energies understanding the moves of the electrons to find what our energies are. So our alpha energy is going to be the energy of the L shell minus the energy of the K shell. And from that problem or from the problem, those given values are negative 11,000 electron volts minus a negative which is a positive 68,500 electron volts. So our alpha energy comes out to 57 1005 ev. And we'll do the same for our beta uh characteristic x-ray. So we'll have the energy from the m shell electron minus the energy of the K shell electron. And so that will be negative ev minus a negative again positive. So plus 68,500 EV and that comes out to 66,000 400 EB. And so they're asking the problem is asking for the wavelengths of these x-rays. And so we can recall that the relationship between energy and wavelength is given by the equation E equals HC divided by lambda, where H is our planks constant C is the speed of light. And then lambda is that wavelength that we're solving for. So we can rearrange this equation to isolate Lambda. On the left hand side and we get Lambda equals HC divided by E. And so now all we have to do is plug in those energy values that we uh solved above and find our wavelengths. So lambda sub alpha is going to be equal to our planks constant, which we can call is 6.626 times 10 to the negative 34. And that's units of jewels multiplied by seconds and multiplied by our speed of light, which we can recall is three times 10 to the 8 m per second. All of that divided by our alpha energy, which we calculated to be 57,500 electron volts. Now, we need to be careful with our units because we have jewels on top and our ev on the bottom. So we need to convert and we can recall the conversion factor. There is one electron volt divided by 1.602 times 10 to the negative 19 jules. And so we plug all of that into our calculator and we get 2.16 times 10 to the negative m. Going back up to our answers. These answers are all in PICO meters which is a 10 to the negative 12 m. And so we can rewrite that as 21.6 PICO meters. And now we can do the same thing for our Yeah, sorry. A beta yeah wavelength. And so our beta wavelength is going to be the same two constants on the top. So we have planks constant multiplied by the speed of light divided by our beta energy, which we calculated to be 66,400 EV. And then we have to um use the same conversion factor as well. And so we plug that into our calculator and we get 1.87 times 10 to the negative 11 m, which again, we can rewrite as 18.7 PICO meters. And so those are, these two answers are the final answers for our problem. We go back up to our multiple choice solutions and we can see that, that aligns with answer choice A so A is the correct answer for this problem. That's all we have for this one. We'll see you in the next video.

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The energies for an electron in the K, L, and M shells of the tungsten atom are -69,500 eV, -12,000 eV, and -2200 eV, respectively. Calculate the wavelengths of the Ka and Kb x rays of tungsten.

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