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Problem 14 Medium Difficulty

The peak of the stability curve occurs at $^{56} \mathrm{Fe},$ which is why iron is prominent in the spectrum of the Sun and stars. Show that $^{56} \mathrm{Fe},$ has a higher binding energy per nucleon has a higher binding energy per nucleon than its neighbors $^{55} \mathrm{Mn}$ and $^{59} \mathrm{Co}$. Compare your results with Figure 29.4.

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Video Transcript

the number 14 were to find the binding energy per new plan for each of these elements. Um, let's remember what that is. You confined the mass defect. How much mass is missing by adding up the mass of the individual pieces. So the mass of the all the protons, plus the massive all the neutrons, those together are always a little bit more than the mass of the overall. Adam combined. I'm calling that mass total. And that's something you can look up and appendix B in the back of the book so that missing mass is the binding energy. So we'll calculate that mass in use atomic mass units and then convert it to energy. Um, so here, every my equation all set up here. This is the mass of a proton in use. This is a massive one neutron in use. So we're given iron 56 so I can tell in a periodic table that the atomic number is 26. So that means it has 26 pro tones and mass number remembers the sum of protons and neutrons. So I subtract and I get that it has 30 neutrons. And then I looked upon the tool? Um, Iron 56. And that is 55 0.934 942 So do the math, and I get that there is a mass defect. Oh, point 52 846 And that is measured in you something else. Unit. I'm gonna convert that. I'm gonna go for it. That's the energy equivalent. I know that one. You is the same as 9 31 0.5 Mega electron volts. So my, you cancels. So my binding energy is 4 92 0.26 mega electron volts. But I'm asked to find the binding energy per nuclear on. So per nuclear on, I have 56. Nuclear owns, some dividing up by 56. So my binding energy per nuclear on for iron 56 is 8.79 I'm just doing the same thing here. For Manganese 55 has atomic number 25. So 25 protons I subtract, I guess that it must have 30 new trans. I look up on the table and I get that Maggie's together has a massive 54 point 93 eight Oh, 50 those from Appendix B in the back I do my mouth, my mouth, I get a mass defect. Oh, wait. 5175 25 And that's in use. I'm going to convert that I get the binding energy ISS 4 82 0.745 I divide that by the number of nuclear Jones, which is 55. Let me get my binding energy. Nuclear on from agonies 55 is 8.7 cents. And now Cobalt 59 couple has a atomic number of 27 so it has 27 protons. Subtract. That means it has 32 neutrons. It looked upon the table on DDE Cobalt 59 This 58 0.9 33 too. Oh, I do my math. I get that the mass defect is 0.555 355 It's news. I'm gonna convert that to energy. So my binding energy is 5 17 quaint. 3132 I'm going to buy the apartment number of nuclear ones, which is 59. My binding energy per nuclear on for Kuba 59 is 8.77 mega electron volts per nuclear on. I'm asked to compare that to figure 29.4 come, and that is just a graph of the different points for each of these come binding energies per nuclear. And yeah, the iron 56. That's the highest 11 of the highest ones in the grass, So my results do agree with that.

University of Virginia
Top Physics 103 Educators
Elyse G.

Cornell University

LB
Liev B.

Numerade Educator

Jared E.

University of Winnipeg

Meghan M.

McMaster University