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A nucleus has nucleon number A a proton number Z and a binding energy EB: The masses of proton and neutron in atomic mass unit (U) are mp and mn respectively: The m...

Question

A nucleus has nucleon number A a proton number Z and a binding energy EB: The masses of proton and neutron in atomic mass unit (U) are mp and mn respectively: The mass of the nucleus in atomic mass unit (U) is given by expression:Zmp +Amn (E/931.5)Zmpt (Atz) mn (E/931,5)mp +Amn (E/931.5)Zmp (A-Z) mn (E/931.5)

A nucleus has nucleon number A a proton number Z and a binding energy EB: The masses of proton and neutron in atomic mass unit (U) are mp and mn respectively: The mass of the nucleus in atomic mass unit (U) is given by expression: Zmp +Amn (E/931.5) Zmpt (Atz) mn (E/931,5) mp +Amn (E/931.5) Zmp (A-Z) mn (E/931.5)



Answers

$.$ co An $\alpha$ particle $\left({ }^{4} \mathrm{He}\right.$ nucleus $)$ is to be taken apart in the following steps. Give the energy (work) required for each step: (a) remove a proton, (b) remove a neutron, and (c) separate the remaining proton and neutron. For an $\alpha$ particle, what are (d) the total binding energy and (e) the binding energy per nucleon? (f) Does either match an answer to (a), (b), or (c)? Here are some atomic masses and the neutron mass. $\begin{array}{llll}{ }^{4} \mathrm{He} & 4.00260 \mathrm{u} & { }^{2} \mathrm{H} & 2.01410 \mathrm{u} \\ { }^{3} \mathrm{H} & 3.01605 \mathrm{u} & { }^{1} \mathrm{H} & 1.00783 \mathrm{u} \\ \mathrm{n} & 1.00867 \mathrm{u} & & \end{array}$

Uh, well, mass defict in C six is given by right six times massive for boat on the six times mass off New journal minus the actual mass equals six times one born double 07 26 you minus six times one born double 08665 You minus two is U equals zero going 0956 you binding energy is given by Judge A. M. Times 9 31. Mega electron voltage So binding energy par nuclear will be Delta M 0.956 times 9 31. Divide by 12 equals 7.4 Do Emmy.

For this problem. On the topic of nuclear physics, we want to take apart an alpha particle in the given steps and find the energy required for each step and then find the total binding energy as well as the binding energy for nuclear. Now the first step can be written as follows. So we have the alpha particle Which you can write as helium four Becoming a proton H one plus hydrogen age three. Now the energy required for this, we'll call delta E one and this is equal to the mess of hydrogen three plus the mass of a proton minus the mess of the alpha particle, helium four. And all of this multiplied by C squared. So using the relative atomic masses, we can write this as three point 01 605. You Plus 1.00 783, you minus four point 00 26. You times the energy equivalent of C squared, which is 931.5 mega electron volts per atomic mass unit. And so we get the binding energy for this For the energy required for this step brother to be 19.8 mega electron volts for part B. The second step is to add energy And so we have hydrogen three becoming a neutron bless yeah hydrogen tube. And so the energy required for this process will call delta E two. And this is the mass of Halogen two plus the mass of a neutron minus the mass of hydrogen three again all times C squared. And so we can write this as 2.01 For 10 Atomic mass units Plus 1.00 867 Atomic mass units minus 3.01 605 atomic mass units times C squared which is 931.5 mega electron volts po atomic mass unit. And this gives The energy required for the step to be six 26 mega electron volts for part C. The third step requires that hydrogen too Becomes a proton. We can write as H one plus a neutron in. And so the work required for the step will call Delta E three and that E three is the mass of the proton. Bless the mess of the neutron minus the mass of age too, times C squared, putting in our masses again, we get this to be 1.00783 Atomic mass units Plus 1.00 867 Atomic mass units minus two point 01 for one zero Atomic mass units. You Multiplied by 931.5 mega electron volts per atomic mass unit. And so This energy is 2.23 mega electron volts. Now we want to find the total binding energy. The total binding energy will call delta B. E. It's simply the some of the energy is calculated in parts A. B and C. So it's down to E one plus delta E two Plus Delta E three. So summing the energies from the previous three parts, we get this total binding energy to be 19.8 Plus 6.26 less 2.23 mega electron volts. Which gives us the total binding energy to be 28.3 mega electron balls. Now that we have the total binding energy, we can find the binding energy per nuclear. For part E. We'll call this delta E. Subsequent B. E. N. And this is equal to the total binding energy down to E, divided by The mass number eight. And so this is 28.3 mega electron volts divided by for nucleons for the alpha particle, which gives the binding energy of 7.07 mega electron evolves. And lastly, for part F, we want to know if either match an answer to a B or C. And we can see that neither the binding energy not dividing and deeper nuclear and match any other answers from A B or C.

Uh, masal A meeting Kim Times C Square is equal to Marcel's Neptune IAM plus most of alpha particle into C square. Less kinetic energy of elf A particle well, from here. Moss off, actually, um is equal to Marcel's, um Amanda Kim, um, minus mass off alpha particle, but minus Connecticut. And to develop a barnacle divided by ah C Square. Now let's plug in the values with massive communication is, uh, 24 1.56 82 atomic mass units minus 4.0 260 It don't make mass units. That's the massive El Barco minus um five 0.5 Mega Electron Walt, divided by ah C Square into one atomic mass unit divided by 9 31 find 49 49 Maga Electron won't Divided by C Square. On therefore muscle napped union is a call to 2 37 0.0 forward eat tree, two atomic mass and it's

By this problem, we're going to find the binding energy at these two. Adam. So where this 1st 1 would stay in Inchon grab and then the next thing you can do it in a m. U. Just to show the difference between the two methods. So for this one, we have 28 pro towns because of the atomic number of, um, nickel is 28 for every proton, there is one point 67 26 2171 times 10 to the negative. 27 a. M u. Uh, kilograms spur proton and we add that to 58 minutes. 28 is 30 times the mass of a neutron, which is 1.6 seven for nine. Two u seven to rate times 10 to the negative, 27 kilograms. And then if we take this value, he comes out to be about 99.7 times 10 to the 26. And if you would convert this Miller Mass and am you to, um, kilograms? It would come out to be, but the molar mass Otis, right. More mass. Like this times. The conversion factor, which is 1.66 zero five or times 10 to the minus 27. When that comes out to be about 9.6, the police attract the first, the second value from the first value, you would get 8.77 times 10 to the minus 28 kilograms. So if we put this into mine sent equation, we have this mass times the speed of light squared, and that would come out to be 7.90 times 10 to the minus 11 Jules per nucleus. And if we take that, uh, finding energy and jewels, we have a conversion factor that says there is 1.6 ot to times 10 to the minus 13 jewels and every one. And maybe we also to divide by the number of nuclear eons her nucleus, though in this one there would be 58 because that's the total. And so we do it out. We would get he point y zero 1000 the second problem. But with terms of A and you, we start with 36 36 times 1.0 seven am you for the amount of protons Plus where he times 1.0 nine. And this comes out to be about 85 a m u. And if we subtract the given were mass from that, we would get, um that the mass defect rmd you call it is 0.7 6673 I am you so turning this into kilograms. The conversion factor is 1.66 times 10 to the minus 27 and it comes out to be 1.27 times 10 to the minus 27th kilograms. And so we take our incident equation and we feel listen for the mass value multiplied by the speed of light square, which is three times tenderly. It's where that then the binding energy ends up being 1.15 times on T minus 10 Jules per nucleus and that we're trying to turn into a maybe per nuclear on. So we take this binding energy and we multiply it by the conversion factor again, which is 1.602 times 10 to the minus 13 um, jewels per everyone m e t. And then for everyone nucleus. There is 84 niggly ons. So the answer would be eight point hi two and needy per nuclear on, and they'll tell you it didn't problem


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