PY 351 - Modern Physis Assignment 6 - Otober 19, 2017. Due in lass on Otober 26, 2017. Assignment 6: Do all six problems. After a base of 4 points (to make the maximum sore equal to 100), eah orret solution will be given 16 points, with points taken away for errors, or for absene of explanation, aording to the severity of the error. Please remember that the solutions must be written learly, in pen or typeset. Clarifiation about binding energy. I am offering a larifiation about the notion of binding energy, prompted by the question from a student. When two or more objets, like the two protons and two neutrons, bind together into a state of lower energy, like the alpha partile in problem 1, the differene in energy between the initial and final states is alled binding energy. It is typially released when the objets bind together and it would have to be spent two separate the onstituents from the final omposite state. (Although the situation is generally not so simple. In order to get the two protons and the two neutrons to oalese together it is neessary to overome the eletrostati repulsion of the two protons. That osts energy, whih would eventually be given bak, on top of the binding energy, if one manages to get the protons and the neutrons within the range of the strong nulear binding fore.) In lass I gave the example of a heavy nuleus of mass M whih splits up into two smaller nulei of mass m 1 and m 2. The differene in rest energy, E M 2 m 1 2 m 2 2, would be released as kineti energy of the two lighter nulei. In this ase E should not be onsidered as a binding energy, or, if we wish, should be onsidered as a negative binding energy. No energy would be released if we brought the lighter nulei together. Indeed one would have to spend energy to make them oalese into the heavier nuleus. But, then, why does the heavier nuleus not disintegrate right away? The answer is that it has to overome an barrier to split apart. With suffiient prodding, 1
like if a neutron hits it, it will split apart and then the extra energy that it ost to keep it together (the negative binding energy) will be released. See the differene with the ase of the two protons and neutrons: bringing them together into an alpha partile releases energy, the binding energy; bringing the lighter onstituents nulei together into the heavier nuleus osts energy, like loading a spring, energy whih will be released when the heavy nuleus splits apart. From an atual physial situation: in an atomi bomb or in a nulear reator a nuleus of uranium 235, when hit by a neutron, splits into esium 137, rubidium 95, and 4 neutrons. The energy balane is as follows: m 2 ( 235 92 U) 218.8969 GeV m 2 (n) 0.9396 GeV total initial energy 219.8365 GeV m 2 ( 137 55 Cs) 127.5011 GeV m 2 ( 95 37Rb) 88.3859 GeV m 2 (4n) 3.7584 GeV total final energy 219.6454 GeV energy released 0.1911 GeV (1) 2
Problems 1 to 4: From the textbook, problems 17, 26, 31, and 36 at the end of Chapter 4. Problems 1 and 2: solution Problems 3 and 4: solution Problem 5 We denote the omponents of the three-dimensional veloity vetor by (u x, u y, u z ). The four-dimensional veloity vetor U µ is defined by U µ dx µ dτ (2) 3
with x 0 t, x 1 x, x 2 y, x 3 z and dτ dt 2 dx2 + dy 2 + dz 2 1 u2 x + u 2 y + u 2 z dt 1 dt (3) 2 2 γ(u) where 1 γ(u) (4) 1 u2 x +u2 y +u2 z 2 (We use upper ase U for the four-dimensional veloity to avoid onfusion with u and the Greek index µ to label its omponents.) We have seen in lass that U 0 γ(u) U 1 γ(u)u x U 2 γ(u)u y U 3 γ(u)u z (5) In a hange of frame U µ transforms in the same way as the spae-time oordinates x µ (beause it is the ratio of dx µ over the invariant dτ), i.e. U 0 γ(v) U 0 + v γ(v) U 1 U 1 v γ(v)u 0 + γ(v) U 1 U 2 U 2 U 3 U 3 (6) Use these Eqs. 5 and 6 to prove the relation between the veloities u x, u y, u z and u x, u y, u z in the two frames. Hint: Substitute U 0, U 1, U 2, U 3 from Eq. 5 into Eq. 6, divide lines 2,3,and 4 in the resulting equation by line 1) and simplify. Solution Substituting U 0, U 1, U 2, U 3 from Eq. 5 into Eq. 6 we get γ(u) γ(v)γ(u ) + v γ(v)γ(u ) u x γ(u)u x v γ(v)γ(u ) + γ(v)γ(u )u x γ(u)u y γ(u )u y γ(u)u z γ(u )u z (7) 4
Dividing lines 2,3,and 4 in the above equation by line 1) and simplifying we find u x v + u x + vu x/ u y u z or, multiplying by aross the equations u y γ(v)( + vu x/) u z γ(v)( + vu x/) (8) u x u y v + u x 1 + vu x/ 2 u y γ(v)(1 + vu x/ 2 ) u z u z γ(v)(1 + vu x/ 2 ) whih give the relation between the three-dimensional veloities in the two frames. Problem 6 Compton sattering: A photon of wavelength λ 1 hits an eletron at rest and is sattered at an angle θ from the inident diretion. Prove that the wavelength λ 2 of the sattered photon is related to the wavelength of the inident photon by λ 2 λ 1 h (1 os θ) (10) m2 where m is the eletron s mass. Note: this problem is solved in the textbook on pp. 125 to 127. To get redit for this problem you must reprodue arefully the solution of the textbook: essentially, you should opy it but with omplete understanding of all the steps. Please also note: beause this problem basially just involves diligene, a poorly written solution will not be redited with any point. Solution: See the solution given in the textbook. 5 (9)