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Fall 07 Phys 143l HW 8 Solutions

Fall 07 Phys 143l HW 8 Solutions - Physics 143 Fall Term...

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Unformatted text preview: Physics 143 Fall Term 2007 HOMEWORK 8 Due: November 5 Problem 1: Estimate the time that a virtual turnip-antiturnip pair could exist. If you somehow had a real turnip-antiturnip pair, about how much energy would be released when they annihilated? Assume each turnip weights 1kg. Therefore the total energy of the pair is: E 2mc Based on the uncertainty relation: E t /2 The time it could exist can be estimated as /2 /2 /2 t~ ~ 3 10 s E E 2mc When they annihilated, the total energy released is 1.8 10 J E E 2mc (You can assume any reasonable mass of turnip and your calculation should consist with your assumption.) (a) Write down an approximate expression for |T| for low-energy electrons in the potential of Problem 12. (b) Calculate|T| for E A/2 . (c) Using the electron mass, estimate your result for part(b) numerically if A=10eV and a=0.05nm. Problem 2: (BFG Ch.8 Problem 13) The expression of the potential: x a V x a 0 others Therefore, the expression for |T| is: (BFG Eq. 8-21) A 1 |T| e V E (a) Assume the energy of electrons is low that E<<A , Plug in |T| |T| exp V x 2 E ~ V x x a exp 8a 2mA 3 dx 2mA 1 (b) The energy of electrons is E=A/2 |T| 2 x a A 2 exp dx 2m A 1 P a g e | 14 The limits on the integral are given by, V(x ) = E = A/2 x = a/2 We obtain |T| exp 2 / dx 2m A 1 x a A 2 exp 4a mA 3 (Notice of that the limits on the integral is NOT (-a,a) but where V>E) (c) Plug in the numbers, |T| exp exp 4 0.05nm 0.511MeV/c 10eV 3 197MeV fm/c 0.76 47% That is the probability of transmission. Problem 3: (BFG Ch.8 Problem 22) Bound-state problem for a square well: write out the matching conditions at the boundary of the well. u x and u x should be continuous at x = a: u a du x dx u x u x u a du x dx A sin qx B exp x B exp a B exp a /2 Here the wave functions are: Plug in the matching conditions A sin qa qA cos qa Divide the 2nd equation by the 1st, we obtain A cot qa or we can change the expression q cot qa q tan qa Problem 4: (BFG Ch.9 Problem 1) Radial Schrdinger equation Express the Schrdinger equation in polar coordinates and separate the variables, then we obtain the equation for R(r): 2d 1 d Ze R r ER r R r 2m dr 4 r r dr r P a g e | 24 rR r . Before plugging in, we shall do some calculation first. u r R r r u u R r r r 2u 2u u R r r r r plug in the equation, 2u 2u 2 u u u 1 u r Ze u r 2m r r 4 r r r r r r r r u 2m r Multiply both sides by r u 1 r u r Ze u r 4 r Eu r 1 u r r r Ze u r 4 r r E u r r Let u r E u r r 2m Rewrite it Ze 1 u r Eu r 2m 4 r 2mr It is a 1D-like Schrdinger equation. However, notice that r should not be negative. That is to say the particle have no probability to be in the r < 0 region which means there is an infinite potential. Therefore the effective potential can be written as x 0 1 Ze V x x 0 4 x 2mx u The fig below shows the effective potential above (thick blue line). And the purple line below the x-axis is for the case l = 0. We can see that when 0, there is a potential barrier which keeps the electron away from the origin. The electron (unlike the classical case, for which it can't go near the origin at all) can only be near the origin by tunneling, i.e. with exponentially decreasing probability. 2 1 0.5 -1 1.0 1.5 2.0 2.5 3.0 3.5 -2 -3 P a g e | 34 Problem 5: (BFG Ch.9 Problem 7) Consider two masses, connected by a rigid rod, rotating about their centre of mass. The angular momentum of this system is: L I mL 2 0.1g 0.5cm 10 J s 800rad/s/2 Compare with the expression in QM L 1 ~ We do the approximation because l must be a very large number. / The energy of the system is: E Where I L 9.5 L 2I 10 and L I E 1 E 1 1 8.4 10 J 2I I E, so if we quantize this macroscopic problem, the quanta is too small to be considered. November 5, 2007 P a g e | 44 ...
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