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### 6.1 6.2 and 6.6

Course: PHYSICS 301, Fall 2011
School: Rutgers
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Word Count: 956

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H7C Physics Fall 1999 Solutions to Problem Set 7 Derek Kimball Above the front door of Niels Bohr's cottage was nailed a horseshoe. A visitor who saw it exclaimed: &quot;Being as great a scientist as you are, do you really believe that a horseshoe above the entrance to a home brings good luck?&quot; &quot;No,&quot; answered Bohr, &quot;I certainly do not believe in this superstition. But you...

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H7C Physics Fall 1999 Solutions to Problem Set 7 Derek Kimball Above the front door of Niels Bohr's cottage was nailed a horseshoe. A visitor who saw it exclaimed: "Being as great a scientist as you are, do you really believe that a horseshoe above the entrance to a home brings good luck?" "No," answered Bohr, "I certainly do not believe in this superstition. But you know," he added with a smile, "they say that it does bring luck even if you don't believe in it!" n1+ and N e2 . 2 2m0 (0 - 2 )2 + 2 2 2 0 - 2 N e2 2 2m0 (0 - 2 )2 + 2 2 equations break down in the vicinity of the resonance, which is where we must work to solve this problem. So here we bring back the 2 2 terms, but continue to assume there are few electrons. In this case our formulas for and n are given by: (4) - George Gamow, excerpted from Thirty Years that Shook Physics. If you have any questions, suggestions or corrections to the solutions, don't hesitate to e-mail me at dfk@uclink4.berkeley.edu! (5) If we take the derivative of n with respect to and set it equal to zero, we find two positive roots yielding the values for the max and min of the the function n, This problem makes sense only if you make some rather poorly motivated ap- namely proximations. In particular, we must assume that we are far away from resonance = 0 1 /0 . 2 (namely that 0 - 2 ). We also can assume that there are very few electrons 2 (namely that N e0 1, which is well-motivated by the fact that is much less than It is safe to assume, since damping is small, that this value can be approximated m n, i.e. few absorbers). If we make these approximations, the results follow almost by the first order Taylor expansion: immediately. If you don't make these assumptions, then the results are clearly = 0 /2. incorrect (see Figure 6.1 in Fowles, which is nothing like the equations Fowles asks us to derive). Thus, we'll make these assumptions! If we plug these values into our expression for (Eq. 5), we see that these values are those where attains half its maximum value. Then we can apply these approximations to equations 6.34 and 6.35 in Fowles. We find that: Problem 3 1 N e2 (1) n2 - 2 n2 1 + 2 We are given that = 6.8 107 mho/m and that Ne = 1.5 1028 electrons/m3 . m0 0 - 2 Using these values in the appropriate Fowles formulas gives us the desired anUsing a first order Taylor expansion, we then find that: swers... 1 N e2 2 2m0 0 - 2 (2) (a) Plasma frequency p = (3) (b) Relaxation time = October 21, 1999 0 c2 = 1.6 10-13 s 2 p N e2 = 6.9 1015 s-1 m0 Problem 1 n1+ Since n is approximately 1, is given by: N e2 . 2 2m0 (0 - 2 )2 Problem 2 Once again Fowles attempts to confuse us by implying the above results can be applied in the solution to this problem when they can't. This is because above Physics the H7C Fall 1999 Solutions to Problem Set 7 Derek Kimball (c) Our frequency with a wavelength of 10-6 m is given by our confidence in the value of a we have measured and then convolve it with the probability for seeing a second event. The probability density function, in this case that for Poisson statistics, is given by: fp (x) = e-a ax x! (6) = 2c = 1.9 1015 s-1 . Real and imaginary parts of the index of refraction can be derived from from Fowles Eqs. 6.55 and 6.56: 2 p 2 + -2 n2 - 2 = 1 - 2 p 1 2 + -2 where x is a non-negative integer and a is the average value of x. The probability density function (with known parameter a) allows us to predict the frequency with which random data x will take on some particular value. 2n = Clearly, p , -1 , so we get 2 n2 - 2 1 - = -12.2 2 1 p = 0.044. 2 p 2n We first want to calculate a value for a = np (where n is the number of interactions and p is the probability for an event), a distribution function based on the most likely values for a and our confidence in those values. The most likely value of p from the data is 10-6 . For a conservative upper limit (without assuming very much about the prior probability distribution), we can estimate that the lower limit of p (at a 95% confidence level) must be that for which the probability of seeing one event in 106 interactions is at least 5%: fp (1) = e-np np 0.05, (7) We can then solve these equations for n and , and with a little algebra we get: n = 0.006 = 3.5. which tells us that np 0.05, or p 5 10-8 . Furthermore, we know that the upper limit at 95% confidence level on p can be found from: fp (1) = e-np np 0.95, which tells us that p 5.14 10-6 . (8) (d) The reflectance is given by the Hagen-Rubens formula, 80 1. R=1- Problem 4 We want to be 90% sure we'll see a second event. We could guess that if we're 95% sure p is bigger than pmin = 5 10-8 and 95% sure that we'll see at least one more event after n2 interactions using this value for p, we'll be 90% sure to see a second event. We're 95% sure we'll see at least one more event if the probability to see zero events is less than 0.05: fp (0) = e-n2 pmin 0.05, This problem, as Prof. Strovink pointed out, is a little bit tricky. Our experimenter finds 1 event in 106 interactions. Now we want to be 90% sure we find a second which gives us event. How many interactions do we need? The basic problem is that we don't n2 6 107 interactions. really know the average number of events we should see per 106 interactions, which is needed to calculate how many more interactions are necessary to be 90% sure This is a conservative upper limit on the number of interactions we need before we'll see a second event. So we'll have to try to figure out some function describing we'll see another event. October 21, 1999
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