PY Modern Physics

Transcription

1 PY 35 - Modern Physics Final exam - December 8, 208. Solve all four problems. Give a clear explanation of your work. Each correct solutions will be credited with 25 points. Points will be taken away for errors, according to the severity of the error, or for lack of explanation. The solutions to the problems must be written clearly and, if they involve calculations, they must contain a few lines of explanation of the work. Please note: 5 points will be taken away for solutions which just show equations with no interconnecting words indicating the reasoning followed in solving the problem. Useful data and equations: (The numerical values have been rounded off to simplify the calculations.) Avogadro number: N A = One mole of atoms (or molecules) is the quantity consisting of N A atoms (or molecules.) The atomic weight A of an atom is (approximately) the mass of one mole of the element in grams. The absolute value of the electron s charge: e = C The Lorentz transformation equations for the change of space-time coordinates from a frame R to a frame R moving with velocity v along the x-axis of R are ct = γct βγx () x = βγct + γx (2) y = y (3) z = z (4)

2 Problem The energy released in the explosion of ton of TNT is about J. A) Where does this energy come from? B) The 235 U nucleus may be broken apart by bombarding it with neutrons, a process called nuclear fission. In the fission process an energy of about 200 MeV is released. Where does this energy comes from? C) What mass of 235 U is needed to produce the equivalent of one megaton of TNT by the fission process? (This was problem 5 in assignment 2.) Solution A) The energy comes from mass; the total mass of the molecules before the explosion is greater than the total mass of the molecules after the explosion. B) The energy comes from mass; the mass of 235 U is larger than the masses of its fission products. C) The energy E released by the fission of one atom of 235 U in Joules is E = 200MeV CV = (2 0 8 ) ( J) = J ev (5) where we used the value of the electron s charge e = Coulomb. The energy released by the explosion of one megaton of TNT will be (0 6 ) (4 0 9 J) = J. This will require the fission of J J = (6) nuclei of 235 U. Dividing by Avogadro s number we obtain the number n of moles of 235 U that will be needed: n = N A = = (7) The mass of a mole of 235 U is kg, so the mass needed to produce the equivalent of one megaton of TNT will be ( ) kg = 50 kg (8) 2

3 Problem 2 The electromagnetic radiation inside a cavity at temperature T consists of an infinite superposition of simple waves, each one behaving like a harmonic oscillator with a frequency ν determined by the geometry of the cavity. The distribution of frequencies is not relevant for this problem. The only important fact is that the frequencies go up to infinity, so that there is an infinite number of oscillators inside the cavity. According to classical physics the mean energy of a harmonic oscillator at temperature T is kt, independently of its frequency. This leads to the paradox that the thermal energy of the radiation inside the cavity is infinite (the so-called ultraviolet catastrophe.) Max Planck found a solution to the paradox with the hypothesis that the energy of the oscillators cannot take a continuum of value, as in classical physics, but that it is quantized in units of hν, h being Planck s constant. A) Starting from Planck s hypothesis that the energy of a harmonic oscillator of frequency ν can only take one of the values E n = nhν, n being a non-negative integer, and using the Maxwell-Boltzmann distribution for the probability that a state of energy E n is occupied at temperature T, give a mathematical expression of the thermal average E of the oscillator energy in the form of the ratio of two infinite series. B) Sum the two series to provide the value of E in closed form as function of ν, h, T and the Boltzmann constant k. Solution A) According to the Maxwell-Boltzmann distribution the probability that a state of energy E n is occupied is given by where the normalizing factor is P n = Z e En/kT (9) Z = e En/kT (0) With E n = nhν the two above equations become P n = Z e nhν/kt () 3

4 where the normalizing factor is Z = e nhν/kt (2) The thermal average of the oscillator energy is given by or E = Z E = E n P n (3) nhνe nhν/kt = hνe nhν/kt e nhν/kt (4) B) It is convenient to define x = hν kt so that the two series in Eq. 4 become (5) and We note that nhνe nhν/kt = hν Z = ne nx (6) e nx (7) ne nx = d dx e nx = dz dx (8) So, if we can obtain the expression of Z as function of x, the calculation of E is reduced to taking a derivative. About Eq. 7, with the substitution e x = y (9) the series for Z becomes y n = y = e x (20) 4

5 Putting everything together we find E = hν dz/dx = hν d log Z = hν d log( e x ) Z dx dx e x hν = hν = e x e x = hν e hν/kt (2) Problem 3 An observer measures the velocity of two electrons and finds that one has a speed c/2 in the x direction and the other has a speed c/2 in the y direction. What is the relative speed of the two electrons? (This was problem 2 in the second midterm exam.) Solution We go to the rest frame R of the electron moving in the x-direction. This frame moves with velocity v = c/2 along the x-axis of the original frame and its space-time coordinates, which we denote by a prime, are related to the coordinates in the original frame R by ct = γct βγx (22) x = βγct + γx (23) y = y (24) z = z (25) where β = v/c = (c/2)/c = /2, γ = / β 2 = / 3/4 = 2/ 3. The relative velocity of the two electrons will be given by the velocity of the second electron in the frame R. We obtain the components of this velocity by taking the differentials of Eqs : c dt = γc dt βγ dx (26) dx = βγc dt + γ dx (27) dy = dy (28) (The z-components are irrelevant.) By dividing Eqs. 27 and 28 by Eq. 26 we 5

6 get or dx = u x c dt c dy = u y c dt c = = βc + dx/dt c β dx/dt = βc + u x c β u x = (29) dy/dt γ(c β dx/dt) = u y γ(c β u x ) (30) u x = v + u x (3) vu x /c 2 u u y y = (32) γ( vu x /c 2 ) where we denoted by u x, u y and u x, u y the components of the velocity of the second electron in frames R and R respectively. With u x = 0, u y = c/2 these equations reduce to u x = v = c 2 (33) u y = u y γ = c/2 2/ 3 = c 3 4 Finally we find for the relative velocity v rel = u x 2 + u y 2 = c = Alternatively one can solve the problem as follows: (34) 7 4 c (35) We denote by u and u the velocities of the second electron in the frames R and R, and by u and u the magnitude of these velocities. In R u = (0, c/2, 0) and thus u = c/2. In R the value of u, which is the relative speed of the two electrons, remains to be determined. Let us consider now the four-component velocities U and U of this electron in the two frames. We have U 0 = γ(u)c = (c/2)2 /c c = c (36) 2 3/4 U = 0 (37) U 0 = γ(u)c = u 2 /c c 2 (38) 6

7 The other components of the four velocity are irrelevant for this problem. Now, from the Lorentz transformation equations we get U 0 = γ(v)u 0 vγ(v) c U = γ(v)u 0 (39) where v is the velocity of frame R in frame R. Since v is also equal to c/2, γ(v) is equal to γ(u) namely γ(v) = 3/4 (40) Inserting values in Eq. 39 we find u 2 /c c = c = 2 3/4 3/4 3/4 c (4) or, simplifying the common factor c, taking the reciprocal of both sides of the equation, and squaring with the solution u 2 /c 2 = 9/6 = 7/6, or u 2 /c 2 = 9/6 (42) u = 7 4 c (43) Problem 4 A) Find the energy(ies) of the bound state(s) of a quantum mechanical particle of mass m in a finite square-well potential of height V and width 2d. Take the potential to be given by V (x) = 0 for x <= d, and V (x) = V for x > d. Consider only the state(s) even under x x (i.e. with ψ( x) = ψ(x).) Give your answer in the form of an equation that must be satisfied by the energy E of a bound state. B) The equation for the bound state energy(ies) cannot be solved analytically. Provide and sketch a graphical determination of their value. (This problem was considered in the discussion session of November 29 and its solution was posted on the course website.) 7

8 Solution 0 tan(sqrt(2me)d/hbar) sqrt((v-e)/e) Figure : Graphical solution of the equation for the bound states energies. With the chosen parameter values, V = 0, 2m d/ h =.5 in arbitrary units, there are two bound states. For x d the Schrödinger equation is h2 2m and the solution even under x x is with d 2 ψ = Eψ(x) (44) dx2 ψ(x) = A cos kx (45) k = For x > d the Schrödinger equation is or h2 2m h 2 2m 2mE h (46) d 2 ψ + V ψ(x) = Eψ(x) (47) dx2 d 2 ψ = (V E)ψ(x) (48) dx2 8

9 and the solution is with 2m(V E) q = h We impose continuity of ψ and dψ/dx at d: Dividing Eq. 52 by Eq. 5 we get ψ(x) = Be qx (49) (50) A cos kd = Be qd (5) Ak sin kd = Bqe qd (52) k tan kd = q (53) or ( 2mE ) V E tan d = (54) h E This equation cannot be solved analytically, but a graphical solution can be obtained by plotting the two sides of Eq. 54. The intersection of the two curves gives the energies of the bound state(s). The graphical solution is illustrated in Figure. 9