Matter Waves. Solutions of Selected Problems


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1 Chapter 5 Matter Waves. Solutions of Selected Problems 5. Problem 5. (In the text book) For an electron to be confined to a nucleus, its de Broglie wavelength would have to be less than 0 4 m. (a) What would be the kinetic energy of an electron confined to this region? (b) On the basis of this result, would you expect to find an electron in a nucleus? Explain. Solution (a) The momentum of the electron is defined by its de Broglie wavelength as: The energy of the electron is: p h λ pc hc λ (ev nm) 0 5 (nm) ev
2 CHAPTER 5. MATTER WAVES. SOLUTIONS OF SELECTED PROBLEMS E (pc) + m ec 4 ( ) + (5 0 3 ) ev 4.0 Mev The kinetic energy K of the electron is: K E m e c M ev (b) The kinetic energy of the electron is too large for the electron to be confined is such small space. Physics 05:Modern Physics I, Chapter 5 Fall 004 Ahmed H. Hussein
3 5.. PROBLEM 5.7 (IN THE TEXT BOOK) 3 5. Problem 5.7 (In the text book) The dispersion relation for free relativistic electron waves is ω(k) c k + (m e c / ) Obtain expressions for the phase velocity v p and group velocity v g of these waves and show that their product is a constant, independent of k. From your result, what can you conclude about v g if v p > c? The total energy of the electron is: Solution Since E ω and p k, we get: E p c + m ec ω ( kc) + (m e c ) ω(k) (kc) + (m ec ) The phase velocity v p is given by: and the group velocity v g is given by: v p ω k (kc) + (mec ) k ( ) me c c + k v g dω dk k kc k c + ( m ec kc k c + ( m ec ) ) Physics 05:Modern Physics I, Chapter 5 Fall 004 Ahmed H. Hussein
4 4 CHAPTER 5. MATTER WAVES. SOLUTIONS OF SELECTED PROBLEMS The product v p v g is; v p v g c (kc) + (mec ) k kc k c + ( m ec ) Therefore, if v p > c, then v g < c Physics 05:Modern Physics I, Chapter 5 Fall 004 Ahmed H. Hussein
5 5.3. PROBLEM 5. (IN THE TEXT BOOK) Problem 5. (In the text book) A beam of electrons is incident on a slit of variable width. If it is possible to resolve a % difference in momentum, what slit width would be necessary to resolve the interference pattern of the electrons if their kinetic energy is MeV,..0 MeV, and MeV? Solution Using the uncertainty principle, with x a and p 0.0p, where a is the slit width, we get: x p a 0.0p a 0.0p c 0.0pc (5.) since we are given the kinetic energy of the electron, we can find pc from: E p c + m ec 4 pc E m ec 4 (K + m e c ) m ec 4 K + Km e c + m ec 4 m ec 4 K + Km e c (5.) Now using Equation (5.) in Equation (5.) and take m e c 0.5 MeV we get: Physics 05:Modern Physics I, Chapter 5 Fall 004 Ahmed H. Hussein
6 6 CHAPTER 5. MATTER WAVES. SOLUTIONS OF SELECTED PROBLEMS a c 0.0pc (MeV nm) 0.0 K (MeV ) + K(MeV )m e c (MeV ) K + 0.5K K +.0K (5.3) (a) Using Equation (5.3) and K 0.0 MeV we get: a K +.0K (0.0) nm (b) Using Equation (5.3) and K.00 MeV we get: a K +.0K (.00) nm (c) Using Equation (5.3) and K 00 MeV we get: a K +.0K (00) nm Physics 05:Modern Physics I, Chapter 5 Fall 004 Ahmed H. Hussein
7 5.4. PROBLEM 5.9 (IN THE TEXT BOOK) Problem 5.9 (In the text book) A twoslit electron diffraction experiment is done with slits of unequal widths. When only slit is open, the number of electrons reaching the screen per second is 5 times the number of electrons reaching the screen per second when only slit is open. When both slits are open, an interference pattern results in which the destructive interference is not complete. Find the ratio of the probability of an electron arriving at an interference maximum to the probability of an electron arriving at an adjacent interference minimum. (Hint: Use the superposition principle). Solution With slit open and slit closed, we have: With slit open and slit closed, we have: P Ψ P Ψ where P and P are the probabilities that the electrons reach the screen though slit and slit respectively. Ψ and Ψ are the wave functions of electrons going through slit and slit respectively. When both slits are open we get: P Ψ + Ψ Ψ + Ψ + Ψ Ψ cos φ where φ is the phase angle between the waves arriving at the screen from the two slits. So P P max when cos φ + and P P min when cos φ. Where P max is the probability that there will be a maximum intensity on the the screen and P min is the probability that there will minimum intensity on the screen. Note that all probabilities are functions of position on the screen, we then have: Since, P max ( Ψ + Ψ ) P min ( Ψ Ψ ) Physics 05:Modern Physics I, Chapter 5 Fall 004 Ahmed H. Hussein
8 8 CHAPTER 5. MATTER WAVES. SOLUTIONS OF SELECTED PROBLEMS we then get: P P Ψ 5 Ψ Ψ 5 P max P min ( Ψ + Ψ ) ( Ψ Ψ ) (5 Ψ + Ψ ) (5 Ψ Ψ ) (6Ψ ) (4Ψ ) Physics 05:Modern Physics I, Chapter 5 Fall 004 Ahmed H. Hussein
9 5.5. PROBLEM 5.35 (IN THE TEXT BOOK) Problem 5.35 (In the text book) A matter wave packet. (a) Find and sketch the real part of the matter wave pulse shape f(x) for a Gaussian amplitude distribution a(k), where a(k) Ae α (k k ) Note that a(k) is peaked at k and has a width that decreases with increasing α. (Hint: In order to put into the standard form complete the square in k.) f(x) π e αz dz a(k)e ikx dk (b) By comparing the result for the real part of f(x) to the standard form of a Gaussian function with width x, f(x) Ae (x/ x) show that the width of the matter wave pulse is x α. (c) Find the width k of a(k) by writing a(k) in standard Gaussian form and show that x k, independent of α. Solution (a) The spectral distribution function a(k) is given by: a(k) Ae α (k k ) Ae α (k kk +k ) Ae α k e α (k kk ) The matter wave pulse shape f(x) becomes: Physics 05:Modern Physics I, Chapter 5 Fall 004 Ahmed H. Hussein
10 0 CHAPTER 5. MATTER WAVES. SOLUTIONS OF SELECTED PROBLEMS f(x) π π π Ae α k π Ae α k π Ae α k π Ae α k a(k) e ikx dk Ae α k e α (k kk ) e ikx dk e α (k kk ) e ikx dk e α (k kk )+ikx dk e α (k kk ikx/α ) dk e β dk where β α (k kk ikx α ). ( β α k kk ikx ) α ( α [k k k + ix )] α ( α [k k k + ix ) + α α [k + f(x), then becomes: ( k + ix α ( k + ix α )] ( + α k + ix α ) ( k + ix α ) ) ] f(x) π Ae α k e α (k + ix α ) π Ae α k e α (k + ix α ) e α [k+(k + ix α )] dk e α z dz where z [ k + ( k + ix α )] and dz dk. Since, e α z π α Physics 05:Modern Physics I, Chapter 5 Fall 004 Ahmed H. Hussein
11 ikx A α ( k k ) α α ( ) ( ) ( ( α 0 ikx A k k k0 ix 0 ) k f x ) a k e dk e e dk e e dk. π π π Now complete the square in order to get the integral into the standard form + az e where 5.5. PROBLEM 5.35 (IN THE TEXT BOOK) dz: then f(x) becomes: e ( ( ) ) e ( ) e ( ( )) α k k0+ ix α k + α k0+ ix α α k k0+ ix α f(x) A e α k e α (k + ix ) π α π + α A A α k α ( ) ( k α 0+ ix α ) α k ( k0+ ix α 0 ) k e α e α [k +(ik x/α ) (x /4α 4 )] π k A α k e α e α k+ e ik x e x /4α A α α e x 4 e ik0x e z dz π A α /4α e x e z ik x A α /4α + e x (cos k x + i sin k x) ix α z π A x 4α 0 e dz, ( ) α z α α The real part of f(x), Ref(x), is: A x 4α Re f x is Re ( ) Ref(x) A cos α α 0 /4α e x cos k x z k k +. Since part of f ( x ), ( ) f x e e e ( ) dk 0 f x e e. The real f x e k x and is a gaussian envelope multiplying a harmonic wave with wave number k 0. A plot of Re f ( x ) is shown below: and is composed of a Gaussian envelope multiplied by a harmonic wave with a wave number k. A plot of Ref(x) is shown in Figure (5.). ik x Re f( x) α A e x 4α x Comparing α cos kx 0 Figure 5.: A e x 4α to ( x x) (b) Comparing Ae implies x α. A α e x /4α Physics 05:Modern Physics I, Chapter 5 Fall 004 Ahmed H. Hussein
12 CHAPTER 5. MATTER WAVES. SOLUTIONS OF SELECTED PROBLEMS to Ae (x/ x) implies that x α. (c) Given that a(k) is: a(k) Ae α (k k+ ) putting k /α, than a(k) can be written as: a(k) Ae (k k ) /( k) the last equation makes a(k) takes the standard Gaussian form, so we then have: x k α α Physics 05:Modern Physics I, Chapter 5 Fall 004 Ahmed H. Hussein
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