Physics 43 Chapter 4 Homework # Key π sin. A particle in an infinitely deep square well has a wave function given by ( ) for and zero otherwise. Determine the epectation value of. Determine the probability of finding the particle near /, by calculating the probability that the particle lies in the range.49.5. (c) What If? Determine the probability of finding the particle near /4, by calculating the probability that the particle lies in the range.4.6. (d) Argue that the result of part does not contradict the results of parts and (c). = π 4π = sin = cos 4π 4π 4π = + = π sin cos 6.5.5 π 4π Probability = sin = sin π 4.49 Probability ( π π) 4π.49 =. sin.4 sin.96 = 5.6 5 (c).6 4π Probability sin = π 3.99 4.4 (d) In the n =, it is more probable to find the particle either near = or = 3 than at the center, where the 4 4 probability density is zero. Nevertheless, the symmetry of the distribution means that the average position is.. Show that the wave function = Ae i(k ωt) is a solution to the Schrödinger equation where U =.
3. Consider a particle moving in a one-dimensional bo for which the walls are at = / and = /. Write the wave functions and probability densities for n =, n =, and n = 3. Sketch the wave functions and probability densities. Model: A particle is confined in a rigid one-dimensional bo of length. Visualize: Solve: () is zero because it is physically impossible for the particle to be there because the bo is rigid. The potential energy within the region / / is U() = J. The Schrödinger equation in this region is where β = me. ( ) d m = E = ( ) β( ) (c) Two functions () that satisfy the above equation are sinβ and cosβ. A general solution to the Schrödinger equation in this region is () = Asinβ + Bcosβ where A and B are constants to be determined by the boundary conditions and normalization. (d) The wave function must be continuous at all points. = just outside the edges of the bo. Continuity requires that also be zero at the edges. The boundary conditions are ( = /) = and ( = /) =. (e) The two boundary conditions are β β β β ( ) = Asin + Bcos = Asin + Bcos = β β ( ) = Asin + Bcos = These are two simultaneous equations. Unlike the boundary conditions in the particle in a bo problem of Section 4.4, there are two distinct ways to satisfy these equations. The first way is to add the equations. This gives To finish satisfying the boundary conditions, β Bcos = B = () = Asinβ β sin = β = π, 4π, 6π, = nπ with n =,, 3, With this restriction on the values of β, the wave function becomes ( ) Asin( nπ) from part, the energy is ( nπ ) h ( n) En = = n =,, 3, m 8m The second way is to subtract the second equation from the first. This gives β Acos = A = () = Bcosβ =. Using the definition of β
To finish satisfying the boundary conditions, β cos = β = π, 3π, 5π, = (n )π n =,, 3, With this restriction on the values of β, the wave function becomes ( ) Bcos( ( n ) π ) β, the energy is ( n ) π h ( n ) En = = n =,, 3, m 8m Summarizing this information, the allowed energies and the corresponding wave functions are ( ) ( ) n π h Bcos En = ( n ) = E, 9 E, 5 E, 8m = ( n) π h Asin En = ( n) = 4 E, 6 E,36 E, 8m =. Using the definition of where E = h /8m. (f) The results are actually the same as the results for a particle located at. That is, the energy levels are the same and the shapes of the wave functions are the same. This has to be, because neither the particle nor the potential well have changed. All that is different is our choice of coordinate system, and physically meaningful results can t depend on the choice of a coordinate system. The new coordinate system forces us to use both sines and cosines, whereas before we could use just sines, but the shapes of the wave functions in the bo haven t changed. 4 A one-dimensional harmonic oscillator wave function is = Ae b Find b and the total energy E. Is this a ground state or a first ecited state? Problem 4 in Chapter 6 helps students to understand how to draw conclusions from an identity. b d b b = Ae so = Ae b Ae and d b = b 3 b + bae 4bAe 4b e = 6b + 4b Substituting into the Schrödinger equation, me mω 6b + 4b = + For this to be true as an identity, it must be true for all values of. me So we must have both 6b = and 4b mω = Therefore mω b = 3b 3 and E = = ω m (c) The wave function is that of the first ecited state.
5. An electron has a kinetic energy of. ev. The electron is incident upon a rectangular barrier of height. ev and thickness. nm. By what factor would the electron s probability of tunneling through the barrier increase assuming that the electron absorbs all the energy of a photon with wavelength 546 nm (green light)? The original tunneling probability is T = e C where ( mu ( E) ) ( ) 3 9 9. kg.6 J C = = =.448 m 6.66 J s π ( ) hc 4 ev nm The photon energy is hf = = =.7 ev, to make the electron s new kinetic energy λ 546 nm +.7 = 4.7 ev and its decay coefficient inside the barrier π ( 3 9 ( ) ) 9. kg 4.7.6 J C = =.5 5 m 6.66 J s Now the factor of increase in transmission probability is C e C ( C ) = = 9 m.3 m 4.45 e e = e = 85.9 C e 6. An electron having total energy E = 4.5 ev approaches a rectangular energy barrier with U = 5. ev and = 95 pm as. Classically, the electron cannot pass through the barrier because E < U. However, quantum-mechanically the probability of tunneling is not zero. Calculate this probability, which is the transmission coefficient. 3 9 ( )( )( ) 9. 5. 4.5.6 kg m s C = = 3.6 m.55 J s 9 ( )( ) ( ) C 9 = = T e ep 3.6 m 95 m = ep 6.88 3 T =.3 7. A particle is described by the wave function Determine the normalization constant A. What is the probability that the particle will be found between = and = /8 if its position is measured? (c) What is the most probable position in the bo? (d) What is the epectation value in the bo? = becomes
4 4 π π 4π π A cos = A + = = π sin A π 4 4 4 or A = and A =. 4 The probability of finding the particle between and 8 is 8 8 π = A cos = + =.49 4 π 8. What is the probability that an electron will tunnel through a.45nm gap from a metal to a STM probe if the work function is 4.eV? The work function E = 4. ev is the energy barrier U E that an electron must either go over (photoelectric effect) or tunnel through. From Equation 4.4, the electron s penetration distance is.5 J s η = = = = mu 9. kg 4. ev.6 J/eV 3 9 ( E) ( )( ) The tunneling probability through the barrier is w η (.5 nm) (.97 nm) 5 Ptunnel = e = e = 3.4 9.7 m.97 nm 9. Find the epectation value for the for the first two states of a harmonic oscillator.. Find the most probable position for the the first two states of a harmonic oscillator. a a = e =, since the integrand is an odd function of. π 3 4a a = e =, since the integrand is an odd function of. π (c) = ( + ) = + + ( ) ( ) The first two terms are zero, from and. Thus: