Exam (elaborations) TEST BANK FOR Applied Quantum Mechanics By A. F. J. Levi (Solution manual)

Exam (elaborations) TEST BANK FOR Applied Quantum Mechanics By A. F. J. Levi (Solution manual) Applied Quantum Mechanics Chapter 1 Problems and Solutions LAST NAME FIRST NAME Useful constants MKS (SI) Speed of light in free space Planck’s constant Electron charge Electron mass Neutron mass Proton mass Boltzmann constant Permittivity of free space Permeability of free space Speed of light in free space Avagadro’s number Bohr radius Inverse fine-structure constant c 2. ´ 108 m s–1 = h 6.(26 ) 10 –16 = ´ eV s h = 1. (82) ´ 10–34 J s e 1. (63) 10 –19 = ´ C m0 = 9.(72) ´ 10–31 kg mn = 1.(13) ´ 10–27 kg mp = 1.(13) ´ 10–27 kg kB = 1.(24 ) ´ 10–23 J K–1 kB = 8.(15) ´ 10–5 eV K–1 e0 = 8. ´ 10–12 F m–1 m0 = 4p ´ 10–7 H m–1 c = 1 ¤ e0m0 NA 6. (79) 1023 ´ mol –1 = aB 0.(19 ) –10 = ´10 m aB 4pe0h2 m0e2 = ---------------- a–1 = 137.(50) a–1 4pe0hc e2 = ----------------- 2 PROBLEM 1 A metal ball is buried in an ice cube that is in a bucket of water. (a) If the ice cube with the metal ball is initially under water, what happens to the water level when the ice melts? (b) If the ice cube with the metal ball is initially floating in the water, what happens to the water level when the ice melts? (c) Explain how the Earth’s average sea level could have increased by at least 100 m compared to about 20,000 years ago. (d) Estimate the thickness and weight per unit area of the ice that melted in (c). You may wish to use the fact that the density of ice is 920 kg m-3, today the land surface area of the Earth is about 148,300,000 km2 and water area is about 361,800,000 km2. PROBLEM 2 Sketch and find the volume of the largest and smallest convex plug manufactured from a sphere of radius r = 1 cm to fit exactly into a circular hole of radius r = 1 cm, an isosceles triangle with base 2 cm and a height h = 1 cm, and a half circle radius r = 1 cm and base 2 cm. PROBLEM 3 An initially stationary particle mass m1 is on a frictionless table surface and another particle mass m2 is positioned vertically below the edge of the table. The distance from the particle mass m1 to the edge of the table is l. The two particles are connected by a taught, light, inextensible string of length L l. (a) How much time elapses before the particle mass m1 is launched off the edge of the table? (b) What is the subsequent motion of the particles? (c) How is your answer for (a) and (b) modified if the string has spring constant k? PROBLEM 4 The velocity of water waves in shallow water may be approximated as where g is the acceleration due to gravity and h is the depth of the water. Sketch the lowest frequency standing water wave in a 5 m long garden pond that is 0.9 m deep and estimate its frequency. PROBLEM 5 (a) What is the dispersion relation of a wave whose group velocity is half the phase velocity? (b) What is the dispersion relation of a wave whose group velocity is twice the phase velocity? (c) What is the dispersion relation when the group velocity is four times the phase velocity? PROBLEM 6 A stationary ground-based radar uses a continuous electromagnetic wave at 10 GHz frequency to measure the speed of a passing airplane moving at a constant altitude and in a straight line at 1000 km hr-1. What is the maximum beat frequency between the out going and reflected radar beams? Sketch how the beat frequency varies as a function of time. What happens to the beat frequency if the airplane moves in an arc? v = gh Applied quantum mechanics 3 PROBLEM 7 How would Maxwell’s equations be modified if magnetic charge g (magnetic monopoles) were discovered? Derive an expression for conservation of magnetic current and write down a generalized Lorentz force law that includes magnetic charge. Write Maxwell’s equations with magnetic charge in terms of a field . PROBLEM 8 The capacitance of a small metal sphere in air is . A thin dielectric film with relative permittivity uniformly coats the sphere and the capacitance increases to . What is the thickness of the dielectric film and what is the single electron charging energy of the dielectric coated metal sphere? PROBLEM 9 (a) A diatomic molecule has atoms with mass m1 and m2. An isotopic form of the molecule has atoms with mass m'1 and m'2. Find the ratio of vibration oscillation frequency w / w' of the two molecules. (b) What is the ratio of vibrational frequencies for carbon monoxide isotope 12 ( ) and carbon monoxide isotope 13 ( )? PROBLEM 10 (a) Find the frequency of oscillation of the particle of mass m illustrated in the Fig. The particle is only free to move along a line and is attached to a light spring whose other end is fixed at point A located a distance l perpendicular to the line. A force F0 is required to extend the spring to length l. (b) Part (a) describes a new type of child’s swing. If the child weighs 20 kg, the length l = 2.5 m, and the force F0 = 450 N, what is the period of oscillation? G = eE + i mH C0 1.1 10 –18 = ´ F er 1 = 10 2.2 10 –18 ´ F C O 12 16 C O 13 16 Fixed Displacement, -x Mass, m Length, l point A Spring 4 SOLUTIONS Solution 1 (a) The water level decreases. If ice has volume V then net change in volume of water in the bucket is . (b) Again, the mass of the volume of liquid displaced equals the mass of the floating object. Assuming the metal has a density greater than that of water, the water level decreases when the ice melts. (c) If there is just ice floating in the bucket, the water level does not change when the ice melts. This fact combined with the results from part (a) and (b) allows us to conclude that only the ice melting over land contributes to increasing the sea level. (d) Today 71% water, 29% land, ratio is 2.45. The average thickness of ice on land is simply 100 m ´ 2.45 = 245 m. If one assumes half the land area under ice, then average thickness of ice is 490 m. If this is distributed uniformly from thin to thick ice, then one might expect maximum ice thicknesses near 1000 m (i.e. ~ 3300 ft high mountains of ice). Ice weighs one metric tone per cubic meter so the weight per unit area is the thickness in meters multiplied by tones. Solution 2 We are asked to sketch and find the volume of the largest and smallest convex plug manufactured from a sphere of radius r = 1 cm to fit exactly into a circular hole of radius r = 1 cm, an isosceles triangle with base 2 cm and a height h = 1 cm, and a half circle radius r = 1 cm and base 2 cm. The minimum volume is 1.333 cm3 corresponding to a geometry consisting of triangles placed on a circular base as shown in the Fig. To calculate this minimum volume of the plug consider the triangle at position x from the origin has height k, base length 2l, and area kl. DV = V(rice ¤ rwate r – 1 ) 2 cm r = 1 cm h = 1 cm 2r = 2 cm Applied quantum mechanics 5 Volume of the plug is Since r = 1 cm the total volume is exactly 4/3 = 1.333 cm3. To find the maximum volume of the plug manufactured from a sphere of radius r = 1 cm we note that the geometry is a half sphere with two slices cut off as shown in the following Fig. The geometry is found by passing the sphere along the three orthogonal directions x, y, and z and cutting using a circle, a triangle, and a half circle, and a triangle respectively. As will be shown, the volume is 1.608 cm3. This is the maximum convex volume of the plug manufactured from a sphere of radius r = 1 cm. To calculate the volume we first calculate the volume sliced off the half sphere as illustrated in following Fig. x r r r h l l k (r - x) l2 k2 ( r + x) ( r – x) r2 x2 = = = – Vol 2 k2dx 0 r ò 2 ( r2 – x2)dx 0 r ò 2 r2x x3 3 – ---- 0 r 4 3 = = = = - r3 2 cm r = 1 cm h = 1 cm 2r = 2 cm 6 so that area of disk radius l at position x is The volume of the disk is the area multiplied by the disk thickness dx. The total volume is the integral from to . Remembering to multiply by 2 because there are two slices gives total volume sliced off The volume of the plug is just the volume of a half sphere minus . Since r = 1 cm the total volume is 1.608 cm3. If we lift the restriction that the plug is manufactured from a sphere of radius 1 cm, then calculating the maximum convex volume turns out to be a bit complicated as may be seen in the following Fig. The geometry is found by cutting along the three orthogonal directions x, y, and z using a circle, a triangle, and a half circle, and a triangle respectively. The volume is 1.659 cm3. x r r/21/2 r l (r - x) l x r r (r - x) l l2 (r + x) (r – x) r2 x2 = = – Area p r2 x2 = ( – ) x = r ¤ 2 x = r Voloff 2p ( r2 – x2 )dx r ¤ 2 r ò 2p r2x x3 3 – ---- 0 r 2pr 3 1 1 3 è – --ø æ ö 1 2 ------- 1 6 2 è – ----------ø – æ ö è ø = = = æ ö Voloff 2pr3 2 3 - 5 6 2 è – ----------ø = æ ö Voloff Vol 2pr3 3 ----------- 2pr3 2 3 --- 5 6 2 è – ---------ø – æ ö 2pr3 1 3 --- 2 3 --- 5 6 2 è – + ---------ø æ ö 2pr3 5 6 2 ---------- 1 3 è – --ø = = = æ ö 2 cm r = 1 cm h = 1 cm 2r = 2 cm x y z Applied quantum mechanics 7 Solution 3 (a) and (b). Force due to gravity acts on particle mass m2. This force is F = gm2 where g is the acceleration due to gravity. The light string transmits the force to the second particle mass m1 that is free to slide horizontally on the table surface. If x is the vertical position of the particle with mass m2 then and, starting from rest, speed with displacement . Eventually, particle m1 is launched with velocity v1 from the end of the table, at which point gravity causes it to accelerate in the x-direction under its own weight. If the distance of particle mass m1 to the edge of the table is l, then it takes time for m1 to reach the edge. The launch velocity is . The initial angular momentum of the system when particle mass m1 is launched from the edge of the table is Lv1m1m2/(m1+m2). (c) If the string has a spring constant then part of the energy of the system can be stored in the string during its trajectory. For example, this will happen during the initial acceleration phase. Solution 4 We are given wave velocity, . This is a constant because h and g are fixed. Hence, group velocity and phase velocity are the same and . For the lowest frequency standing wave and using g ~ 10, then the frequency of the wave is . Solution 5 (a) . (b) Because the group velocity is twice the phase velocity we have and so . Integrating both sides of this equation allows one to write so that the dispersion relation is where A is a constant. (c) . Solution 6 For a source moving at velocity v, the Doppler shifted frequency f’ is where f0 is the original frequency. The ± sign refers to the source moving towards or away from the detector. Hence, the maximum change in frequency relative to f0 is . In our case, v = 3.6 ´ 108 m s-1, c = 3 ´ 108 m s-1, and f0 = 1010 Hz giving Df = 9.26 kHz. The component of velocity in the direction of the detector is v ´ cos(q) where q is the angle between the ground and the plane flying at height h. A simple expression for the time dependence of the angle q = q(t) is found knowing that the aircraft is moving at a constant velocity (and assuming the airplane flies right over the ground station). If the dist 2 2 d d x gm2 m1 + m2 = ------------------- d t dx gm2t m1 + m2 = ------------------ x gm2t2 2 (m1 + m2 ) = -------------------------- t 2l(m1 + m2) m2g = ---------------------------- v1 gm2 m1 + m2 ------------------- 2l(m1 + m2) m2g = ---------------------------- v = gh w = 2pf = gh ´ k = g h ´ 2p ¤ l l = 2 ´ 5 m f = 10 ´ 0.9 ¤ 10 = 0.3 Hz w Ak0.5 = vg k ¶ = ¶w vp = w ¤ k dk dw 2 w k = --- dw w ------- 2dk k = ----- lnw = 2lnk + lnA = lnAk2 w = Ak2 w Ak4 = f¢ = f0 ¤ (1 ± v ¤ c) Df = f0(1 – 1 ¤ (1 ± v ¤ c) ) = ±f0(v ¤ (c + v) ) 8 tance to the plane (range) is l then (assuming we set t = 0 when the plane is overhead). Then . The Doppler shift as a function of time is Df ´ cos(q(t)). If the plane does not fly overhead it is necessary to introduce another angle f to describe its position. If the plane is flying in an arc the Doppler shift can decrease. For example, it will be zero if the plane is flying in a circle centered on the ground station. Assuming the plane flys at a constant speed, whatever trajectory the plane describes, it cannot have a Doppler shift that is greater than the maximum value we calculated Df for that velocity. The reason why a typical radar antenna at an airport is highly directional and rotates is to find the angular position of the aircraft with respect to the radar detector. Pulsed radar can be used to find the range (distance) to the aircraft. The trajectory of the aircraft can be obtained by comparing sequential position updates. Doppler radar is not required. Solution 7 If magnetic charge g existed Maxwell’s equations would have to include magnetic charge density and magnetic current density as well as the usual electron charge density and electron current density . The new equations would be where and . Conservation of magnetic current is expressed as For a particle carrying both electric charge e and magnetic charge g, the Lorentz force is Reminding ourselves of the SI units used for magnetic intensity H[A m-1], magnetic flux density B[T] where B = mH, electric field strength E[V m-1],the displacement vector field D = eE, permiability m[N A-2] or [H m-1], permittivity e[A2 s2 N-1 m-2] or [F m-1], and where c is the speed of light in free-space. If corresponds to energy density then it makes sense to write since and have dimensions of energy density U[J m-3]. Maxwell’s equations in terms of E and H with magnetic charge can be written l2 = h2 + v2t2 cos (q(t) ) vt ¤ l v t h2 v2t2 ¤ + 1 h2 v2t2 = = = ¤ ¤ + 1 rm Jm re Je Ñ × D = re Ñ × B = rm E ¶B ¶ t Ñ´ = – ------- – Jm H Je ¶D ¶t Ñ´ = + ------- D = eE B = mH Jm = rmv ¶ t ¶rm Ñ Jm 0 + × = F = e(E + v ´ B) + g(B – v ´ E) e0 1 ¤ m0c2 = G*G U = (E × D + B × H) ¤ 2 G e 2 ---E i m 2 = + ---H e E2 m H 2 e 2 ---Ñ × E re 2e = --------- m 2 ---Ñ H rm 2m × = ---------- e 2 - E e 2 – -- m ¶H ¶ t è ------- + Jmø Ñ´ = æ ö Applied quantum mechanics 9 It follows that and which can be rewritten as since The complex field G can be used to simplify the usual four Maxwell equations to just two equations. Solution 8 We are given the information that a small metal sphere has capacitance in air. First, we calculate the radius r0 of the sphere using the formula It follows that Now consider the case when the metal sphere radius r0 with charge Q is coated with a dielectric of relative permittivity to radius r1 and from r1 radius to r2. The voltage drop from r0 through the dielectrics to a much larger metal sphere of radius r2 is found by integrating For this particular problem we now take the limit . Capacitance m 2 --- H m 2 --- e¶E ¶t è ------- + Jeø Ñ´ = æ ö Ñ × G re 2e ---------- i rm 2m = + ---------- Ñ ´ G e 2 --- E i m 2 Ñ´ + ---Ñ´H e 2 – --m ¶H ¶t ------- e 2 – - Jm i m 2 ---e ¶E ¶t ------- i m 2 = = + + --- Je 1 em ----------Ñ ´ G i dt dG Jm 2m – ---------- iJe 2e = + ---------- i dt dG m 2 – --- ¶H ¶t ------- i e 2 --- ¶E ¶t = + ------- C0 = 1.1 ´ 10–18 F C0 = 4pe0r0 r 0 C0 4pe0 ----------- 1.1 10 –18 ´ 4p 8.85 10 –12 ´ ´ = = ----------------------------------------- = 10 nm er 1 er2 V Er r Erdr r2 r1 d – ò r1 r0 – ò Q 4pe0er 1 r2 ---------------------- r Q 4pe0er 2 r2 ---------------------- dr r2 r1 d – ò r1 r0 = = – ò V Q 4pe0 ----------- 1 er1 r0 ---------- 1 er 1 r1 – ---------- 1 er 2 r1 ---------- 1 er2 r 2 è + – ----------ø = æ ö r2 ® ¥ C Q V ---- 4pe0 1 er 1 r0 ---------- 1 er1 r 1 – ---------- 1 er2 r1 è + ---------- – 0ø æ ö ---------------------------------------------------------- 4pe0 1 er1 r0 ---------- 1 r1 ---- er2 er1 – er 1er2 ----------------- è ø ç ÷ æ ö – è ø ç ÷ æ ö = = = ------------------------------------------------ C er1 r0 ---------- C r 1 ---- er2 er1 – er1 er2 ----------------- è ø ç ÷ æ ö – = 4pe0 10 Putting in the numbers. We have , , , , and , so that and we may conclude that the thickness of the dielectric film is . Charging energy . This value is similar to ambient thermal energy . Solution 9 (a) The molecule consists of two particles mass and mass with position and respectively. The center of mass coordinate is and relative position vector is . We assume that the potential depends only on the difference vector and if the origin is the center of mass then , so that and Since, for example, Now, combining center of mass motion and relative motion, the Hamiltonian is the sum of kinetic and potential energy terms where the total kinetic energy is or C er1 r0 ---------- – 4pe0 C r1 ---- er2 er 1 – er1 er 2 ----------------- è ø ç ÷ æ ö = r1 C er2 er1 – er1 er 2 ----------------- è ø ç ÷ æ ö er1 r0 C 4pe0er1 r0 – -------------------------------- er2 er1 – er2 è-----------------ø æ ö r 0 1 er1C0 C è – -----------ø æ ö = = -------------------------- er1 = 10 er 2 = 1 C0 = 1.1 ´ 10–18 F C = 2.2 ´ 10–18 F r0 = 10 nm r1 1 – 10 1 è-------------- ø æ ö r0 1 10 2 è – ----- ø æ ö -------------------- 9 4 = = --- r0 = 22.5 nm r1 r0 –= 12.5 nm DE e2 ¤ 2C 1.6 10 –19 4.4 10 –18 = = ´ ¤ ´ = 36 meV kBT = 25 meV m1 m2 r1 r2 R r r = r2 – r1 m1r1 + m2r2 = 0 r1 m2 (m1 + m2) = -----------------------r r2 m1 (m1 + m2) = -----------------------r r1 –m2 m1 --------- r2 –m2 m1 = = --------- (r1 – r ) r1 1 m2 m1 è + ------ ø æ ö m2 m1 = ------ r r1(m1 + m2) = m2r r1 m2 (m1 + m2) = -----------------------r H T + V 1 2 -m1 R · m2 (m1 + m2) -----------------------r · è – ø æ ö2 1 2 ---m2 R · m1 (m1 + m2) ----------------------- + r · è ø æ ö2 = = + + V( r ) T 1 2 --- m1 m2 + ( )R · 2 m1m2 2 (m1 + m2)2 --------------------------r · 2 m1 2 m2 (m1 + m2) 2 --------------------------r · 2 + + 1 2 --- m1 m2 + ( )R · 2 1 2 --- m1m2 (m1 + m2) -----------------------r · 2 = = + T 1 2 ---MR · 2 1 2 ---mr · 2 = + Applied quantum mechanics 11 where and the reduced mass is Because the potential energy of the two interacting particles depends only on the separation between them, . The frequency of oscillation of the molecule is determined by the potential and given by where k is the spring constant and m is the reduced mass . Since the isotope form of the molecule interacts in the same way, , and the ratio of oscillation frequency is given by We could get the same result quite simply if we assume motion is restricted to one dimension. In that case the force on mass m1 and mass m2 is given by their relative displacement multiplied by the spring constant k. which has solution of the form giving giving a characteristic equation M = m1 + m2 m m1m2 (m1 + m2) = ----------------------- R m1 m2 r V = V( r2 – r1 ) = V( r ) w = k ¤ m 1 ¤ m = 1 ¤ m1 + 1 ¤ m2 k = k¢ w¢ w ----- m1m2(m¢1 + m¢2) m¢1m¢2(m1 + m2) = ---------------------------------------- m1 m2 u v k m1 t2 2 d d u = k(v – u) m2 t2 2 d d v = k(u – v) e –iwt k m1w2 ( – )u – k v = 0 – ku + (k – m2w2 )v = 0 12 and and, as before, since the isotope form of the molecule interacts in the same way, , and the ratio of oscillation frequency is given by (b) To find the ratio of vibrational frequencies we use the result in part (a) and put in the atomic masses to give . As expected, the lighter molecule vibrates at a higher frequency (by about 2%) than the molecule. Solution 10 (a) For small displacement x we assume F0 doesn’t change much so for extension the work done is . Hence, the potential energy of the spring is equal to the force F0 multiplied by the extension . For we have , so that the potential energy is . This is the potential energy of a harmonic oscillator with spring constant . This oscillator has frequency . (b) The frequency of oscillation in radians per second is given by , so the period of oscillation is , i.e., about 2 seconds. k – m1w2 –k –k k m2w2 – = (k – m1w2) (k – m2w2) – k2 = 0 k m1m2 m1 + m2 è------------------ø æ öw2 mw2 = = w k m1 + m2 m1m2 è------------------ø = æ ö = k ¤ m k = k¢ w¢ w ----- m1m2(m¢1 + m¢2) m¢1m¢2(m1 + m2) = ---------------------------------------- w C 12 w C 13 --------- 13 ´ 16 ´ (12 + 16) 12 ´ 16 ´ (13 + 16) = ----------------------------------------------- = 1.02 C O 12 16 C O 13 16 Dl F0Dl Dl x « l Dl l2 x2 + – l x 2 = = ¤ 2l V F0x2 ¤ 2l kx2 = = ¤ 2 k = F0 ¤ l w = k ¤ m = F0 ¤ ml w F0 ¤ ml 450 ¤ 20 ´ 2.5 9 3 rad s –1 = = = = t = 1 ¤ f = 2p ¤ w = 2.09 s Applied quantum mechanics 1 Applied Quantum Mechanics Chapter 2 Problems and Solutions LAST NAME FIRST NAME Useful constants MKS (SI) Speed of light in free space Planck’s constant Electron charge Electron mass Neutron mass Proton mass Boltzmann constant Permittivity of free space Permeability of free space Speed of light in free space Avagadro’s number Bohr radius Inverse fine-structure constant c 2. ´ 108 m s–1 = h 6.(26 ) 10 –16 = ´ eV s h = 1.(82 ) ´ 10–34 J s e 1. (63) 10 –19 = ´ C m0 = 9.(72 ) ´ 10–31 k g mn = 1.(13 ) ´ 10–27 k g mp = 1.(13 ) ´ 10–27 k g kB = 1.(24 ) ´ 10–23 J K–1 kB = 8. (15) ´ 10–5 eV K–1 e0 = 8. ´ 10–12 F m–1 m0 = 4p ´ 10–7 H m–1 c = 1 ¤ e0m0 NA 6.(79 ) 1023 ´ mol –1 = aB 0.(19 ) –10 = ´10 m aB 4pe0h2 m0e2 = ---------------- a–1 = 137.(50) a–1 4pe0hc e2 = ----------------- 2 PROBLEM 1 (a) The Sun has a surface temperature of 5800 K and an average radius 6.96 ´ 108 m. Assuming the mean Sun-Mars distance is 2.28 ´ 1011 m, what is the total radiative power per unit area incident on the upper Mars atmosphere facing the Sun? (b) If the surface temperature of the Sun was 6800 K, by how much would the total radiative power per unit area incident on Mars increase? PROBLEM 2 If a photon of energy 2 eV is reflected from a metal mirror, how much momentum is exchanged? Why can the reflection not be modeled as a collision of the photon with a single electron in the metal? PROBLEM 3 Consider a lithium atom (Li) with two electrons missing. (a) Draw an energy level diagram for the Li++ ion. (b) Derive the expression for the energy (in eV) and wavelength (in nm) of emitted light from transitions between energy levels. (c) Calculate the three longest wavelengths (in nm) for transitions terminating at . (d) If the lithium ion were embedded in a dielectric with relative permittivity er = 10, what would be the expression for the energy (in eV) and wavelength (in nm) of emitted light from transitions between energy levels. PROBLEM 4 (a) There is a full symmetry between the position operator and the momentum operator. They form a conjugate pair. In real-space momentum is a differential operator. Show that in k-space position is a differential operator, , by evaluating expectation value in terms of f(kx) which is the Fourier transform of y(x). (b) The wave function for a particle in real-space is . Usually it is assumed that position x and time t are continuous and smoothly varying. Given that particle energy is quantized such that , show that the energy operator for the wave function is . PROBLEM 5 A simple model of a heterostructure diode predicts that current increases exponentially with increasing forward voltage bias. Under what conditions will this predicted behavior fail? PROBLEM 6 Write down the Hamiltonian operator for (a) a one-dimensional simple harmonic oscillator, (b) a Helium atom, (c) a Hydrogen molecule, (d) a molecule with nn nuclei and ne electrons. n = 2 ih px ¶ ¶ áxˆ ñ dxy*(x )xy(x) –¥ ¥ = ò y(x, t) E = hw y(x, t) ih ¶t ¶ Applied quantum mechanics 3 PROBLEM 7 Calculate the classical velocity of the electron in the n-th orbit of a Li++ ion. If this electron is described as a wave packet and its position is known to an accuracy of Dx = 1 pm, calculate the characteristic time DtDx for the width of the wave packet to double. Compare DtDx with the time to complete one classical orbit. PROBLEM 8 What is the Bohr radius for an electron with effective electron mass in a medium with low-frequency relative permittivity corresponding to the conduction band properties of single crystal InAs? PROBLEM 9 Because electromagnetic radiation possesses momentum it can exert a force. If completley absorbed by matter, the absorbed electromagnetic radiation energy per unit time per unit area is a pressure called radiation pressure. (a) If the maximum radiative power per unit area incident on the upper Earth atmosphere facing the Sun is 5.5 kW m-2, what is the corresponding radiation pressure? (b) Estimate the photon flux needed to create the pressure in (a). (c) Compare the result in (a) with the pressure due to one atmosphere. m* = 0.021 ´ m0 er0 = 14.55 4 SOLUTIONS Solution 1 (a) 2.4 kW m-2. (b) 4.5 kW m-2. Solution 2 Assume the photon of energy E = 2 eV is incident from free-space at angle q normal to a flat metal mirror. The magnitude of photon momentum is where frequency w is given by the quanta of photon energy . The momentum change upon reflection is which, for , gives a value . The reason why this reflection can not be modeled as a collision of the photon with a single electron in the metal is that in such a situation it is not possible to conserve both energy and momentum and have the photon maintain its wavelength. Collision with an electron would take away energy from the photon and cause a change in photon wavelength. Since, by definition, reflected light has the same wavelength as the incident light, photon energy cannot change. Another way to express this is that the difference in dispersion relation for free electrons and photons is such that light cannot couple directly to a free electron. Solution 3 For Li++ Z = 3 and so . Emission wavelength is , giving , , . For we have , so divide energy differences by Solution 4 (a) The expectation value of the position operator is pph = hk = hw ¤ c E = hw Dp = (2hw ¤ c) cosq q = 0 2E ¤ c 2 2 1.6 10 –19 3 108 ´ ´ ´ ¤ ´ 2.1 10 –27 kg m s –1 = = ´ En RyZ2 n2 = ----- En 122.4 n2 = ------------- eV l 2phc En2 En1 – -------------------- –10.1 1 n2 ----- 1 n1 – ----- = = ---------------- nm l3, 2 = 72.7 nm l4, 2 = 53.8 nm l5, 2 = 48.1 nm er = 10 En2 En1 – –m0Z2e4 (4pe0er) 2h2 ---------------------------- 1 n2 ----- 1 n1 è – ----ø = æ ö er 2 = 100 áxˆ ñ dxy* (x )xy(x) –¥ ¥ ò 1 2p ------ dx dk ¢f* (k¢)e –i k¢xx d kf(k )eikx –¥ ¥ ò –¥ ¥ ò –¥ ¥ = = ò áxˆ ñ 1 2p ------ dx d k¢f* (k ¢)e –ik ¢xx dkf(k) ¶k ¶ eikx ix è -------ø æ ö –¥ ¥ ò –¥ ¥ ò –¥ ¥ = ò Applied quantum mechanics 5 Hence, in k-space position is a differential operator, . (b) The expectation value of the energy operator is Hence, in t-space energy is a differential operator, . Solution 5 Series resistance. Space charging at current density . For bipolar devices with direct a band gap can have stimulated emission. Solution 6 We are asked to write down the Hamiltonian operator for (a) a one-dimensional simple harmonic oscillator, (b) a Helium atom, (c) a Hydrogen molecule, (d) a molecule with nn nuclei and ne electrons. (a) The Hamiltonian for a particle mass m restricted to harmonic oscillatory motion of frequency w in the x-direction is (b) and (c) may be found from the general solution (d) for a molecule with nn nuclei and ne electrons in which the Hamiltonian is áxˆ ñ 1 2p ------ d x d k¢f*(k¢ )e–ik¢ xx f(k )eikx ix ------- –¥ ¥ dk ¶ k ¶ f(k ) è ø æ öei kx ix ------- –¥ ¥ è – ò ø æ ö –¥ ¥ ò –¥ ¥ = ò áxˆ ñ –1 2pi -------- d x dk ¢f* (k¢ )e –ik¢ x dk ¶k ¶ f(k ) è ø æ öei kx –¥ ¥ ò –¥ ¥ ò –¥ ¥ ò i 2p ------ dxei(k – k¢)x dk –¥ ¥ ò d k¢f * (k ¢) ¶ k ¶ f(k ) –¥ ¥ ò –¥ ¥ = = ò á xˆñ i d k –¥ ¥ ò dk ¢f* (k¢ ) ¶k ¶ f(k )d(k – k ¢) –¥ ¥ ò i d kf* (k¢ ) ¶k ¶ f(k ) –¥ ¥ ò ih d kf* (k¢ ) ¶p ¶ f(k ) –¥ ¥ = = = ò ih ¶p ¶ Eˆ á ñ dwy* (w)hwy(w) –¥ ¥ ò 1 2p ------ dw d t¢f * (t¢)e –iwt ¢hw d tf( t)eiwt –¥ ¥ ò –¥ ¥ ò –¥ ¥ = = ò Eˆ á ñ 1 2p ------ dw d t¢f * (t¢)e –iwt ¢hw d tf( t) ¶t ¶ eiwt¢ iw è--------ø æ ö –¥ ¥ ò –¥ ¥ ò –¥ ¥ = ò Eˆ á ñ 1 2p ------ dw dt¢f*( t¢)e –iwt ¢hw f( t)eiwt iw ------- –¥ ¥ d t ¶t ¶ f( t) è ø æ öeiwt iw ------- –¥ ¥ è – ò ø æ ö –¥ ¥ ò –¥ ¥ = ò Eˆ á ñ –h 2pi -------- dw dt¢f* ( t¢)e –iwt ¢ dt ¶t ¶ f( t) è ø æ öeiwt –¥ ¥ ò –¥ ¥ ò –¥ ¥ ò ih 2p ------ dweiw(t – t ¢) d t –¥ ¥ ò d t¢f * (t¢) ¶t ¶ f( t) –¥ ¥ ò –¥ ¥ = = ò Eˆ á ñ ih dt –¥ ¥ ò d t¢f* ( t¢) ¶t ¶ f( t)d(t – t¢ ) –¥ ¥ ò ih d tf* ( t¢) ¶t ¶ f( t) –¥ ¥ ò ih d tf* ( t¢) ¶t ¶ f( t) –¥ ¥ = = = ò ih t ¶ ¶ j nev H T + V h2 – 2m -------- x2 2 d d mw2 2 ---------- x2 = = + H h2 – 2Mj --------- Ñ2 h2 – 2m0 ---------Ñ2 + V( r ) i = 1 i = ne + å j = 1 j = nn = å 6 Here, the first term is the contribution to the kinetic energy from the nn nuclei of mass Mj, the second term is the kinetic energy contribution from the ne electrons of mass m0, and V(r) is the potential energy given by The first term in the potential is the electron-electron coulomb repulsion between the electrons i and i'. The second term is the nucleus-nucleus coulomb repulsion between the nuclei j and j' with charges Zj and Zj' respectively. The third term is the coulomb attraction between electron i and nucleus j. Solution 7 For n = 1, orbit time ~ 10-17 s, dispersion time ~ 10-20 s. Solution 8 Solution 9 (a) The maximum radiative power per unit area incident on the upper Earth atmosphere facing the Sun is Stotal = 5.5 kW m-2. The radiation pressure us just . (b) From Exercise 1 Chapter 2, so the number of photons per second per unit area is . (c) The radiation pressure in (a) is a small value compared to one atmosphere which is about . Never-the-less, electromagnetic radiation from the Sun can be absorbed, exert a force, and change the direction of small particles in space. For example, it is responsible for continuously sweeping dust particles out of the solar system. As another example, uncharged dust particles from a comet can form a dust tail whose direction and shape is determined by electromagnetic radiation pressure from the Sun. Interestingly, comets can also shed charged dust particles to form what is called a gas tail. The gas tail is swept along by a stream of charged particles and magnetic field lines emitted by the Sun that is called the solar wind. Often a comet will have two separate tails because the solar wind often does not point radially outward from the Sun. V(r ) e2 4pe0ri i¢ ------------------ ZjZj ¢e2 4pe0rjj ¢ ------------------ Zje2 4pe0ri j ---------------- i j , å – j j ¢ å + i i¢ å = aB * 4pe0erh2 m0meffe2 --------------------- aB er me ff = = -------- = 0.529 ´ 10–10 ´ 693 = 36.7 nm Stotal ¤ c = 1.8 ´ 10–5 N m–2 hwaverage = 1.92 eV Stotal ¤ ehwaverag e = 1.8 ´ 1022 s –1 m–2 105 N m–2 Applied quantum mechanics 1 Applied Quantum Mechanics Chapter 3 Problems and Solutions LAST NAME FIRST NAME Useful constants MKS (SI) Speed of light in free space Planck’s constant Electron charge Electron mass Neutron mass Proton mass Boltzmann constant Permittivity of free space Permeability of free space Speed of light in free space Avagadro’s number Bohr radius Inverse fine-structure constant c 2. ´ 108 m s–1 = h 6.(26 ) 10 –16 = ´ eV s h = 1.(82 ) ´ 10–34 J s e 1. (63) 10 –19 = ´ C m0 = 9.(72 ) ´ 10–31 k g mn = 1.(13 ) ´ 10–27 k g mp = 1.(13 ) ´ 10–27 k g kB = 1.(24 ) ´ 10–23 J K–1 kB = 8. (15) ´ 10–5 eV K–1 e0 = 8. ´ 10–12 F m–1 m0 = 4p ´ 10–7 H m–1 c = 1 ¤ e0m0 NA 6.(79 ) 1023 ´ mol –1 = aB 0.(19 ) –10 = ´10 m aB 4pe0h2 m0e2 = ---------------- a–1 = 137.(50) a–1 4pe0hc e2 = ----------------- 2 PROBLEM 1 Prove that particle flux (current) is zero if the one-dimensional exponential decaying wave function in tunnel barrier of energy V0 and finite thickness L is , where k is a real positive number and particle energy . PROBLEM 2 (a) Use a Taylor expansion to show that the second derivative of a wavefunction sampled at positions , where j is an integer and h0 is a small fixed increment in distance x, may be approximated as (b) By keeping additional terms in the expansion, show that