Modern Physics for Scientists and Engineers Instructors Solutions. Joseph N. Burchett

Transcription

1 Modern Physics for Scientists and Engineers Instructors Solutions Joseph N. Burchett January 1, 011

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3 Introduction - Solutions 1. We are given a spherical distribution of charge and asked to calculate the potential energy of a point charge due to this distribution. There are two situations to consider which deliver quite different results. The problem does not specify whether the point charge of interest is inside or outside the charge distribution so we must take both possibilities into consideration. Before we get started, let s set up an appropriate description of our charge distribution so that we may proceed with each of the above situations. It is most useful to express the charge distribution in terms of a charge density ρ: ρ = Q 4 3 πr3 This is the charge per unit volume as found by merely dividing the total charge Q by the total volume of the sphere. We now handle the outside the sphere configuration. All of the charge inside the sphere may be considered as though it is concentrated at a single point. We may then use the Coulomb s Law result for the force acting on the point charge q: F = 1 Qq 4πɛ 0 r We integrate to find the potential energy: V = Qq 4πɛ 0 r 1 Qq 1 dx = r 4πɛ 0 r This result is also identical to the one obtained in the text in the case of the two-point charge configuration. We see that this should in fact be the case since we started with the same quantity of force. Now, to handle the inside the sphere situation, we shall use the above stated charge density. We easily see that the distance r from the center of the 3

4 4 distribution of charge to the point charge is less than the radius R. Imagine now a sphere of radius r that is inside the larger spherical charge distribution but centered at the same point. Only the charge within this inner sphere will act on the point charge. We may now use our charge density to find the charge contained in the smaller sphere: Q in = 4 3 πr 3 ρ The force on a charge q due to the inner sphere is then given by Coulomb s Law: F = q 4πr 3 ρ = qρr 4πɛ 0 3r 3ɛ 0 Before we integrate to find the potential energy, we must make an important distinction from the outside the sphere procedure above. We were able to integrate directly from infinity to the location of our point charge because the amount of source charge taken into account did not change (Q). However, as we enter the sphere, the amount of source charge begins to decrease until we reach our point at distance r from the center. We split the integration into two parts corresponding to both regions to find the potential energy: V = [ R Qq 1 4πɛ 0 r dr + r R qρr 3ɛ 0 dr ] = Qq 1 4πɛ 0 R ρq (r R ) 6ɛ 0 Note that the first integral above was identical to the one for the outside configuration, just evaluated at r = R. Now, we just substitute our charge density ρ and simplify: V = Qq 4πɛ 0 1 R Qqr 8πɛ 0 R 3 + = 3Qq 8πɛ 0 R Qqr 8πɛ 0 R 3 Qq 8πɛ 0 R. First, convert to SI (1 ev = J) then calculate speed from the definition of kinetic energy: KE = 1 mv

5 J = 1 ( kg)v v = m s Multiply by the mass of the electron (from Appendix A) to find the momentum: p = ( kg) ( m s = kg m s 3. Looking at the definition of kinetic energy: KE = 1 mv Let s solve for velocity: Let KE = 4KE. v = (KE) m v = (KE ) (4KE) (KE) m = m = m = v So, the speed will increase by a factor of if the kinetic energy is increased by a factor of 4. By the definition of momentum: p = mv m(v) = p We should also expect the momentum of the particle above to increase by a factor of. In fact, as long as the mass does not change, the momentum should increase or decrease by the same factor as the changing speed. 4. We see that the given function is identidcal in form to that of I.19, comparing the two: ψ(x, t) = Ae 10 i(kx ωt) = Ae 10 (αx+βt)

6 6 Therefore, we may relate α and β to their physical parameters k and ω by: k = α and ω = β. FRom (I.8), k = π/λ and: λ = π α 5. The frequency and wavelength of light are related by equation I.3: f = c λ The constant c refers to the speed of light which we know to be m/s. Taking care to convert our wavelength, which is given in nanometers, to meters: f = m s m = Hz 6. We may use (I.6) which relates the wavelength and frequency via the velocity of the wave (the speed of light): λf = v λ( s 1 ) = m/s λ = m 7. Equation I.5 relates the photon energy to the wavelength: E = hc λ Substituting the value hc = 140 ev nm: E = 140 ev nm 500nm =.48 ev 8. I.5 relates the wavelength and energy: E = hc λ

7 7 λ = hc E = 140 ev nm 5.4 ev = 9.6 nm 9. The energy of a quantum of light (the photon) must be equal to the difference in energies of the two states: E photon = E E 1 According to equation I-5, the energy also obeys the following: E photon = hc λ Equating the two and solving for λ, we get: λ = hc E E Once again, we use equation I.5, however to ensure the correct units for the result, we convert 1Å =.1 nm. E = 140 ev nm.1 nm = ev

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9 1 The Wave-Particle Duality - Solutions 1. The energy of photons in terms of the wavelength of light is given by eq Substituting the given wavelength into eq. 1.5: E photon = hc λ = 140 ev nm 00 nm = 6. ev. The 100 Watt beam of light will carry 100 J of energy every second. We then want to find the number of 00 nm photons required to deliver 100 J of energy. The energy per photon in electron volts is given by (1.5): E = Now, convert the energy to Joules: 140 ev nm 00 nm E = 6. ev J 1 ev Then, find the number of photons: = 6. ev = J 100 J J/photon = photons 9

10 10 1. THE WAVE-PARTICLE DUALITY - SOLUTIONS 3. Considering that we are given the power of the laser in units of milliwatts, the power maybe expressed as: 1 mw = J/s. This gives us a strong hint how to proceed. We are hoping to find the rate of emission of photons, so that if we can find the energy of a single photon, we can use the power as stated above to calculate the number of photons. The energy of a single photon can be calculated as in Problem 1 by substituting the wavelength of the light into equation 1.5: E photon = hc λ We now convert to SI units: ev Now, using the given laser power: = 140 ev nm 63.8 nm = ev 1 J ev = J Rate of emission = J s photon power energy per photon = J = photons s 4. The maximum kinetic energy of emitted photoelectrons may be calculated via (1.6) and work functions of the metals may be found in table 1.1: a)na: (KE) max = hc λ W 140 ev nm =.8 ev 00 nm = 3.9 ev 140 ev nm b)al: (KE) max = 4.08 ev 00 nm =.1 ev 140 ev nm c)ag: (KE) max = 4.73 ev 00 nm = 1.47 ev

11 11 5. This problem concerns the photoelectric effect. Given the work function of the material and the emitted electron kinetic energy, we wish to calculate the wavelength of the light incident to the surface. Equation 1.6 provides the following: (KE) max = hc λ W where W is the work function of the material. The hc/λ term describes the energy supplied by the incoming photons. By viewing the work function as an energy threshold for producing photoelectric current, we see that the amount by which the photon energy (hc/λ) exceeds the work function is the resultant maximum kinetic energy of the emitted electrons. We may thus write: Using hc = 140 ev nm, λ = hc λ = (KE) max + W =.3 ev ev = 3. ev 140 ev nm 3. ev = nm 6. Here, we are given the stopping potential of a photoelectric experiment and wish to determine the work function of the metal. Since 0.7 ev is the necessary potential energy to cease the flow of electrons, the maximum kinetic energy of the electrons being emitted must equal 0.7 ev. Our problem is then reduced to solving eq. 1.6 for the work function: W = hc λ (KE) max = 140 ev nm 460 nm 0.7 ev = 1.98 ev 7. Given that the work function from problem 6 is 1.98 ev, we find the maximum kinetic energy of the photoelectrons by (1.6): (KE) max : 140 ev nm 40 nm 1.98 ev = 3.19 ev

12 1 1. THE WAVE-PARTICLE DUALITY - SOLUTIONS Therefore a stopping potential of 3.19 ev must be applied as a stopping potential to prohibit the photoelectrons from reaching the anode. This corresponds to the increase: 3.19 ev 0.7 ev =.47 ev 8. Section gives the equation for just this situation: W = hc λ 0 = 3.44 ev 9. Given that the work function for sodium is.8 ev, we find the maximum kinetic energy of the emitted electrons by (1.6): hf = J s s 1 = J = 4.97 ev (KE) max = hf.8 ev = 4.97 ev.8 ev =.69 ev 10. Equation 1.7 is the expression for transitions where m =. Substituting into (1.8), ( 1 1 λ = R 4 1 ) n and thus: Taking the reciprocal: ( 1 1 λ = cm ) n λ = ( n ) 1

13 13 We may then rewrite, And we then get (1.7): n = n 4 4n The equations are equivalent. n λ n = n cm 11. For this atomic transition, the energy of the emitted photon must equal the difference in energy of the two states of hydrogen (n = and n = 5). Equation 1. gives us those energies, thus: From eq. 1.1: 13.6 ev 13.6 ev E photon = E 5 E = ( ) =.86 ev 5 λ = hc E = 140 ev nm.86 ev = 434. nm 1. We may use (1.7) to find the wavelength of the photons, then use (1.1) to find their corresponding energies. 4 λ = cm = cm = nm E = hc λ =.55 ev 13. Using 1.7: λ = cm = cm

14 14 1. THE WAVE-PARTICLE DUALITY - SOLUTIONS 14. From figure 1.6, we see that the energy of an n= electron is -3.4 ev. Therefore, 3.4 ev is the required energy to just ionize the atom. Now, use (1.1) to find the wavelength: λ = hc E 140 ev nm = 3.4 ev = nm 15. We must first find the energy of the UV photons, which may be done via eq. 1.5: E photon = hc 140 ev nm = = 7.6 ev λ 45 nm If electrons are to be emitted, we need to overcome the energy that is keeping them in a bound state. For hydrogen atoms, equation 1. gives us that energy. We need only supply the n quantum number for the state of the atom (for ground state, n = 1): 13.6 ev E = n 13.6 ev = = 13.6 ev 1 The kinetic energy of these emitted electrons should then be equal to the difference between the energy provided by the incident light and the ground state electron energy: KE = 7.6 ev 13.6 ev = 14.0 ev For the velocity calculation, it will be useful to convert from ev to J: 14.0 ev J 1 ev = J (1.1) We can then find the velocity using the definition of kinetic energy and the mass of the electron: KE = 1 (KE) mv v = m = J kg =.1 m 106 (1.) s

15 16. We may use (1.8) where each transition will have its own values of m and n: ( 1 1 a.) m=1, n= : λ = cm ) b.) m=, n=5 : c.) m=3, n=5 : λ = cm ( 1 1 λ = cm 1 1 ) 5 λ = cm ( 1 1 λ = cm ) 5 λ = cm It should be first noted that the wavelength of the emitted light increases as the photon energy decreases. Thus, the longest acceptable wavelength will correspond to a transition to the nearest excited state. Transitions to higher-lying states will require more energy and thus correspond to a shorter wavelength. We then examine the energy of the transition from the n = 3 state of the hydrogen atom. Equation 1. gives: 13.6 ev E 3 = n 13.6 ev = = 1.51 ev 3 The next available state is the state corresponding to n=4 whose energy may be found by eq. 1. again: 13.6 ev E 4 = n 13.6 ev = = 0.85 ev 4 So, in order for the absorption to take place, the incident photons must contain at least as much energy as the difference in these energies ( 0.85 ev ( 1.51 ev ) =.66 ev ). We may now solve eq. 1.5 for the wavelength: λ = hc E = 140 ev nm.66 ev = 1878 nm Thus, any light of wavelength greater than 1878 nm would not possess sufficient energy to be absorbed by the hydrogen atom.

16 16 1. THE WAVE-PARTICLE DUALITY - SOLUTIONS 18. Using (1.6), p = h λ = J s m = N s The kinetic energy may then be calculated: KE = p m e = J Convert to ev: J = ev 19. Using a procedure such as that of Example 1.5, we can relate the kinetic energy to the de Broglie wavelength by the following equation: λ = h m(ke) We need only convert the kinetic energy from ev to J and substitute the appropriate masses. KE = 0 ev J 1 ev = J Electron : λ = P roton : λ = J s ( kg)( J) = m J s ( kg)( J) = m An α-particle is the most commonly occurring isotope of the helium atom consisting of two protons and two neutrons. We can thus approximate its mass: m = ( ) + ( ) = kg. α particle : λ = J s ( kg)( J) = m

17 17 0. From (1.1), we may find the corresponding wavelength for 40 kev X-rays: λ = hc E = 140 ev nm ev =.031 nm We then find the momentum possessed by an electron with the same wavelength from (1.5): p = h λ = J s m = kg m s The kinetic energy of the electron may be calculated as follows: E = p = kg m s m e ( kg = J To find the necessary potential for such electrons, we merely convert this energy quantity to ev: J 1 ev J = 1566 ev Therefore, 1566 V would be needed as an accelerating potential difference. 1. We may write the kinetic energy as KE = p /m, the momentum as p = k (1.7), and k = π/λ. Substituting k into (1.7): Substituting our given wavelength: p = π λ = h λ p = J s m = kg m s

18 18 1. THE WAVE-PARTICLE DUALITY - SOLUTIONS This is now simply a matter of using the mass of each particle in our first equation above: Electron : KE = ev Proton : KE = 8. ev Neutron : KE = 8. ev. Visible light has corresponding wavelength between about 400 nm and 700 nm. Therefore, from 1.5: p (400 nm) = J s m = kg m s p (700 nm) = J s m = kg m s We then divide each momentum by the electron mass to obtain the speeds: λ = 400 nm : v = 1815 m s λ = 700 nm : v = 1035 m s 3. We first convert to SI and then use E = p /m to find the momentum: 40, 000 ev = J p = m(ke) = ( kg)( J = kg m s

19 Since k = π/λ, the proton must have the same value of k if it is to have the same wavelength. From (1.7), p = k where k = π/λ. Therefore, a proton having the same de Broglie wavelength should also have the same momentum. We now use the mass of the proton to find its kinetic energy: KE = ( kg m s ) ( kg) = J = 1.8 ev The Davisson-Germer experiment measured the scattering of electrons by a crystal during which the interference patterns characteristic of light were shown to also occur with particles. We can solve Bragg s Law (eq. 1.4) for the inter-atom spacing of the crystal in terms of the de Broglie wavelength and the scattering angle: d sin(θ) = nλ d = nλ sin(θ) We may use the relation derived in example 1.5 to find the de Broglie wavelength from kinetic energy after first converting our kinetic energy to SI units: KE = 54 ev 1 ev λ = h m(ke) = m ( kg)( J) = m We let n=1 and solve Bragg s Law for d, the spacing in the crystal: d = λ sin(θ) = m sin(50 ) = m

20 0 1. THE WAVE-PARTICLE DUALITY - SOLUTIONS

21 The Schrödinger Wave Equation - Solutions 1. Equation.17 gives the energies for the particles in an infinite potential well: E = n h ml Part (a) asks for the size of the region, which we can find by solving eq..17 for L: n h L = 8mE We need only supply some information such as the energy (1.0 ev as given), the mass of the electron ( kg from Appendix A), and the appropriate value of n. The lowest energy will be found when the electron is at the ground state (n=1). We must also convert 1.0 ev to J. Now we are ready: L = 1 ( J s) 8( kg)( J) = m While the derivation of eq..17 is correct, let s do a little dimensional analysis to better see through our length calculation above. The quantum number n is dimensionless as is of course the constant 8, so let s examine the operations of units of the rest of the quantities: n h 8me J s kg J = J s kg 1

22 . THE SCHRÖDINGER WAVE EQUATION - SOLUTIONS Now substitute 1 J = 1 kg m s s : J s kg = kg m s s kg = m = m It should be emphasized that the m here is the abbreviation for the unit of length meters, not the quantity of mass. Part (b) can be approached in a couple of different ways. The essential quantity we must obtain now is the energy of the first level above ground state (also called the first excited state). The energy required to excite the electron must be: E = E n E n1 = E E 1 The quantum numbers n= and n=1 correspond to the first excited state and ground state respectively. So, for E from equation.17: E = h 8mL = ( J s) ( kg)( m) = ev J J = 4.00 ev Here we used the result of the size of the well from part (a). The other method would involve solving eq..17 for everything except n and using our given quantities as follows: Now for our desired result: E = n h ml h 8mL = E n h 8mL = (1.0 ev )(1 ) = 1.0 ev E n = n (1.0 ev ) E = 4.0 ev E = 4.0eV 1.0eV = 3.0eV

23 3. Equation (.0 states that the wavefunction for odd n is: ( nπx ) Ψ(x) = L cos L ( ) 3πx = 10 nm cos where L 10 nm x L Now, evaluate at the given x-values: Ψ(0) =.447 Ψ() =.139 Ψ(4) =.364 Ψ(8) = 0 Ψ(10) = 0 The last two are equal to zero because x > L/ in each of these cases. 3. As with the atomic transitions we dealt with in chapter 1, the emitted photon energy will equal the difference of the energies of the two states corresponding to n = 3 and n =. So our task becomes finding each of these energies via equation.17: E 3 = 3 h 8mL = 9( J s) 8( kg)( m) = J 1 ev J =.034 ev E = h 8mL = 4( J s) 8( kg)( m) = J 1 ev J =.015 ev E =.034 ev.015 ev =.019 ev This value of E will then be equal to the energy of the photon and we may use eq. 1.5 to calculate the wavelength of the light: KE = hc λ λ = hc KE

24 4. THE SCHRÖDINGER WAVE EQUATION - SOLUTIONS λ = 140 ev nm.019 ev = 6563 nm 4. Since E = p /m, the momentum squared may be written as: where the energy is given by.17: p = me p = m n h 8mL = n h 4L The average value of a function is found via the method prescribed on page 30. f(x) = f(x) Ψ(x) dx Using (.0) for Ψ(x), p = n h 4L p = n h 4L L/ L/ L/ L/ ( nπx ) L cos L ( nπx L sin L ) dx dx for odd n for even n By a similar argument to that used in example., the integral over sin (nπx/l) is equal to the integral over cos (nπx/l) and the average value of the square of momentum is the same for the even and odd n: p = n h 4L = n h 8L L/ L/ 1 dx 5. Note the typo, Draw the wave function... should be Find the wave function...

25 We are given the following information: finite well of depth 0.3 ev and thickness L = 10 nm, we are dealing with conduction electrons if GaAS in their lowest state, and these electrons have an effective mass of times the electron mass. From section.3, we see the process of solving the finite square well which culminates in the numerical solution of tanθ = (θ 0 /θ ) 1, for the even solutions, and cotθ = (θ0/θ ) 1 for the odd case. Our variables θ and θ 0 correspond to the following expressions: θ = kl, θ 0 = mv 0L (from eq..30) Notice that the latter of the two is a squared quantity where the former is not, but we ll deal with that a little later on. We are interested in the ground state of the electron, which will correspond to n=1, leading us to an even solution which may be found by equation.9: tanθ = θ 0 θ 1 In the text near the end of section.3, the value of θ 0 is given to be 13.. Our equation becomes: 13. tanθ = 1 θ By using a computer algebra system or calculator to graph the left-hand and right-hand sides, we find the first intersection at θ Thus, we may solve θ = kl/ for k: k = θ L = (1.46) m = m 1 For the equations outside the well, we must solve for κ as well. Let s first solve equation.4 for E: k = me E = k m = ( m 1 ) ( J s) ( kg) E = J 1 ev J =.049 ev 5

26 6. THE SCHRÖDINGER WAVE EQUATION - SOLUTIONS Now for κ (from eq..6): ( kg)(.3 ev.049 ev )( J ev κ = ) = m 1 ( J s) We know have our necessary information for the wave equations: B exp( ( m 1 )x) : x 5 nm Ψ(x) = A cos(( m 1 )x) : 5 nm x 5 nm B exp( ( m 1 )x) : x 5 nm 6. We may use a similar method to that for problem 5 however we must first calculate a new value for θ 0 corresponding to the new well depth. By (.30): θ 0 = mv 0L =.067( kg)(. ev )( J/eV )( m) ( J s) = 8.8 Eq (.9) then becomes: tanθ = 8.8 θ 1 which yields a solution of θ We then find k, E, and κ: k = θ L = m 1 E = k =.045 ev m [ ] 1 m(v0 E) κ = = m 1 The wave equations are then: B exp( ( m 1 )x) Ψ(x) = A cos(( m 1 )x) B exp( ( m 1 )x) : x 5 nm : 5 nm x 5 nm : x 5 nm

27 7 8. The methodology of example.1 holds until we apply the boundary conditions which are now: Ψ(0) = 0 Ψ(L) = 0 First examining the even solutions, we apply the boundary conditions: Ψ(0) = A cos 0 = 0 Ψ(L) = A cos kl = 0 Upon examining Ψ(0), we reject these solutions. Now turning to the odd solutions, Ψ(x) = B sin(kx), we again apply our boundary conditions: Ψ(0) = A sin 0 = 0 Ψ(L) = A sin kl = 0 We see that the first of these is automatically satisfied while second is satisfied provided that kl = nπ. Therefore, ( nπx ) Ψ(x) = B sin where 0 x L L The energy levels are found from (.1) where k = nπ/l E = n π ml Comparing these with the results of example.1, we see that the wavefunctions are all odd while the energy levels remain the same. 9. Outside the finite potential well, V = V 0, thus our time-independent Schrodinger equation is: If we multiply both sides by m/ : d ψ m dx + V 0ψ = Eψ d ψ dx mv 0 ψ = me ψ

28 8. THE SCHRÖDINGER WAVE EQUATION - SOLUTIONS Moving me/ to the LHS: We make the following substitution: d ψ dx + m(e V 0) ψ = 0 ( m(e V0 ) k = Since E is greater than V 0, k is in fact a real number and our Schrodinger equation becomes: d ψ dx + k ψ = 0 And, we may readily confirm by substitution that the general form of the solution of this equation is a linear combination of the functions, A cos(kx) and B sin(kx), and is oscillatory in nature. 10. First evaluating the second derivative, dψ dx = Amωx ) 1 e mωx / d ψ dx = Amω / e mωx + A m ω x Substituting into the left hand side of (.33): m e mωx / d ψ dx + 1 mω x ψ = A ω / e mωx Setting this equal to the right-hand side, ω Ae mωx / = Eψ This is in fact satisfied given E = ω, which is consistent with (.34) when n = The normalized wave function must satisfy the normalization condition (equation.18): ψ(x) dx = 1

29 Therefore, Ae mωx \ Ae mωx \ dx = 1 A = e mωx \ dx = 0 A e mωx \ e mωx \ dx This looks very similar to the integral we are given if we let a = mω/. Then: 1 A = π mω A = = ( mω ) 1 4 π π mω 9 1. b.) The normalization condition is given by (.18). For our given wavefunction, we need only integrate the piece where x > 0: 0 A x e ax dx = 1 To solve the integral, use integration by parts twice and solve for A: 1 A = 1 4a 3 A = a 3 c.) In finding the most probable position of the particle, we maximize the probability as given by (.9). This leads to: Setting the derivative equal to zero: Now, solve for x: ψ(x) = 4a 3 x e ax d dx ψ(x) = 8a 3 xe ax 8a 4 x e ax = 0 x = 1 a

30 30. THE SCHRÖDINGER WAVE EQUATION - SOLUTIONS d.) By (.1), the average value of the position of a particle is given by: < x > = = 0 After three rounds of integrating by parts: 0 x ψ(x) < x >= 3 a 4a 3 x 3 e ax dx 13. a) From 0 x L, V = 0, and our Schrodinger equation becomes: For x L, V = V 0 and: d ψ m dx = Eψ or d ψ dx + me ψ = 0 d ψ m dx + V 0 = Eψ or d ψ dx (V 0 E)m ψ = 0 b) If we let: me m(v0 E) k = and κ = Then, d ψ dx + k ψ = 0 ; 0 x L d ψ dx k ψ = 0 ; x L Similar to the situation of the finite well discussed in the text, the above equations are thus satisfied by the following: ψ(x) = A cos(kx) and ψ(x) = A sin(kx) ; 0 x L ψ(x) = Be κx ; x L Once again, the negative argument in the exponential is to insure our wave function decays as x.

31 c) If the potential is to be infinite at x = 0, then the wave function must go to zero at x = 0. Therefore, we must impose new boundary conditions to insure the continuity of the wave function. Above we had both even and odd solutions for the wave function, so let s examine their behavior at x = 0 given the new conditions: Even : ψ(0) = A cos(k0) = 0 Odd : ψ(0) = A sin(k0) = 0 We see immediately that the fist of the equations is not physically acceptable, so we must reject the even solutions, thus we are left with: ψ(x) = A sin(kx) for 0 x L. We may now proceed with our boundary conditions at x = L: ψ(l) = A sin(kl) = Be κl Imposing the continuity of the first derivative: Ak cos(kl) = Bκe κl 31 Divide eq. by eq. 1: k cot(kl) = κ cot(kl) = κ k We insert our κ as defined in part (b): mv0 cot(kl) = k me k Substitute k into the second term of the RHS: mv0 cot(kl) = k 1 Let θ = kl: mv0 L cot(θ) = 1 θ We can transform this into a form similar to that of eq..31 if: θ 0 = mv 0L

32 3. THE SCHRÖDINGER WAVE EQUATION - SOLUTIONS Our equation to solve then becomes: cot(θ) = θ 0 θ 1 This can be solved graphically for the values of θ for which the LHS and RHS expressions intersect. From here we obtain the corresponding values of k: And solve eq..4 for E: k = θ L E = k m Note that the value for θ 0 which we derived above is slightly different from that of eq..30. This occurs because our potential well for this problem is actually oriented from x = 0 to x = L, as opposed to x = L/ to x = L/ in the example of section.3. This in turn gave us a different argument of the cotangent function and thus necessitated we choose a different θ 0 in order to keep the same clean form of eq..31 to solve numerically. Procedurally, the process we just followed is identical to that of section Equation (.37) states: ψ(x, t) = Ae ikx e iωt. We first evaluate the left-hand side of (.4): d ψ m dx = m ( Ak e ikx e iωt ) [ ] + V (x, t) ψ(x, t) = A k m x m eikx e iωt + V (x, t)e ikx e iωt Now for the right-hand side: ψ(x, t) i t = ωae ikx e iωt If we assume the free particle situation, V=0. And, since k = p and p /m = KE, the left hand side becomes: k m Aeikx e iωt = Eψ(x, t)

33 33 Similarly, for the right-hand side, ω = E and: ωae ikx e iωt = Eψ(x, t) Therefore, the time-dependent Schrodinger equation does in fact hold forthe traveling wave. 15. The infinite potential well represents a situation where the potential energy does not evolve with time. The time-dependent solutions must then satisfy eq..45: Ψ(x, t) = u E (x)e iωt where w = E From eq..0, the normalized spatial wave functions of a particle in the infinte well are given by: ( nπx ) ψ(x) = L sin L The corresponding energies are given by eq..17: E = n h 8mL We may then calculate ω from eqs..39 and.17: Our total wave function is then: Ψ(x, t) = ω = E = n πh 4mL ( nπx ) L sin exp L ( ) in πht 4mL

34 34. THE SCHRÖDINGER WAVE EQUATION - SOLUTIONS

35 3 Operators and Waves - Solutions 1. We may obtain the momentum by multiplying the momentum operator as given in equation 3. ˆp = i d dx times the wave function given in the problem ˆpψ(x) = i d dx Aeiαx = i α Ae iαx ˆpψ(x) = α Ae iαx Our wavefunction is therefore seen to be an eigenfunction of the momentum operator corresponding to the eigenvalue α. Thus, a measurement of the momentum of the electron in this state would yield a value of α. The kinetic energy may be found via the relation: KE = p m The square of the momentum operator that appears here in the numerator will result in the operator begin applied twice, therefore: ˆp = ( i ) d d = dx dx We substitute this expression into the expression of the kinetic energy: KE = ˆp m = m 35 d dx

36 36 3. OPERATORS AND WAVES - SOLUTIONS and operate on the wavefunction: d ψ m dx = m d dx Aeiαx = m Aα e iαx KE = α m. From (3.): ˆp = i d dx ˆp φ 1 (x) = i k cos kx ˆp φ (x) = i k sin kx These wavefunctions are not eigenfunctions of the momentum operator. 3. Our task is to find ψ = Aφ 1 (x) + Bφ (x) such that: ˆpψ = pψ Where p is the eigenvalue of the momentum operator. So, let s try operating on the following wavefunction: ψ = A cos(kx) + B sin(kx) ˆpψ = i dψ dx = i A d d cos(kx) i B dx dx sin(kx) = i k Asin(kx) i k cos(kx) This choice of ψ is not generally an eigenfunction of ˆp, as we can not factor out our original ψ = A cos(kx) + B sin(kx). Let us now try the following: ψ = cos(kx) + i sin(kx) ˆpψ = i k( sin(kx)) i k(cos(kx)) = k cos(kx) + i k sin(kx) = k(cos(kx) + i sin(kx)) Thus, we have found an eigenfunction of the momentum operator which corresponds to the eigenvalue k. Note, that using the Euler formula: e ikx = cos(kx) + i sin(kx)

37 37 Our wavefunction may be written: ψ = e ikx Using this form of the wavefunction and the result of Problem 1, we may thus see that the wavefunction is an eigenfunction of the momentum operator corresponding to the eigenvalue k. 4a. In general, from (3.3): ˆp ψ(x) = pψ(x) i d ψ(x) = p ψ(x) dx b. We may solve the differential equation in (a) by first dividing both sides by i : ψ(x) = Ae i p h x Using (.), p = k and: c. Using the Euler equation: ψ(x) = Ae ikx ψ(x) = Acos kx + iasin kx Since ψ(0) = ψ(a): Once again, from (.): ψ(0) = A ψ(a) = Acos ka cos ka = 1 ka = π k = πn a p = k = π n a 5. Notice that if the energy is less than the minimum value of V (x), the [V (x) E] portion of the RHS is always positive. Therefore, if this equation

38 38 3. OPERATORS AND WAVES - SOLUTIONS holds, the second derivative of the wavefunction will have the same sign as the wavefunction itself, and the function will increase or decrease in value monotonically. 6. The first two excited states should correspond to a ψ 0 and (a ) ψ 0, respectively. Therefore, using (3.) and (3.31): mω a ψ = A 0 xe (mω/ )x The next excited state may be found by applying the a operator again: ( a (a mω ψ) = A 0 x e (mω/ )x ) mω e (mω/ )x 7. Sections and 3.3. provide the initial setup for this problem. We must solve the appropriate Schrodinger equation for each region, then use boundary conditions to find our desired ratios. In the first region, where V = 0, we have the following from section 3.3.1: and given by eq. 3.3: This yields solutions: d ψ 1 dx + k 1 ψ 1 = 0 k 1 = me ψ 1 = Ae ik 1x + Be ik 1x Region produces equations 3.39 through 3.40: d ψ dx k m(v0 E) ψ = 0 where k = With physically acceptable solutions: ψ (x) = De k x Now, we impose continuity of the wave equations at x = 0 and of their first derivatives at x = 0: ψ 1 (0) = ψ (0) A + B = D

39 ψ 1 (0) = ψ (0) ik 1 (A B) = k D Divide both side of the second equation by ik 1 and our two equations become: 39 A + B = D (1) A B = k ik 1 D = i k k 1 D () Here, the fact that 1/i = i was used to bring i out of the denominator and cancel the negative. If we add these two equations, we solve for A in terms of D: A = D(1 + i k k 1 ) A = 1 D(1 + ik k 1 ) (3) If we subtract (1) from (), we can solve for B in terms of D: B = D(1 i k k 1 ) From here, we divide each (3) and (4) by D: B = 1 D(1 ik k 1 ) (4) A D = 1 (1 + ik k 1 ) B D = 1 (1 ik k 1 ) The reflection coefficient R is found by Using our ratios above: R = B v 1 A = B v 1 A R = ( B D ) ( B ) 1 D ( A D ) ( A ) = (1 i k 4 k 1 )(1 + i k 1 D (1 + i k 4 k 1 )(1 i k k 1 ) = 1 k 1 )

40 40 3. OPERATORS AND WAVES - SOLUTIONS 8. Since region extends from x = 0 to infinity, the probability that the particle penetrates into this region is: P = 0 ψ(x) dx Where, from (3.40), ψ (x) = De k x and: P = 0 = D k D e k x dx Substituting (3.39): P = D ( m(v0 E) ) 1 9. Following figure 3.6, we are given the general solutions for each region: ψ 1 (x) = Ae ik1x + Be ik1x, x 0 ψ (x) = Ce kx + De kx, 0 x L ψ 3 (x) = Ee ik1x, x 0 We impose the boundary conditions at x = 0 and x = L: ψ 1 (0) = ψ (0) ψ 1 (0) = ψ (0) ψ (L) = ψ 3 (L) ψ (L) = ψ 3 (L) The first two yield the following system of equations: A + B = C + D ik 1 (A B) = k (C D)

41 We divide the second of these by ik 1 and add the result to the first to obtain an expression for A in terms of C and D: 41 A = C + D i k k 1 (C D) (1) Now, let s turn to the interface between regions and 3. conditions produce the following system: The boundary Ce k L + De k L = Ee k 1L k Ce k L k De k L = ik 1 Ee k 1L Dividing the second of these equations by k reduces to the system to a more manageable one: Ce k L + De k L = Ee k 1L Ce k L + De k L = i k 1 k Ee k 1L Add the two equations and we can solve for C in terms of E: Ce k L = Ee k 1L + i k 1 k Ee k 1L C = E ( 1 + i k ) 1 e (ik 1 k )L k Subtract those same equations to solve for D in terms of E: De k L = Ee k 1L i k 1 k Ee k 1L D = E Insert () and (3) into (1): A = E ( 1 + i k ) 1 e (ik 1 k )L + E k i k [( E 1 + i k 1 k 1 k ( 1 i k ) 1 e (ik 1+k )L k ) e (ik 1 k )L + ( 1 i k ) 1 e (ik 1+k )L k ( 1 i k 1 k ) ] e (ik 1+k )L () (3)

42 4 3. OPERATORS AND WAVES - SOLUTIONS Divide both sides by A and simplify: 4 = E [( 1 + i k ) ( 1 i k )] 1 1 e ik 1L k L + E [( A k k A = E [( ) ( + i k 1 + k e ik 1L k L + + i k k1 A k 1 k k 1 k Solve for E/A: 1 i k 1 k ) e ik 1L+k L ) + ] [( ) ( ) ] E A = 4 + i k 1 + k e ik 1L k L + + i k k 1 1 e ik 1L+k L k 1 k k 1 k ( i k )] e ik 1L+k L k 11. The Heisenberg uncertainty principle states: Therefore we solve for E: Now, converting to electron volts: E = J E t E = t J s = ( s) = J 1 ev J = ev 1. By the normalization condition: Using the boundary of the problem: 1 0 ψ(x) dx = 1 B e x dx = 1

43 [ ] 1 1 B = 1 e 1 B = [1 1e ] Equation 3.48 gives the average value of an observable: < Q >= ψ (x) ˆQψ(x) dx Where ˆQ is the corresponding operator, in this case the kinetic energy operator, which is given by: ˆp m = d m dx The integral becomes: 1 ( ) < KE > = Be x d Be x dx m dx 0 = m B 1 0 e x dx = 4m B [ e 1 ] where B is the normalization constant found in problem 1.

44 44 3. OPERATORS AND WAVES - SOLUTIONS

45 4 The Hydrogen Atom - Solutions 1. From table 4., we see that: P 10 (r) = 1 a 0 r a 0 e r/a 0 where z=1 corresponds to hydrogen. Then the radial probability density is given by: P10(r) = 4r e r/a a Per equation 4.9, the radial probability density for the 1s state hydrogen is found by: ( ) 1 P 1s (r) r = e r/a 0 = r e r/a a 0 a 0 a The 1s radial wave function used here is listed in table 4. in the row corresponding to n=1 and l=0. Since hydrogen is the atom of interest, we used the appropriate atomic number Z=1. We shall maximize the above expression by taking the first derivative and equating to zero: d dr P 1s(r) = 4r ( e r/a 0 + r a 3 0 a 3 0 a 0 = 4r e r/a 0 4r e r/a 0 = 0 a 3 0 a ) e r/a 0 = 0

46 46 4. THE HYDROGEN ATOM - SOLUTIONS Divide both sides by e r/a 0 and solve for r: 4r a 3 0 4r a 4 0 = 0 ) 4r (1 ra0 = 0 a 3 0 Therefore we have one solution at r = 0 and one at r = a 0. Let s test the intervals (0, a 0 ) and (a 0, ) to reveal the nature of these points. Let r = a 0 / for the former and r = a 0 for the latter: d dr P 1s(r) x=a0 / d dr P 1s(r) = 4 x=a0 a 3 0 = 4 ( a0 ) e 1 4 ( ) a 0 e 1 a 3 0 a ( = e 1 1 ) > 0 a 0 a 0 (a 0 ) e 4 4 ( ) 4a a 3 0 e 4 0 = 8 a 0 e 4 (1 ) < 0 Since the first derivative of the radial probability density takes a positive values for (0, a 0 ) and negative values for (a 0, ), we may conclude that r = a 0 is in fact the maximum value. This is consistent with the description of the Bohr radius as the innermost orbital radius of the electron in the hydrogen atom. 4. Starting with equations 4.7 and 4.8, dp = ψ(r) r sin(θ) dr dθ dψ We construct our wavefunction by equation 4.4: ψ(r, θ, φ) = P nl(r) r Θ lm (θ) Φ m (θ, φ) P nl (r), Θ lm (θ), and Φ m (θ, φ) for the 1s state may be found in tables 4.1 and 4. in rows corresponding to n=1, l=0, and m l =0 (Note that m l =0 is the only

47 possibility for l=0 thus only one row for l=0 appears in table 4.1). Carrying on, ψ(r, θ, φ) = r = 1 1 a 3/ 0 a 0 ( r a 0 e r/a 0 1 π ) e r/a π We now integrate to find the probability: P = ψ(r, θ, φ) r sin(θ) dr dθ dφ 47 = 1 π = 1 a 3 0 a0 0 a0 0 4r a 3 0 r e r/a 0 dr e r/a 0 dr π 0 sin(θ) dθ π 0 dφ Grouping 1/a 0 with r and dr: a0 P = 0 ( r a 0 ) ( ) e r/a 0 r d a 0 Substitute ρ = r a 0 : 1 P = 0 ρ e ρ dρ This integral may be solved analytically using integration by parts: 1 0 ρ e ρ dρ = 1 1 e ρ ρ 0 1 = 1 ( e + 1 ) e ρ ρ = 1 ( ) 1 5e 4 0 ρ ( 1 ) e ρ dρ ( 1 ) e ρ dρ 5. For a given n-state, the possible values of l are n-1, n-, Therefore,

48 48 4. THE HYDROGEN ATOM - SOLUTIONS n=5 allows l=0,1,,3,4. For any given l value, the possiblities for the magnetic quantum number are m l = -l, -l+1,... 0, 1,... l. The results are as follows: l = 0 : m = 0 l = 1 : m = 1, 0, 1 l = : m =, 1, 0, 1, l = 3 : m = 3,, 1, 0, 1,, 3 l = 4 : m = 4, 3,, 1, 0, 1,, 3, 4 6. The radial wave function P 5d (r) corresponds to n=5 and l=, therefore we calculate from (4.6): v nodes = n l 1 = 5 1 = 7. Recall from equation 4.4 our construction of a wave function as a product of radial and angular parts: ψ(r, θ, φ) = P nl(r) Θ lml (θ)φ(φ) r Our strategy will be to express the given wave function in terms of radial and angular parts, then use their properties to check whether the wavefunction is an eigenfunction of the Schrodinger equation and find the corresponding energy. For our given wavefunction, we see that we may separate it in terms of dependence on θ and dependence on r as cos(θ) and r e Zr/a 0. Note that the presence of r is due to fact the radial part is to be divided by r in eq. 4.4 to construct the total wave function. We need not incorporate the constant C as it will not affect whether the wave function is a solution of the Schrodinger equation. For the angular part, note that in table 4.1, cos(θ)

49 49 appears in the spherical harmonic corresponding to l = 1 and m l = 0. When we distribute the left-hand side of eq. 4.4, we get: m d dr P nl(r) + l(l + 1) mr P nl (r) 1 4πɛ 0 Ze r P nl(r) Taking the second derivative of the radial wave function: d dr P nl(r) = e Zr/a 0 rz e Zr/a 0 + Z r e Zr/a 0 a 0 4a 0 Upon substitution and using l = 1, the left-hand side of eq. 4.4 becomes: m d dr P nl(r) + l(l + 1) P mr nl (r) 1 Ze 4πɛ 0 r P nl(r) = m e Zr/a 0 + rz ma 0 e Zr/a 0 + Z r 8ma 0 e Zr/a 0 + m e Zr/a 0 Ze r 4πɛ 0 e Zr/a 0 = rz ma 0 e Zr/a 0 Ze r 4πɛ 0 e Zr/a 0 + Z r 8ma 0 e Zr/a 0 Recall from equation 1.0 that the Bohr radius a 0 may expressed as follows: a 0 = 4πɛ 0 me If we substitute this expression for a 0 into the first term of the previous equation, the first two terms cancel and equation 4.5 becomes: Z 8ma 0 r e Zr/a 0 = EP nl (r) The radial wave function is in fact an eigenfunction of the Schrodinger equation with corresponding eigenvalue (energy): E = Z 8ma 0

50 50 4. THE HYDROGEN ATOM - SOLUTIONS 8. Hydrogen Ne +9 E 1 = 13.6 ev E = 3.4 ev E 3 = 1.5 ev E 1 = 1360 ev E = 340 ev E 3 = 151 ev 9. We shall start with eq. 4.5: ( And 1.0: m d dr + l(l + 1) 1 Ze mr 4πɛ 0 r ) P nl (r) = EP nl (r) a 0 = 4πɛ 0 me The substitution as directed is ρ = r/a 0, so we can directly substitute for F: r = a 0 ρ. We let Z=1 as is the case hydrogen and carry out the substitution: ( m d d(ρa 0 ) + l(l + 1) m(ρa 0 ) 1 Ze 4πɛ 0 ρa 0 ) P nl (r) = EP nl (r) Since a 0 is a constant, we may bring it outside the derivative in the first term on the left-hand side and substitute eq. 1.0 for one a 0 in each of the first two terms. ( 1a0 me d m 4πɛ 0 dρ + 1 l(l + 1) me a 0 m(ρa 0 ) 4πɛ 0 1 Ze a 0 ρa 0 ) P nl (r) = EP nl (r) After making the appropriate cancellations and factoring we have: ( e 1 1 d a 0 4πɛ 0 dρ + 1 l(l + 1) 1 ) P ρ nl (r) = EP nl (r) ρ ( 1 d l(l + 1) + 1 ) E P dρ ρ nl (r) = ρ (1/4πɛ 0 )(e /a 0 ) P nl(r)

51 51 10 a.) For l = 1, our expression for V l (ρ) becomes: V 1 (ρ) = 1 ρ 1 ρ b.) From problem 8, E 3 = 1.5 ev. Per the problem statement, make this positive and set equal to V 1 (ρ): 1 ρ 1 ρ = 1.5 ev 1.5ρ + ρ 1 = 0 ρ = 1., ρ =.55 c.) We find from table 4.: ( 1 4 r P 31 (r) = a a 0 ) ( 1 r ) e r/3a 0 6a 0 Now, substitute ρ = r/a 0 and evaluate the derivative: dp 31 1 dρ = 4 [ρe a 0 7 ρ/3 1 3 ρ e ρ/3 13 ] ρ e ρ/3 + ρ 18 e ρ/3 Using sample values ρ = 1 and ρ = 1: dp 31 dρ /P 31 = 53 ρ= 1 1 dρ /P 31 = 4 ρ=1 3 dp Equation 4.15 gives the following: [ ] m + V (r, t) Ψ = i Ψ t

52 5 4. THE HYDROGEN ATOM - SOLUTIONS Given that Ψ(r, t) = φ(r)t (t) and letting H equal the operator within square brackets: T (t)hφ(r) = i φ(r) T (t) dt Divide through by Ψ(r, t) = φ(r)t (t): 1 (t) i dt T (t) dt = Hφ(r) φ(r) Setting the left-hand side equal to the separation constant E: 1 (t) i dt T (t) dt = E Integrate both sides: Let ω = E/ : dt (t) T (t) = ie dt ln T (t) = i E t T (t) = e iωt Using the fact that φ(r) is a solution to the time-independent case, the timedependent solution may be written as: Ψ(r, t) = φ(r)e iωt 1. The transition integral for the general case is given by φ n 1 l 1 m 1 l ( z a 0 ) φ n l m l We are interested in the transition 3d 1 p 1, so the transition integral becomes: ( ) z φ,1,1 dv φ 3,,1 a 0 dv

53 From here on, we assume that the radial distance will be given in terms relative to a 0, so we take a 0 = 1 and constructs the wavefunctions by tables 4.1 and 4.. φ 3,,1 = P 3,(r) r = r e r/3 φ,1,1 = P,1(r) r = 1 6 re r/ Y,1 (θ, φ) ( ) 15 8π sin(θ)cos(θ)eiφ Y 1,1 (θ, φ) ( ) 3 8π sin(θ)eiφ Now, we take the complex conjugate of φ 3,,1, substitute z = rcos(θ), and proceed with the integration: φ 3,,1zφ,1,1 dv = 1 r 4 e 5r/6 sin (θ)cos (θ) dv 648π = 1 π π r 6 e 5r/6 sin 3 (θ)cos (θ) dr dθ dφ 648π The integration may be separated as follows: π 0 π r 6 e 5r/6 dr 0 π sin 3 (θ)cos (θ) dθ dφ = π 1565 = π 13. In order to find the transition coefficient for hydrogen, we will use the results found in Appendix FF (online) along with eqs. 4.0 and 4.1. We According to the selection rules, summarized in table 4.3, m l = ±1 for x- and y-polarized light. Therefore, we must calculate the transition integrals for both the p 1 1s 0 and p 1 1s 0 cases. As directed in Appendix FF, we use the following in place of the z operator in eq. 4.1: x = 1 (r + + r ) y = 1 i (r + r )

54 54 4. THE HYDROGEN ATOM - SOLUTIONS We first solve the case for x-polarized light. I 1 = ( φ 1s 0 xφ p1 + φ 1s 0 xφ p 1 ) Which gives us the following when we substitute the above expression for x: φ 1s 0 xφ p1 = 1 (φ 1s 0 r + φ p1 + φ 1s 0 r φ p1 ) φ 1s 0 xφ p 1 = 1 (φ 1s 0 r + φ p 1 + φ 1s 0 r φ p 1 ) Since φ 1s 0 r + φ p1 and φ 1s 0 r φ p 1 will equal zero, we may use the results given in Appendix FF for the remaining terms, giving us: ( 1 I 1 = 6 R i + 1 ) 6 R i = 3 R i Using the results of eqs. 4.3 and 4.6 into eq. 4.: A 1 = (11.6) 3 6 = per atom per second We may follow the same procedure for y-polarized light. First, the transition integral: I 1 = ( φ 1s 0 yφ p1 + φ 1s 0 yφ p 1 ) Using the operator substitution: φ 1s 0 yφ p1 = 1 i (φ 1s 0 r + φ p1 φ 1s 0 r φ p1 ) φ 1s 0 yφ p 1 = 1 i (φ 1s 0 r + φ p 1 φ 1s 0 r φ p 1 ) Once again, the nonzero terms lead to: ( 1 I 1 = 6 R i + 1 ) 6 R i = 3 R i

55 55 The result for y-polarized light is then: A 1 = (11.6) 3 6 = per atom per second 14. The p-state corresponds to l = 1. Per the selection rules, the 4p electron may decay to the following states: 3s, 3d, s, 1s. 15. a) Let s imagine that the angular momentum vector l points from the origin and lies on the surface of a cone, then the radius of the circle at the base of the cone is: r = l sinθ And, for a change of the azimuthal angle dφ, the distance that the tip of the angular momentum vector moves is given by: d l = l sinθ dφ b) We start with the relationship of the torque to angular to momentum: d l dt = τ This equation together with eq immediately leads to the equation: d l dt = e l B m The definition of the vector product then gives: d l dt = e l B sin(θ) m Substituting d l = l sinθ dφ, we may then find our Larmor frequencies: l sin(θ) dφ dt = e l B sin(θ) m ω L = dφ dt = eb m

56 56 4. THE HYDROGEN ATOM - SOLUTIONS 16. The f-state corresponds to l = 3, therefore the possible values of m l and m s are: m l = 3,, 1, 0, 1,, 3 m s = 1 and The total angular momentum of the 4f electron will contain contribution form both the electron s spin and its orbital angular momentum. Section shows that the total angular momentum may take on the values: J = j 1 + j, j 1 + j 1,... j j 1 We let j 1 be our orbital angular momentum and j the spin. The electron occupies an f-state, therefore j 1 = 3, and the intrinsic spin of the electron means j = 1/. We may now list the possible values for J: J = 7, 5 The spin-orbit coupling energy is then found by: E s o = ζ l = 3ζ for j = l + 1 E s o = ζ (l + 1) = ζ for j = l We take into consideration two angular momenta: the orbital angular momentum and the spin. The p electron has l = 1 and spin s = 1/. Therefore, the possible values of the combined angular momentum will be: j = 3, 1 From (4.57) and the equations that follow: g 3/ = 4 3

57 57 g 1/ = 3 The splitting of the m-levels is given by: E = g j µ B Bm j where µ B = J T 1. For j = 3/, the possible m-values are m = 3/, 1/, 1/, 3/ while the possibilities for j = 1/ are m = ±1/. Now, to calculate the splitting of energies: j = 3 : m = 3 E = ev m = 1 m = 1 m = 3 E = ev E = ev E = ev j = 1 : m = 1 E = ev m = 1 E = ev 19. Using the two equations following eq. 4.50, the separation in energies due to spin-orbit coupling is: ( ) ( ) E = ζ ζ ζ l (l + 1) = (l + 1) Since we are using the atomic system of units, = 1 and = ζ (l + 1) (1) We then let l=1 (due to the occupation of a p-orbital): E = 3 ζ = ζ = The spin-orbit constant ζ for the levels neon may be calculated directly by eq. 4.48, however we may use equation 4.51 to set up a relationship between

58 58 4. THE HYDROGEN ATOM - SOLUTIONS ζ He and ζ Ne. Since He + and Ne 9+ are both hydrogenic atoms (containing only one electron), the quantity ( a0 ) 3 Zr should be equal in both cases as the grouping of a 0 and z with r will account for the scaling of the radial distance with changing nuclear charge. We now use the appropriate atomic numbers of helium and neon to generate a ratio between ζ He and ζ Ne.: ζ Ne ζ He = = 65 ζ Ne = 65ζ He = For the separation between the 3p 3/ and 3p 1/ levels, we use equation (1) above: E = (l + 1) = A comparison with the separation of the corresponding helium states reveals the vastly increased impact of spin-orbit coupling in the case of neon.

59 5 Many-Electron Atoms - Solutions 1. An electron has an intrinsic spin s = 1/, therefore the possible values of m s are ±1/. The 4f state of the electron has an orbital angular momentum quantum number l = 3 and therefore may have values m l = 3,, 1, 0, 1,, 3. In the case of two electrons occupying the 4f state, we first see from our results above, that there are 7 possible m l quantum numbers m s quantum numbers. There are thus 14 possible states in which to put the first electron. Once we assign first electron, there are now 13 possible states. Because the order in which placed them does not matter in our configuration, we then divide the product by two: # of distinct states = = 91. If we interchange the coordinates of (5.5), we are left with: Ψ = 1 [φ a ()φ b (1) φ a (1)φ b ()] = 1 1 [φ a (1)φ b () φ a ()φ b (1)] This expression is identical to (5.5) but with the opposite sign. 59

60 60 5. MANY-ELECTRON ATOMS - SOLUTIONS 3. Using the form of eq. 5.6, we attribute each row of the determinant to a particular state and each column to a particular electron occupying a corresponding state (row). Our system contains these electrons, therefore N=3 in our coefficient and Φ = 1 3 ψ 10 α(1) ψ 10 α() ψ 10 α(3) ψ 10 β(1) ψ 10 β() ψ 10 β(3) ψ 0 α(1) ψ 0 α() ψ 0 α(3) 4. For each of these elements, we shall fill their shells in the following order: 1s, s, p, 3s, 3p, 4s, 4p, and so on. The number of electrons in a neutral atom of any element will equal the number of protons in its nucleus, therefore we use the atomic number of each element as the number of electrons we use to fill these shells: Flourine(Z = 9) : 1s s p 5 Magnesium(Z = 1) : 1s s p 6 3s [Ne] 3s Silicon(Z = 14) : 1s s p 6 3s 3p [Ne] 3s 3p Potassium(Z = 19) : 1s s p 6 3s 3p 6 4s 1 [Ar] 4s 1 Cobalt(Z = 7) : 1s s p 6 3s 3p 6 4s 3d 7 [Ar] 4s 3d 7 An alternate (shorter) way to notate these, as I ve done above to the right, is to use the symbol of the inert gas that occurs closest before a particular element to denote the filled shells through complete rows of the periodic table and merely denote the filling of the remaining shells. 5. a.) The elements with a p 4 ground configuration will reside in the same column of the periodic table: O, S, Se, Te, Po b.) The elements with d 5 ground configuration are located in the fifth column of the transition metals on the periodic table: Mn, Te, Re, Ns 6. The ground state configuration of carbon is 1s s p. The next two low-lying configurations are 1s s p3s and 1s s p3p. 7. The nitrogen atom will have the following ground configuration: 1s s p 3.

61 Since the 1s and s orbitals have less energy than the p orbital, the next higher configurations are 1s s p 3s and 1s s p 3p. 8. ns : Here are two electrons with l=0, therefore, L=0. Their spins produce S=0,1. Since L+S must be even for two equivalent electrons, the only possibility is 1 S (S=0, L=0). nd : Two electrons with l= may produce L=4,3,,1,0. Again, the addition of their spin angular momenta produce S=0,1. Imposing the condition that L+S must be even, we have the following configurations: 1 G (L=4, S=0), 3 F (L=3, S=1), 1 D (L=, S=0), 3 P (L=1, S=1), and 1 S (S=0, L=0). 4f : These two electrons occupy f-states (l=3), thus may contribute to L=6,5,4,3,,1,0. As has been the case above, S=0,1. Imposing the condition that L+S must be even, we have the following LS terms: 1 I (L=6, S=0), 3 H (L=5, S=1), 1 G (L=4, S=0), 3 F (L=3, S=1), 1 D (L=, S=0), 3 P (L=1, S=1), and 1 S (S=0, L=0). 61 Let s apply Hund s rules to the 4f configuration. Of these terms, the ones corresponding to the maximum value of S are 3 H, 3 F, and 3 P. Of these, the one with maximum value of L ( 3 H) should lie lowest. 9. We shall add up the contributions from the various M L and M S values possible for each LS-term: 1 I(L = 6, S = 0) : (13 values M L ) (1 value M S ) = 13 3 H(L = 5, S = 1) : (11 values M L ) (3 values M S ) = 33 1 G(L = 4, S = 0) : (9 values M L ) (1 value M S ) = 9 3 F (L = 3, S = 1) : (7 values M L ) (3 values M S ) = 1 1 D(L =, S = 0) : (5 values M L ) (1 value M S ) = 5 3 P (L = 1, S = 1) : (3 values M L ) (3 values M S ) = 9 1 S(L = 0, S = 0) : (1 values M L ) (1 value M S ) = 1 The sum of states is 91, which is precisely the result obtained in problem The p electron has l = 1 while the 3d electron has l = therefore the possible values of L are L = 1,, 3. The spins may produce S = 0, 1.