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1 Problem 8.1 A particle with mass m moves in a 3D box with edges L1 = L, L2 = 2L, and L3 = 2L. Find the energies of the six lowest states. Which ones are degenerate? 1.1 Solution We get the wavenumbers the usual way, using the boundary conditions. r 2mE1 n1 π k1 = = L ~2 r n2 π 2mE2 k2 = = 2L ~2 r n3 π 2mE3 = k3 = 2L ~2 Solving for the energies gives E1 = ~2 π 2 2 n 2mL2 1 E2 = ~2 π 2 n22 2mL2 4 E3 = ~2 π 2 n23 2mL2 4 Or ~2 π 2 E = E1 + E2 + E3 = 2mL2 n21 n2 n2 + 2+ 3 4 4 The ground state is when n1 = n2 = n3 = 1 leading to ~2 π 2 12 12 3~2 π 2 2 E= 1 + + = 2mL2 4 4 4mL2 There is a twofold degeneracy in the first excited state: n1 = n2 = 1; n3 = 2 or n1 = n3 = 1; n2 = 2 ~2 π 2 E= 2mL2 22 12 + 1 + 4 4 2 ~2 π 2 = 2mL2 12 22 + 1 + 4 4 2 = 9~2 π 2 8mL2 The second excited state is unique and will be n1 = 1; n2 = n3 = 2. 22 22 3~2 π 2 ~2 π 2 2 E= 1 + + = 2mL2 4 4 mL2 1 Again, there is a twofold degeneracy in the third excited state n1 = 1; n2 = 2; n3 = 3 or n1 = 1; n2 = 3; n3 = 2 ~2 π 2 E= 2mL2 22 32 1 + + 4 4 2 ~2 π 2 = 2mL2 32 22 17~2 π 2 2 1 + + = 4 4 8mL2 These are the lowest six states. 1 ground + 2 first excited + 1 second excited + 2 third excited = 6. 2 Problem 8.3 A particle of mass m is in a 3D cube with sides L. It is in the third excited state, corresponding to n2 = 11. (a) Calculate the energy of the particle. (b) The possible combinations of n1 , n2 , and n3 (c) The wavefunctions for these states. 2.1 2.1.1 Solution Part (a) Just plug in n2 = 11 to the 3D box’s energy. E= 2.1.2 11~2 π 2 2mL2 Part (b) and Part (c) We’ll need n2 = n21 + n22 + n23 There are three ways to do this 11 = (32 + 12 + 12 ) = (12 + 32 + 12 ) = (12 + 12 + 32 ) Corresponding to the three states and their wavefunctions n1 = 3; n2 = 1; n3 = 1 → ψ(x, y, z) = 3/2 πy πz 2 3πx sin sin sin L L L L n1 = 1; n2 = 3; n3 = 1 → ψ(x, y, z) = 3/2 πx πz 2 3πy sin sin sin L L L L n1 = 1; n2 = 1; n3 = 3 → ψ(x, y, z) = 3/2 πy πx 2 3πz sin sin sin L L L L 2 3 Problem 8.4 A particle of mass m is stuck in a 2D box of length L. (a) What are the wavefunctions? (b) What are the energies of the ground state and the first excited state? 3.1 Solution 3.2 Part (a) We start with a general wavefunction see if separation of variables works. ψ(x, y) = ψ1 (x)ψ2 (y) Plugging this into the Schrodinger Equation and mixing around a few terms around we get ~2 1 ∂ 2 ψ1 (x) 1 ∂ 2 ψ2 (y) − + =E 2m ψ1 (x) ∂x2 ψ2 (y) ∂y 2 This separates into two equations − ~2 ∂ 2 ψ1 (x) = E1 ψ1 (x) 2m ∂x2 − ~2 ∂ 2 ψ2 (y) = Eψ2 (y) 2m ∂y 2 This will have solutions and energies of r n πx 2 1 sin ψ1 (x) = L L r n πy 2 2 ψ2 (y) = sin L L 3.3 E1 = ~2 π 2 2 n 2mL2 1 E2 = ~2 π 2 2 n 2mL2 2 Part (b) We can see that the total energy will be ~2 π 2 2 (n + n22 ) 2mL2 1 The ground state will be at n1 = n2 = 1 and will have energy E = E1 + E2 = ~2 π 2 mL2 There will be a degeneracy in the first excited state between n1 = 2; n2 = 1 and n1 = 1; n2 = 2 which will have energy E= 3 ~2 π 2 2 5~2 π 2 ~2 π 2 2 (2 + 12 ) = (1 + 22 ) = 2 2 2mL 2mL 2mL2 E= 4 Problem 8.7 (a) Normalize the wavefunction for a particle in a cube of length L. (b) What if it is a rectangular prism with sides L1 6= L2 6= L3 4.1 Solution 4.1.1 Part (a) We know that the wavefunction will be n πy n πz n πx 2 3 1 sin sin ψ(x, y, z) = A sin L L L We apply the normalization rule Z +∞ 1= ψ ∗ (~r)ψ(~r)dV −∞ Z 1 = A2 L Z x=0 1=A 2 L Z y=0 L sin2 z=0 n πy n πz n πx 2 3 1 sin2 sin2 dxdydz L L L Z L n πz n πx Z L 3 1 2 n2 πy dx dy dz sin sin2 sin L L L y=0 z=0 x=0 Z L 2 3 L 2 3/2 2 A= L 1 = A2 4.2 Part (b) The wavefunction (and its square) will be the same as before but with a minor tweak 1=A 2 Z L1 L2 Z L3 2 sin x=0 1 = A2 Z Z y=0 L1 x=0 sin2 z=0 n1 πx L1 n1 πx L1 Z sin L2 dx 2 sin2 y=0 4 n2 πy L2 n2 πy L2 2 sin Z n3 πz L3 L3 dy z=0 sin2 dxdydz n3 πz L3 dz L1 L2 L3 2 2 2 r 8 A= L1 L2 L3 1 = A2 5 Problem 8.11 The orbital angular momentum of the Earth is 4.83 × 1031 kg· m2 /s. Calculate ~ when l → l + 1 l and the fractional change in kLk 5.1 Solution ~ is The quantization rule for kLk ~ = kLk p l(l + 1)~ It will be proven in the second part of this problem that l ≈ l + 1. Using this we get ~ kLk ~ Since ~ = 1.05 × 10−34 kg· m2 /s we see l= 4.83 × 1031 kg · m2 /s ≈ 4.6 × 1065 1.05 × 10−34 kg · m2 /s That is a HUGE quantum number! To get a sense of how big this is, look at the second part of the problem l= ~l k kL~l+1 k − kL = ~l k kL p p √ √ (l + 1)(l + 2)~ − (l)(l + 1)~ l+2− l 2 p √ = ≈ 65 ≈ 10−65 ≈ 0 10 l (l)(l + 1)~ This is essentially zero! This is why quantum mechanics isn’t necessary for examining macroscopic objects and the natural world ”feels” continuous. When you work with numbers that are so big, the quantization of nature isn’t noticeable. 6 Problem 8.18 A hydrogen atom is in the 6g state. (a) What is the principle quantum number? (b) What is the energy of the atom? ~ (c) What are the possible values for l and kLk? (d) What are the possible values of ml ? (e) What are the possible values of Lz ~ make with the z-axis? and for each of these what angle will L 5 6.1 6.1.1 Solution Part (a) n=6 6.1.2 Part (b) In general, the energy for a hydrogen-like atom in the n-th energy state is En = − ke2 Z 2 2a0 n2 In this case Z = 1 and n = 6 so E6 = − 6.1.3 ke2 72a0 Part (c) The rule for l is that it must be an integer between 0≤l<n This makes the allowed values 0, 1, 2, 3, 4, and 5. 6.1.4 Part (d) The rule for ml is that it must be an integer between −l ≤ ml ≤ l This makes the allowed values 0, ±1, ±2, ±3, ±4, and ±5. The z component of the angular momentum is given by Lz = ml ~ So the allowed values will be 0, ±~, ±2~, ±3~, ±4~, and ±5~. The values of ~ are given by kLk p ~ = l(l + 1)~ kLk In general we will have θlm = ± sin−1 kLz k ~ kLk ! = ± sin−1 All possible values of θlm are listed below θ00 = 90◦ 6 km k p l l(l + 1) ! θ1±1 = ± sin−1 θ10 = 90◦ ◦ θ20 = 90 −1 θ2±1 = ± sin θ3±2 = ± sin−1 θ40 = 90 √ θ50 = 90 3 √ 4·5 −1 √ ≈ ±24 3 5·6 √ 1 5·6 = ±45◦ −1 ◦ ≈ ±13 √ 1 3·4 √ −1 θ5±5 = ± sin 5 √ 5·6 4 √ 4·5 −1 ◦ θ5±2 = ± sin θ5±4 = ± sin−1 = ±33◦ ≈ ±55◦ = ±60◦ θ4±2 = ± sin θ4±4 = ± sin ≈ ±11 3 3·4 −1 ◦ 2 √ 2·3 ≈ ±17◦ −1 θ2±2 = ± sin ◦ 1 1·2 θ3±3 = ± sin−1 = ±42 θ5±1 = ± sin θ5±3 = ± sin−1 1 √ 4·5 √ ≈ ±35◦ θ4±1 = ± sin θ4±3 = ± sin ◦ 2 3·4 −1 −1 1 √ 2·3 θ3±1 = ± sin−1 θ30 = 90◦ ◦ √ 4 5·6 2 √ 4·5 ≈ ±27◦ = ±63◦ √ 2 5·6 ≈ ±21◦ = ±47◦ = ±66◦ One cool thing to notice is that θll increases as l increases. This agrees with ~ which our prediction that in the limit l >> 1 we will be able to have Lz = kLk, is usually what we deal with classically. 7 Problem 8.20 Using a semiclassical approximation, the energy of a circular orbit becomes L̂2 kZe2 − 2mr2 r Use this to show that we must have l ≤ n − 1 E= 7 (1) 7.1 Solution The formula for energy is En = − ke2 Z 2 2a0 n2 The formula for angular momentum is p ~ = l(l + 1)~ kLk According to Bohr, the allowed values of r for an atom with Z protons is a0 n2 Z Plugging these into Equation 1 we get: r= − l(l + 1)~2 Z 2 Z ke2 Z 2 = − kZe2 2 2 2 2a0 n 2m (a0 n ) a0 n2 Shuffling a few terms around will yield l(l + 1)~2 −ke2 n2 = − ke2 n2 2 2ma0 Remember that ~2 mke2 Using this will simplify our equation to a0 = −n2 l(l + 1) = − n2 2 2 n2 = l(l + 1) Remember that we were dealing with the largest l possible. With this in mind we can see that we must have l ≤n−1 8 Problem 8.22 Given the radial wavefunction R2p (r) = Are−r/2a0 Calculate hri for an electron in this state. 8 8.1 Solution First we need to normalize the wavefunction. Z +∞ ∗ 1= ψ2p (~r)ψ2p (~r)dV −∞ The question is only asking about the radial part of the wave equation. In reality, there will be an angular dependence as well, but we can still consider solely the radial portion since r̂ = r and the wavefunction is separable. Z +∞ ∗ 1= R2p (r)R2p (r)[r2 dr] r=0 Z +∞ ∗ R2p (r)R2p (r)r2 dr 1= −∞ ∞ Z A2 r4 e−r/a0 dr 1= 0 1 = A2 (24a50 ) 1 24a50 A2 = Now we can calculate the expectation value for r. Z +∞ ∗ hri = R2p (r)r̂R2p (r)[r2 dr] r=0 hri = 1 24a50 hri = ∞ Z r5 e−r/a0 dr 0 1 (120a60 ) 24a50 hri = 5a0 9