Exam (elaborations) TEST BANK FOR Quantum Physics 3rd Edition By Steph

Exam (elaborations) TEST BANK FOR Quantum Physics 3rd Edition By Steph The energy contained in a volume dV is U(ν,T)dV = U(ν,T)r 2 drsinθdθdϕ when the geometry is that shown in the figure. The energy from this source that emerges through a hole of area dA is dE(ν,T) = U(ν,T)dV dAcosθ 4πr 2 The total energy emitted is . dE(ν,T) = dr dθ dϕU(ν,T)sinθ cosθ dA 0 4π 2π ∫ 0 π /2 ∫ 0 cΔt ∫ = dA 4π 2πcΔtU(ν,T) dθ sinθ cosθ 0 π / 2 ∫ = 1 4 cΔtdAU(ν,T) By definition of the emissivity, this is equal to EΔtdA. Hence E(ν,T) = c 4 U(ν,T) 2. We have w(λ,T) = U(ν,T) | dν / dλ |= U( c λ ) c λ 2 = 8πhc λ5 1 ehc/λkT −1 This density will be maximal when dw(λ,T) / dλ = 0. What we need is d dλ 1 λ5 1 eA /λ −1 ⎛ ⎝ ⎞ ⎠ = (−5 1 λ 6 − 1 λ 5 e A /λ eA/λ −1 (− A λ2 )) 1 eA /λ −1 = 0 Where A = hc / kT . The above implies that with x = A /λ , we must have 5 − x = 5e−x A solution of this is x = 4.965 so that λ maxT = hc 4.965k = 2.898 ×10−3 m In example 1.1 we were given an estimate of the sun’s surface temperature as 6000 K. From this we get λ max sun = 28.98 ×10 −4 mK 6 ×103 K = 4.83 ×10−7 m = 483nm 3. The relationship is hν = K + W where K is the electron kinetic energy and W is the work function. Here hν = hc λ = (6.626 ×10−34 J .s)(3×108 m / s) 350 ×10 −9 m = 5.68 ×10 −19 J = 3.55eV With K = 1.60 eV, we get W = 1.95 eV 4. We use hc λ1 − hc λ2 = K1 − K2 since W cancels. From ;this we get h = 1 c λ1λ2 λ2 − λ1 (K1 − K2) = = (200 ×10−9 m)(258 ×10−9 m) (3×108 m / s)(58 ×10−9 m) × (2.3− 0.9)eV × (1.60 ×10−19 )J / eV = 6.64 ×10 −34 J .s 5. The maximum energy loss for the photon occurs in a head-on collision, with the photon scattered backwards. Let the incident photon energy be hν , and the backwardscattered photon energy be hν' . Let the energy of the recoiling proton be E. Then its recoil momentum is obtained from E = p 2 c 2 + m 2 c 4 . The energy conservation equation reads hν + mc 2 = hν'+E and the momentum conservation equation reads hν c = − hν' c + p that is hν = −hν'+ pc We get E + pc − mc 2 = 2hν from which it follows that p 2 c 2 + m 2 c 4 = (2hν − pc + mc 2 ) 2 so that pc = 4h2 ν 2 + 4hνmc 2 4hν + 2mc 2 The energy loss for the photon is the kinetic energy of the proton K = E − mc 2 . Now hν = 100 MeV and mc 2 = 938 MeV, so that pc = 182MeV and E − mc 2 = K = 17.6MeV 6. Let hν be the incident photon energy, hν' the final photon energy and p the outgoing electron momentum. Energy conservation reads hν + mc 2 = hν'+ p 2 c 2 + m 2 c 4 We write the equation for momentum conservation, assuming that the initial photon moves in the x –direction and the final photon in the y-direction. When multiplied by c it read i(hν) = j(hν') + (ipx c + jpy c) Hence px c = hν; pyc = −hν'. We use this to rewrite the energy conservation equation as follows: (hν + mc 2 − hν')2 = m 2 c 4 + c 2 (px 2 + py 2 ) = m 2 c 4 + (hν) 2 + (hν') 2 From this we get hν'= hν mc 2 hν + mc 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ We may use this to calculate the kinetic energy of the electron K = hν − hν'= hν 1− mc 2 hν + mc 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = hν hν hν + mc 2 = (100keV ) 2 100keV + 510keV =16.4keV Also pc = i(100keV ) + j(−83.6keV) which gives the direction of the recoiling electron. 7. The photon energy is hν = hc λ = (6.63×10−34 J.s)(3 ×108 m /s) 3×106 ×10−9 m = 6.63×10 −17 J = 6.63×10−17 J 1.60 ×10 −19 J / eV = 4.14 ×10−4 MeV The momentum conservation for collinear motion (the collision is head on for maximum energy loss), when squared, reads hν c ⎛ ⎝ ⎞ ⎠ 2 + p 2 + 2 hν c ⎛ ⎝ ⎞ ⎠ pηi = hν' c ⎛ ⎝ ⎞ ⎠ 2 + p' 2 +2 hν' c ⎛ ⎝ ⎞ ⎠ p'η f Here ηi = ±1, with the upper sign corresponding to the photon and the electron moving in the same/opposite direction, and similarly for ηf . When this is multiplied by c 2 we get (hν) 2 + (pc) 2 + 2(hν) pcηi = (hν')2 + ( p'c) 2 + 2(hν') p'cη f The square of the energy conservation equation, with E expressed in terms of momentum and mass reads (hν) 2 + (pc) 2 + m 2 c 4 + 2Ehν = (hν')2 + ( p'c) 2 + m 2 c 4 + 2E' hν' After we cancel the mass terms and subtracting, we get hν(E −ηipc) = hν'(E'−ηf p'c) From this can calculate hν' and rewrite the energy conservation law in the form E − E'= hν E − ηi pc E'−p'cη f −1 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ The energy loss is largest if ηi = −1;ηf = 1. Assuming that the final electron momentum is not very close to zero, we can write E + pc = 2E and E'− p'c = (mc 2 ) 2 2E' so that E − E'= hν 2E × 2E' (mc 2 ) 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ It follows that 1 E' = 1 E +16hν with everything expressed in MeV. This leads to E’ =(100/1.64)=61 MeV and the energy loss is 39MeV. 8.We have λ’ = 0.035 x 10-10 m, to be inserted into λ'−λ = h me c (1− cos600 ) = h 2me c = 6.63 ×10−34 J.s 2 × (0.9 ×10−30 kg)(3×108 m / s) = 1.23×10 −12 m Therefore λ = λ’ = (3.50-1.23) x 10-12 m = 2.3 x 10-12 m. The energy of the X-ray photon is therefore hν = hc λ = (6.63×10 −34 J .s)(3 ×108 m / s) (2.3×10 −12m)(1.6 ×10 −19 J / eV) = 5.4 ×105 eV 9. With the nucleus initially at rest, the recoil momentum of the nucleus must be equal and opposite to that of the emitted photon. We therefore have its magnitude given by p = hν / c , where hν = 6.2 MeV . The recoil energy is E = p2 2M = hν hν 2Mc2 = (6.2MeV) 6.2MeV 2 ×14 × (940MeV ) = 1.5 ×10−3MeV 10. The formula λ = 2asinθ / n implies that λ / sinθ ≤ 2a / 3. Since λ = h/p this leads to p ≥ 3h / 2asinθ , which implies that the kinetic energy obeys K = p 2 2m ≥ 9h2 8ma2 sin2 θ Thus the minimum energy for electrons is K = 9(6.63×10 −34 J.s) 2 8(0.9 ×10−30 kg)(0.32 ×10−9 m) 2 (1.6 ×10−19 J / eV) = 3.35eV For Helium atoms the mass is 4(1.67 ×10−27 kg) / (0.9 ×10−30 kg) = 7.42 ×103 larger, so that K = 33.5eV 7.42 ×103 = 4.5 ×10 −3 eV 11. We use K = p 2 2m = h2 2mλ 2 with λ = 15 x 10-9 m to get K = (6.63×10 −34 J.s) 2 2(0.9 ×10−30 kg)(15 ×10−9 m) 2 (1.6 ×10−19 J / eV) = 6.78 ×10−3 eV For λ = 0.5 nm, the wavelength is 30 times smaller, so that the energy is 900 times larger. Thus K =6.10 eV. 12. For a circular orbit of radius r, the circumference is 2πr. If n wavelengths λ are to fit into the orbit, we must have 2πr = nλ = nh/p. We therefore get the condition pr = nh / 2π = n= which is just the condition that the angular momentum in a circular orbit is an integer in units of =. 13. We have a = nλ / 2sinθ . For n = 1, λ= 0.5 x 10-10 m and θ= 5o . we get a = 2.87 x 10-10 m. For n = 2, we require sinθ2 = 2 sinθ1. Since the angles are very small, θ2 = 2θ1. So that the angle is 10o . 14. The relation F = ma leads to mv 2 /r = mωr that is, v = ωr. The angular momentum quantization condition is mvr = n =, which leads to mωr 2 = n=. The total energy is therefore E = 1 2 mv 2 + 1 2 mω 2 r 2 = mω 2 r 2 = n=ω The analog of the Rydberg formula is ν(n → n') = En − En' h = =ω(n − n') h = (n − n') ω 2π The frequency of radiation in the classical limit is just the frequency of rotation νcl = ω / 2π which agrees with the quantum frequency when n – n’ = 1. When the selection rule Δn = 1 is satisfied, then the classical and quantum frequencies are the same for all n. 15. With V(r) = V0 (r/a) k , the equation describing circular motion is m v 2 r =| dV dr |= 1 r kV0 r a ⎛ ⎝ ⎞ ⎠ k so that v = kV0 m r k ⎛ ⎝ ⎞ ⎠ k / 2 The angular momentum quantization condition mvr = n= reads ma 2 kV0 r a ⎛ ⎝ ⎞ ⎠ k+2 2 = n= We may use the result of this and the previous equation to calculate E = 1 2 mv 2 + V0 r a ⎛ ⎝ ⎞ ⎠ k = ( 1 2 k +1)V0 r a ⎛ ⎝ ⎞ ⎠ k = ( 1 2 k +1)V0 n 2 =2 ma2 kV0 ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ k k+2 In the limit of k >>1, we get E → 1 2 (kV0 ) 2 k+2 =2 ma2 ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ k k+ 2 (n 2 ) k k+2 → =2 2ma2 n 2 Note that V0 drops out of the result. This makes sense if one looks at a picture of the potential in the limit of large k. For r< a the potential is effectively zero. For r > a it is effectively infinite, simulating a box with infinite walls. The presence of V0 is there to provide something with the dimensions of an energy. In the limit of the infinite box with the quantum condition there is no physical meaning to V0 and the energy scale is provided by =2 / 2ma 2 . 16. The condition L = n= implies that E = n 2 =2 2I In a transition from n1 to n2 the Bohr rule implies that the frequency of the radiation is given ν12 = E1 − E2 h = =2 2Ih (n1 2 − n2 2 ) = = 4πI (n1 2 − n2 2 ) Let n1 = n2 + Δn. Then in the limit of large n we have (n1 2 − n2 2 ) → 2n2Δn , so that ν12 → 1 2π =n2 I Δn = 1 2π L I Δn Classically the radiation frequency is the frequency of rotation which is ω = L/I , i.e. νcl = ω 2π L I We see that this is equal to ν12 when Δn = 1. 17. The energy gap between low-lying levels of rotational spectra is of the order of = 2 / I = (1 / 2π)h= / MR2 , where M is the reduced mass of the two nuclei, and R is their separation. (Equivalently we can take 2 x m(R/2)2 = MR2 ). Thus hν = hc λ = 1 2π h = MR2 This implies that R = =λ 2πMc = =λ πmc = (1.05 ×10 −34 J.s)(10−3 m) π(1.67 ×10−27 kg)(3×108m / s) = 26nm CHAPTER 2 1. We have ψ(x) = dkA(k)e ikx −∞ ∞ ∫ = dk N k2 + α2 e ikx −∞ ∞ ∫ = dk N k2 + α2 coskx −∞ ∞ ∫ because only the even part of e ikx = coskx + i sinkx contributes to the integral. The integral can be looked up. It yields ψ(x) = N π α e−α |x | so that |ψ(x) |2 = N2 π 2 α2 e−2α |x| If we look at |A(k) 2 we see that this function drops to 1/4 of its peak value at k =± α.. We may therefore estimate the width to be Δk = 2α. The square of the wave function drops to about 1/3 of its value when x =±1/2α. This choice then gives us Δk Δx = 1. Somewhat different choices will give slightly different numbers, but in all cases the product of the widths is independent of α. 2. the definition of the group velocity is vg = dω dk = 2πdν 2πd(1/ λ) = dν d(1/λ) = −λ 2 dν dλ The relation between wavelength and frequency may be rewritten in the form ν 2 −ν 0 2 = c 2 λ2 so that −λ 2 dν dλ = c 2 νλ = c 1− (ν 0 /ν) 2 3. We may use the formula for vg derived above for ν = 2πT ρ λ −3/2 to calculate vg = −λ 2 dν dλ = 3 2 2πT ρλ 4. For deep gravity waves, ν = g / 2πλ −1/2 from which we get, in exactly the same way vg = 1 2 λg 2π . 5. With ω = =k 2 /2m, β = =/m and with the original width of the packet w(0) = √2α, we have w(t) w(0) = 1+ β 2 t 2 2α2 = 1 + =2 t 2 2m2 α2 = 1 + 2=2 t 2 m2 w4 (0) (a) With t = 1 s, m = 0.9 x 10-30 kg and w(0) = 10-6 m, the calculation yields w(1) = 1.7 x 102 m With w(0) = 10-10 m, the calculation yields w(1) = 1.7 x 106 m. These are very large numbers. We can understand them by noting that the characteristic velocity associated with a particle spread over a range Δx is v = =/mΔx and here m is very small. (b) For an object with mass 10-3 kg and w(0)= 10-2 m, we get 2=2 t 2 m2 w4 (0) = 2(1.05 ×10−34 J.s) 2 t 2 (10−3 kg) 2 × (10 −2 m) 4 = 2.2 ×10−54 for t = 1. This is a totally negligible quantity so that w(t) = w(0). 6. For the 13.6 eV electron v /c = 1/137, so we may use the nonrelativistic expression for the kinetic energy. We may therefore use the same formula as in problem 5, that is w(t) w(0) = 1+ β 2 t 2 2α2 = 1 + = 2 t 2 2m2 α2 = 1 + 2=2 t 2 m2 w4 (0) We caclulate t for a distance of 104 km = 107 m, with speed (3 x 108 m/137) to be 4.6 s. We are given that w(0) = 10-3 m. In that case w(t) = (10−3 m) 1 + 2(1.05 ×10 −34 J.s) 2 (4.6s) 2 (0.9 ×10−30 kg) 2 (10−3 m) 4 = 7.5 ×10−2 m For a 100 MeV electron E = pc to a very good approximation. This means that β = 0 and therefore the packet does not spread. 7. For any massless particle E = pc so that β= 0 and there is no spreading. 8. We have φ( p) = 1 2π= dxAe −μ|x| e −ipx/ = −∞ ∞ ∫ = A 2π= dxe(μ−ik )x −∞ 0 ∫ + dxe−(μ+ik )x 0 ∞ { ∫ } = A 2π= 1 μ − ik + 1 μ + ik ⎧ ⎨ ⎩ ⎫ ⎬ ⎭ = A 2π= 2μ μ 2 + k 2 where k = p/=. 9. We want dxA2 −∞ ∞ ∫ e−2μ|x | = A2 dxe2μx + dxe−2μx 0 ∞ ∫ −∞ 0 { } ∫ = A2 1 μ=1 so that A = μ 10. Done in text. 11. Consider the Schrodinger equation with V(x) complex. We now have ∂ψ(x,t) ∂t = i= 2m ∂ 2 ψ(x,t) ∂x 2 − i = V(x)ψ(x,t) and ∂ψ *(x,t) ∂t = − i= 2m ∂ 2 ψ *(x,t) ∂x 2 + i = V *(x)ψ(x,t) Now ∂ ∂t (ψ *ψ) = ∂ψ * ∂t ψ +ψ * ∂ψ ∂t = (− i= 2m ∂2 ψ * ∂x 2 + i = V * (x)ψ*)ψ +ψ * ( i= 2m ∂2 ψ(x,t) ∂x 2 − i = V(x)ψ(x,t)) = − i= 2m ( ∂2 ψ * ∂x 2 ψ −ψ * ∂2 ψ(x,t) ∂x 2 ) + i = (V *−V)ψ *ψ = − i= 2m ∂ ∂x ∂ψ * ∂x ψ −ψ * ∂ψ ∂x ⎧ ⎨ ⎩ ⎫ ⎬ ⎭ + 2ImV(x) = ψ *ψ Consequently ∂ ∂t dx |ψ(x,t) |2 −∞ ∞ ∫ = 2 = dx(ImV(x)) |ψ(x,t) |2 −∞ ∞ ∫ We require that the left hand side of this equation is negative. This does not tell us much about ImV(x) except that it cannot be positive everywhere. If it has a fixed sign, it must be negative. 12. The problem just involves simple arithmetic. The class average 〈g〉 = gng g ∑ = 38.5 (Δg) 2 = 〈g 2 〉−〈g〉 2 = g 2 ng g ∑ − (38.5)2 = 1570.8-1482.3= 88.6 The table below is a result of the numerical calculations for this system g ng (g - <g>)2 /(Δg) 2 = λ e -λ Ce -λ 60 1 5.22 0.0054 0.097 55 2 3.07 0.0463 0.833 50 7 1.49 0.2247 4.04 45 9 0.48 0.621 11.16 40 16 0.025 0.975 17.53 35 13 0.138 0.871 15.66 30 3 0.816 0.442 7.96 25 6 2.058 0.128 2.30 20 2 3.864 0.021 0.38 15 0 6.235 0.002 0.036 10 1 9.70 0.0001 0.002 5 0 12.97 “0” “0” __________________________________________________________ 15. We want 1 = 4N 2 dx sin2 kx x2 −∞ ∞ ∫ = 4N2 k dt sin2 t t 2 −∞ ∞ ∫ = 4πN 2 k so that N = 1 4πk 16. We have 〈x n 〉 = α π ⎛ ⎝ ⎞ ⎠ 1/ 2 dxx n −∞ ∞ ∫ e −αx 2 Note that this integral vanishes for n an odd integer, because the rest of the integrand is even. For n = 2m, an even integer, we have 〈x 2m 〉 = α π ⎛ ⎝ ⎞ ⎠ 1/2 = α π ⎛ ⎝ ⎞ ⎠ 1/2 − d dα ⎛ ⎝ ⎞ ⎠ m dx −∞ ∞ ∫ e−αx 2 = α π ⎛ ⎝ ⎞ ⎠ 1/2 − d dα ⎛ ⎝ ⎞ ⎠ m π α ⎛ ⎝ ⎞ ⎠ 1/ 2 For n = 1 as well as n = 17 this is zero, while for n = 2, that is, m = 1, this is 1 2α. 17. φ( p) = 1 2π= dxe− ipx/ = −∞ ∞ ∫ α π ⎛ ⎝ ⎞ ⎠ 1/4 e−αx 2 /2 The integral is easily evaluated by rewriting the exponent in the form − α 2 x 2 − ix p = = −α 2 x + ip =α ⎛ ⎝ ⎞ ⎠ 2 − p 2 2=2 α A shift in the variable x allows us to state the value of the integral as and we end up with φ( p) = 1 π= π α ⎛ ⎝ ⎞ ⎠ 1/4 e − p2 / 2α= 2 We have, for n even, i.e. n = 2m, 〈p 2m 〉 = 1 π= π α ⎛ ⎝ ⎞ ⎠ 1/ 2 dpp2m e− p2 /α= 2 −∞ ∞ ∫ = = 1 π= π α ⎛ ⎝ ⎞ ⎠ 1/ 2 − d dβ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ m π β ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1/2 where at the end we set β = 1 α=2 . For odd powers the integral vanishes. 18. Specifically for m = 1 we have We have (Δx) 2 = 〈x2 〉 = 1 2α (Δp) 2 = 〈p 2 〉 = α=2 2 so that ΔpΔx = = 2 . This is, in fact, the smallest value possible for the product of the dispersions. 22. We have dxψ *(x)xψ(x) = 1 2π= −∞ ∞ ∫ dxψ * (x)x dpφ( p)eipx/ = −∞ ∞ ∫ −∞ ∞ ∫ = 1 2π= dxψ * (x) dpφ(p) = i −∞ ∞ ∫ −∞ ∞ ∫ ∂ ∂p eipx/= = dpφ * (p)i= ∂φ(p) −∞ ∂p ∞ ∫ In working this out we have shamelessly interchanged orders of integration. The justification of this is that the wave functions are expected to go to zero at infinity faster than any power of x , and this is also true of the momentum space wave functions, in their dependence on p. CHAPTER 3. 1. The linear operators are (a), (b), (f) 2.We have dx' x'ψ(x') = λψ(x) −∞ x ∫ To solve this, we differentiate both sides with respect to x, and thus get λ dψ(x) dx = xψ(x) A solution of this is obtained by writing dψ /ψ = (1/ λ)xdx from which we can immediately state that ψ(x) = Ceλx 2 / 2 The existence of the integral that defines O6ψ(x) requires that λ < 0. 3, (a) O2O6 ψ(x) − O6 O2 ψ(x) = x d dx dx' x'ψ(x') − −∞ x ∫ dx' x' 2 dψ(x') dx' −∞ x ∫ = x2 ψ(x) − dx' d dx' −∞ x ∫ x' 2 ( ) ψ(x') + 2 dx' x'ψ(x') −∞ x ∫ = 2O6ψ(x) Since this is true for every ψ(x) that vanishes rapidly enough at infinity, we conclude that [O2 , O6] = 2O6 (b) O1O2ψ(x) − O2O1 ψ(x) = O1 x dψ dx ⎛ ⎝ ⎞ ⎠ − O2 x 3 ( ) ψ = x 4 dψ dx − x d dx x 3 ( ψ ) = −3x3 ψ(x) = −3O1 ψ(x) so that [O1, O2] = -3O1 4. We need to calculate 〈x 2 〉 = 2 a dxx 2 sin2 nπx 0 a a ∫ With πx/a = u we have 〈x 2 〉 = 2 a a 3 π3 duu2 sin2 nu = a 2 π3 0 π ∫ duu2 0 π ∫ (1− cos2nu) The first integral is simple. For the second integral we use the fact that duu2 cosαu = − d dα ⎛ ⎝ ⎞ 0 ⎠ π ∫ 2 ducosαu = − 0 π ∫ d dα ⎛ ⎝ ⎞ ⎠ 2 sinαπ α At the end we set α = nπ. A little algebra leads to 〈x 2 〉 = a 2 3 − a 2 2π2 n2 For large n we therefore get Δx = a 3 . Since 〈p 2 〉 = =2 n 2 π 2 a2 , it follows that Δp = =πn a , so that ΔpΔx ≈ nπ= 3 The product of the uncertainties thus grows as n increases. 5. With En = =2 π 2 2ma2 n 2 we can calculate E2 − E1 = 3 (1.05 ×10−34 J .s) 2 2(0.9 ×10 −30 kg)(10 −9m) 2 1 (1.6 ×10−19 J / eV) = 0.115eV We have ΔE = hc λ so that λ = 2π=c ΔE = 2π(2.6 ×10−7 ev.m) 0.115eV =1.42 ×10−5 m where we have converted =c from J.m units to eV.m units. 6. (a) Here we write n 2 = 2ma 2 E =2 π2 = 2(0.9 ×10 −30 kg)(2 ×10 −2 m) 2 (1.5eV )(1.6 ×10−19 J / eV ) (1.05 ×10 −34 J .s) 2 π2 = 1.59 ×1015 so that n = 4 x 107 . (b) We have ΔE = =2 π 2 2ma2 2nΔn = (1.05 ×10−34 J.s) 2 π 2 2(0.9 ×10−30 kg)(2 ×10 −2 m) 2 2(4 ×107 ) =1.2 ×10 −26 J = 7.6 ×10−8 eV 7. The longest wavelength corresponds to the lowest frequency. Since ΔE is proportional to (n + 1)2 – n 2 = 2n + 1, the lowest value corresponds to n = 1 (a state with n = 0 does not exist). We therefore have h c λ= 3 = 2 π 2 2ma2 If we assume that we are dealing with electrons of mass m = 0.9 x 10-30 kg, then a 2 = 3=πλ 4mc = 3π(1.05 ×10 −34 J.s)(4.5 ×10 −7 m) 4(0.9 ×10−30 kg)(3×108 m / s) = 4.1×10−19 m 2 so that a = 6.4 x 10-10 m. 8. The solutions for a box of width a have energy eigenvalues En = = 2 π 2 n 2 2ma2 with n = 1,2,3,…The odd integer solutions correspond to solutions even under x → −x , while the even integer solutions correspond to solutions that are odd under reflection. These solutions vanish at x = 0, and it is these solutions that will satisfy the boundary conditions for the “half-well” under consideration. T