http://wenku.baidu.com/view/b785887302768e9951e7388d
http://electron9.phys.utk.edu/phys514/modules/module12/problems12.htm
Problem 1:
Find σk(θ) and σk for the scattering of a particle from a perfectly rigid sphere (an infinitely repulsive potential) of radius a. Choose the energy of the particle such that ka << 1.
Solution:
 | Concepts: Elastic scattering, the method of partial waves, s-wave scattering. |
| Reasoning: For very slow particles or very short-range potentials the method of partial waves is the preferred method of calculating the scattering cross section, because only s-waves need to be considered. |
| Details of the calculation: V(r) = 0, for r > a, and V(r) = ∞ for r < a. Since ka << 1, ka << (l(l + 1))1/2 for all l except l = 0. We can neglect all phase shifts except that of the s-wave. Then σk(θ) = |fk(θ)|2 = (1/k2)sin2δ0 and σk = (4π/k2)sin2δ0. To calculate δ0 we must solve the radial equation for l = 0. (∂2/∂r2 + k2)uk0(r) = 0, uk0(a) = 0 yields uk0(r) = Csin(kr - ka), r > a and uk0(r) = 0, r < a. As r --> ∞ we expect uk0(r) to be of the form Csin(kr + δ0). We therefore have δ0 = -ka. The phase shift is negative, the wave function is "pushed out". Since ka << 1 we now have (1/k2)k2a2 = a2 and σk = 4πa2. The total cross section is independent of energy (as long as ka << 1) and equal to four times the geometrical cross section of the hard sphere. Classical mechanics would have yielded πa2. Low energy scattering means very large wavelength scattering, and we do not necessarily expect a classically reasonable result. [But for high energy scattering off a hard sphere we might expect the classical result. However we obtain σk = 2πa2 twice the geometrical cross section. This is called shadow scattering. If the wavelength of the incident particle is very small compared to a, then the sphere will cast a shadow. Directly behind the sphere we will find no scattered particles, but the shadow will extend only up to a finite distance. Very far away from the sphere we will not see the shadow at all, so it must get filled in by scattering of some of the waves at the edges of the sphere. The scattered flux must have the same magnitude as the flux that is taken out of the incident beam by the geometrical cross section of the sphere in order to fill in the shadow. The total scattering cross section therefore must have twice the magnitude of the geometrical cross section.] |
Problem 2:
A slow particle is scattered by a spherical potential well of the form V(r) = -V0, for r < a, V(r) = 0, for r > a.
(a) Write down the radial wave equation for this potential and boundary conditions that apply at r = 0, r = a, and r = ∞.
(b) Assume that the de Broglie wavelength exceeds the dimension of the well, so that s-wave scattering dominates and write solutions both inside and outside the r = a sphere. Using the continuity conditions at r = a, calculate the phase shift that occurs at this boundary.
Solution:
| Concepts: Elastic scattering, the partial wave method |
| Reasoning: For low energy scattering from a finite-range, central potential, the partial wave method is the preferred method for calculating the scattering cross section. We find the scattering cross section by calculating the phase shifts of the partial waves. |
| Details of the calculation:(a) (∂2/∂r2 + k2 - U(r) - l(l + 1)/r2)ukl(r) = 0 is the radial equation. Here E = ħ2k2/(2m), V(r) = ħ2U(r)/(2m). We need ukl(0) = 0, ukl(r) and ∂ukl(r)/∂r to be continuous at r = a, and ukl(r) to stay finite at infinity. (b) To calculate δ0 we must solve the radial equation for l = 0. (∂2/∂r2 + k2 + U0)uk0(r) = 0, uk0(0) = 0, for r < a. Therefore uk0(r) = C1sin((k2 + U0)1/2r). (∂2/∂r2 + k2)uk0(r) = 0, for r > a. uk0(r) = C2sin(kr + δ0). We need uk0(r) and ∂uk0(r)/∂r to be continuous at r = a. This yields C1sin((k2 + U0)1/2a) = C2sin(ka + δ0) and (k2 + U0)1/2C1cos((k2 + U0)1/2a) = kC2cos(ka + δ0) or tan(ka + δ0) = (k/(k2 + U0)1/2)tan((k2 + U0)1/2a) = C. We have δ0 = nπ + tan-1(C) - ka. |
Problem 3:
A slow particle is scattered by a spherical potential well of the form V(r) = -V0, for r < a, V(r) = 0, for r > a.
What must V0a2 be for a 3-dimensional square well potential in order that the scattering cross section be zero at zero bombarding energy (Ramsauer-Townsend effect)?
Solution:
| Concepts: Elastic scattering, the partial wave method |
| Reasoning:As E --> 0, k --> 0, ka --> 0. If ka << 1 then the scattering is dominated by s-wave scattering and we use the partial wave method to calculate the scattering cross section. Then σk(θ) = (1/k2)sin2δ0 and σk = (4π/k2)sin2δ0. |
| Details of the calculation: When δ0 = π, then σk(θ) = σk = 0. To find what V0a2 has to be for δ0 = π, we need to find the expression for a 3-dimensional square-well potential. To calculate δ0 we must solve the radial equation for l = 0. Let E = ħ2k2/(2m), V0 = ħ2U0/(2m). Then (∂2/∂r2 + k2 + U0)uk0(r) = 0, uk0(0) = 0, for r < a. Therefore uk0(r) = C1sin((k2 + U0)1/2r). (∂2/∂r2 + k2)uk0(r) = 0, for r > a. uk0(r) = C2sin(kr + δ0). We need uk0(r) and ∂uk0(r)/∂r to be continuous at r = a. This yields C1sin((k2 + U0)1/2a) = C2sin(ka + δ0) and (k2 + U0)1/2C1cos((k2 + U0)1/2a) = kC2cos(ka + δ0) or tan(ka + δ0) = (k/(k2 + U0)1/2)tan((k2 + U0)1/2a). We want δ0 = π. Then tan(ka) = (k/(k2 + U0)1/2)tan((k2 + U0)1/2a). As ka --> 0 this becomes ka = (k/U01/2)tan(U01/2a) or U01/2a = tan(U01/2a) U01/2a = nπ + b. We are not interested in the solution n = 0, b = 0 for the free particle. n = 1, b = 1.352, U01/2a = 4.494, tan(U01/2a) = 4.497. V0a2 = ħ2U0a2/(2m) = 20.2 ħ2/(2m). The Ramsauer-Townsend effect is analogous to the perfect transmission found at particular energies in a one dimensional problem. It is the explanation for the extremely low minimum observed in the scattering cross section of electrons by rare gas atoms at about 0.7 eV bombarding energy. The Ramsauer-Townsend effect cannot occur with a repulsive potential, since ka would have to be at least π to make δ0 = π. (See problem 1. The perfectly rigid sphere produces the largest |δ0| for any repulsive potential with range a.) A potential of this large range produces higher l phase shifts. |
Problem 4:
Determine the differential scattering cross section σ(θ) in units of cm2/sr for a particle of mass m = 9.1*10-31kg incident on a spherically symmetric potential
V(r) = 0, 0 < r < a, V(r) = V0, a < r < b, V(r) = 0, r > b,
with a = 0.05 nm and b = 0.1 nm. Let E = 1 eV and V0 = 0.8 eV.Solution:
| Concepts: Elastic scattering, the method of partial waves, s-wave scattering. |
| Reasoning: E = ħ2k2/(2m), k2 = 2mE/ħ2, k = 5.1*109/m, kb = 0.51 < 1. Only s-wave scattering is important. We have σk(θ) = (1/k2)sin2δ0. |
| Details of the calculation: To calculate δ0 we must solve the radial equation for l = 0. (∂2/∂r2 + k2 - U(r))uk0(r) = 0, V(r) = ħ2U(r)/(2m). r < a: U(r) = 0, uk0(r) = C1sin(kr). a < r < b: U(r) = U0, V0 = ħ2U0/(2m), E > V0. Therefore uk0(r) = C2sin((k2 - U0)1/2r + δ') = C2sin((k1r + δ'). r > a: U(r) = 0, uk0(r) = C3sin(kr + δ0). We need uk0(r) and ∂uk0(r)/∂r to be continuous at r = a and r = b. This yields at r = a C1sin(ka) = C2sin(k1a + δ'), kC1cos(ka) = k1C2cos(k1a + δ'), or δ' = tan-1((k1/k)tan(ka)) - k1a. k1/k = ((E-V0)/E)1/2 = 0.447, ka = 0.256, δ' = 0.002. At r = b we have C2sin((k1b + δ') = C3sin(kb + δ0), k1C2cos((k1b + δ') = kC3cos(kb + δ0), or δ0 = tan-1((k/k1)tan(k1b + δ')) - kb. δ0 = -2.8*10-2. σk(θ) = (1/k2)sin2δ0 = 3*10-19 cm2/sr. |
Problem 5:
From measurements of the differential cross section for scattering of electrons off protons (in atomic hydrogen) it was found that the proton had a charge density given by ρ(r) = αexp(-βr) where α and β are constants.
(a) Find α and β such that the proton charge equals e, the charge on the electron.
(b) Show that the protons mean square radius <r2> is 12/β2.
(c) Assuming a reasonable value for <r2>1/2 calculate a in esu/cm3.
Solution:
| Concepts: Fundamental assumptions of Quantum Mechanics |
| Reasoning: We interpret |ψ(r)|2 as the probability density and assume that the probability density is proportional to the charge density. We then normalize the wave function ψ(r) and find the average value of the observable r2. |
| Details of the calculation: (a) 4π∫0∞ρ(r)r2dr = α4π∫0∞exp(-βr)r2dr = (α/β3)4π∫0∞exp(-r')r'2dr' = (α/β3)8π = e. (b) Assume ρ(r) = A2|ψ(r)|2. For a normalized wave function 4π∫0∞|ψ(r)|2r2dr = 1. Therefore A2 = β3/(8πα), |ψ(r)|2 = [β3/(8π)]exp(-βr) <r2> = 4π∫0∞|ψ(r)|2r4dr = [β3/2]∫0∞exp(-βr)r4dr = (1/(2β2))∫0∞exp(-r')r'4dr' = 4!/(2β2) = 12/β2. (c) Assume <r2>1/2 = 10-15 m = 10-13 cm. Then β2 = 12/<r2> = 1.2*1027 cm-2. (α/β3)8π = 4.8*10-10 esu, α = 7.9*1029 esu cm-3. |
Problem 6:
An electron of incident momentum ki is scattered elastically by the electric field of an atom of atomic number Z. The potential due to the nucleus is of the form V0(r) = -Ze2/r. This potential is screened by the atomic electron cloud. As a result, the total potential energy of the incident electron is V(r) = (-Ze2/r)exp(-r/a), where a is the radius of the atom. Let kf be the final momentum of the electron and q = kf - ki be the momentum transfer.
(a) Calculate the differential cross section, dσ/dΩ, in the Born approximation for the scattering of the electron by the atom (using the screened potential V(r)).
(b) Now calculate the differential cross section, dσ/dΩ, in the Born approximation for the scattering of the electron by the nucleus only (using the unscreened potential V0(r)).
(c) Plot the ratio of the two cross sections as a function of x = a|q| and briefly discuss the limits x --> 0 and x --> ∞.
Solution:
| Concepts: The Born approximation |
| Reasoning: We are asked to evaluate the scattering cross section in the Born approximation. |
| Details of the calculation:(a) In the Born approximation we have σkB(θ,φ) = σkB(k,k') = [μ2/(4п2ħ4)]|∫d3r' exp(-iq∙r')V(r')|2, where q = k' - k, k = μv0/h, k' = μv0/ħ (k'/k'), and μ is the reduced mass. Here μ = me. k is the incident wave vector, and (k'/k') is a unit vector pointing in the direction (θ,φ). With V(r) = V(r) = (-Ze2/r)exp(-r/a) we have ∫d3r' exp(-iq∙r')V(r') = -Ze2∫∫∫r'dr'sinθ'dθ'dφ'exp(-r'/a)exp(-iqr'cosθ') = -Ze22π∫0∞r'dr'exp(-r'/a)∫-11exp(-iqr'cosθ')dcosθ' = -(Ze22π/q)∫0∞dr'exp(-r'/a)∫-qr'qr'exp(-ix)dx = -(Ze22π/q)∫0∞dr'exp(-r'/a)[-i(eiqr' - e-iqr')] = -(Ze24π/q)∫0∞dr'exp(-r'/a)sin(qr') = -(Ze24π/q2)q2a2/(1 + q2a2) Therefore σkB(θ,φ) = [4μ2Z2e4/ħ4][1/((1/a2) + q2)]2. With q = 2ksin(θ/2) we have σkB(θ,φ) = [4μ2Z2e4/ħ4][1/((1/a2) + 4k2sin2(θ/2))]2. (b) Let a --> ∞, then V(r) = (-Ze2/r)exp(-r/a) --> V0(r) = -Ze2/r. Then σkB(θ,φ) = [4μ2Z2e4/ħ4]/(16k4sin4(θ/2)) = [Z2e4]/(16E2sin4(θ/2)) This is the Rutherford cross section. (c) Ratio: σkB(θ,φ)part a/σkB(θ,φ)part b = (q2a2/(1 + q2a2))2 = (x2/(1 + x2))2 with x = |qa|.  As x --> 0, the ratio approaches 0. As a --> 0, x --> 0, the nuclear charge is completely screened, we have a neutral object. As x --> ∞, the ratio approaches 1. As a --> ∞, x --> ∞, the screening vanishes, we have a bare nucleus. |
Problem 7:
Consider an electron of energy E0 and velocity v0 in the z-direction incident on an ionized Helium atom He+, with just one electron in its ground state. Compute the differential cross section dσ/dΩ for the incident electron to scatter into the solid angle dΩ about the spherical angles (θ,φ). Explain your assumptions and approximations.
Solution:
| Concepts:The Born approximation |
| Reasoning:For fast electrons we can use the Born approximation to calculate the scattering cross section. |
| Details of the calculation:In the Born approximation we have σkB(θ,φ) = σkB(k,k') = [μ2/(4п2ħ4)]|∫d3r' exp(-iq∙r')V(r')|2, where q = k' - k, k = μv0/h, k' = μv0/ħ (k'/k'), and μ is the reduced mass. Here μ = me, k is the incident wave vector, and (k'/k') is a unit vector pointing in the direction (θ,φ).    [   Take the limit as λ --> 0.]  where  is the atomic scattering form factor.   If we assume that F(q) = 1, i.e. that the incoming electron sees a screened nucleus of charge 1, we obtain the Rutherford scattering cross section. |
Problem 8:
Evaluate, in the Born Approximation, the differential cross section for the scattering of a particle of mass m by a delta-function potential V(r) = Bδ(r). Comment on the angular and velocity dependence. Find the total cross section.
Solution:
| Concepts:The Born Approximation |
| Reasoning: We are asked to evaluate the scattering cross section in the Born approximation. |
| Details of the calculation: In the Born approximation σkB(θ,φ) = σkB(k,k') = [μ2/(4п2ħ4)]|∫d3r' exp(-iq∙r')V(r')|2, ,  σkB(θ,φ) = [Bμ2/(4π2ħ4)]|∫d3r' exp(-iq∙r')δ(r')|2 = Bμ2/(4π2ħ4). The differential scattering cross section is independent of scattering angle and velocity. σkB = Bμ2/(πħ4) is the total scattering cross section. 22.54 中子与物质的相互作用及应用（2004年春季） 第七讲（ 2004 年 2 月 26 日） 中子弹性散射——热运动及化学键效应 参考文献—— J. R. Lamarsh, Introduction to Nuclear Reactor Theory (Addison-Wesley, Reading,1966), chap 2. S. Yip, 22.111 Lecture Notes (1975), chap 7. G. I. Bell and S. Glasstone, Nuclear Reactor Theory (Van Nostrand Reinhold, New York, 1970), chap 7. 所有的截面都是相互作用空间位置的点函数。核相互作用的范围远小于任何常见能量的中 子波长，因此可以认为中子相互作用是发生在一个点上而非发生在一个有限范围的区域上。 本讲侧重于讨论弹性散射截面 ) ( E σ 与入射中子能量 E 之间的依赖关系， E 指实验室系 （ LCS ） 下的中子能量。在上次课程里，已经推导出能量转移核 形式的出射中子能量分布， 但没有提及方程（ 6.1 ） 对能量的依赖关系，其原因就是 ) ( ' E E F → ) ( E σ 远复杂于 。对于处于 热中子能区的中子，研究 ) ( ' E E F → ) ( E σ 必须考虑到热运动及靶原子化学键结合能的影响。关于这些影 响有更多内容可以讨论，不仅是 ) ( E σ ，还需要考虑二阶微分散射截面 。本讲只 讨论总截面，对于双微分散射截面的讨论将在以后进行。关于 ' 2 / dE d d Ω σ ) ( E σ 随能量变化的定性理解可 以参考第二讲。 我们在 2003 年第三讲里提到过，当入射中子的速度远大于靶核速度时，从简化运动学分析 的角度来说，假定靶核静止不动是一个很好的近似。在第四讲关于截面的讨论中，采用这种近 似把一个二体碰撞问题转化为等效的单体问题——在势场 V(r) 中的粒子散射。这里，矢量 r 指中 子对靶核的相对位置。这样，需要在质心系（ CMCS ）里求解薛定谔方程，由此得到的截面也 是指质心系（ CMCS ）的截面。前面已经知道，对于低能散射来说，只需要考虑 s 波的贡献，这 样，角分布微分截面是球对称的，总截面是一个常数（ ）。 2 4 a π = 热运动效应 在热中子能区，中子能量与靶核能量是相当的，后者服从由温度决定的麦克斯维尔分布。 这时，假定靶核静止就不合理了。为了将靶核热运动的影响考虑进去，需要指明靶核的物理状 态，如是晶体靶还是液体靶。关于靶核动力学方面的讨论将在后面的一讲中进行，并只考虑一 种较为简单的情况：中子在处于热平衡状态的气体中的弹性散射。此时，靶核的运动轨迹为直 线，其速度服从于一个与靶核温度 T 有关的麦克斯维尔分布。 必须注意到，在实验室中测到的截面与理论计算出的截面是有区别的。为讨论明确起见， 前者用 obs σ 表示，后者用 theo σ 表示。则 obs σ 是实验室系（ LCS ）下中子能量的函数（既然测量 是在实验室内进行的）；而 theo σ 是质心系（ CMCS ）下中子能量的函数，或者更精确地说，是 方程 （ 4.6 ） 中的相对能量 E 的函数。 即： ) ( v obs obs σ σ = ， ( theo theo v V σ σ = − ) ， 其中 v 和 V 分 别是中子和靶核在实验室系（ LCS ）下的速度。两种截面之间的关系如下： 3 ( ) ( ) obs theo v v V v V P V σ σ = − − ∫ d V (7.1) 其中， ( ) P V 是靶核速度服从的麦克斯维尔分布。方程（ 7.1 ）是实验室系（ LCS ）下和质心系 （ CMCS ）下散射率之间的基本关系。 （严格来说，方程（ 7.1 ）两边都乘上靶核密度才能得到散 射率。 ） |
No comments:
Post a Comment