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Mathematical Methods for Physics and Engineering - Matematica.NET

Mathematical Methods for Physics and Engineering - Matematica.NET

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4.9 HINTS AND ANSWERSfind a closed-<strong>for</strong>m expression <strong>for</strong> α, the Madelung constant <strong>for</strong> this (unrealistic)lattice.4.35 One of the factors contributing to the high relative permittivity of water to staticelectric fields is the permanent electric dipole moment, p, of the water molecule.In an external field E the dipoles tend to line up with the field, but they do notdo so completely because of thermal agitation corresponding to the temperature,T , of the water. A classical (non-quantum) calculation using the Boltzmanndistribution shows that the average polarisability per molecule, α, isgivenbyα = p E (coth x − x−1 ),where x = pE/(kT) <strong>and</strong>k is the Boltzmann constant.At ordinary temperatures, even with high field strengths (10 4 Vm −1 or more),x ≪ 1. By making suitable series expansions of the hyperbolic functions involved,show that α = p 2 /(3kT) to an accuracy of about one part in 15x −2 .4.36 In quantum theory, a certain method (the Born approximation) gives the (socalled)amplitude f(θ) <strong>for</strong> the scattering of a particle of mass m through an angleθ by a uni<strong>for</strong>m potential well of depth V 0 <strong>and</strong> radius b (i.e. the potential energyof the particle is −V 0 within a sphere of radius b <strong>and</strong> zero elsewhere) asf(θ) = 2mV 0(sin Kb − Kbcos Kb). 2 K3 Here is the Planck constant divided by 2π, the energy of the particle is 2 k 2 /(2m)<strong>and</strong> K is 2k sin(θ/2).Use l’Hôpital’s rule to evaluate the amplitude at low energies, i.e. when k <strong>and</strong>hence K tend to zero, <strong>and</strong> so determine the low-energy total cross-section.[ Note: the differential cross-section is given by |f(θ)| 2 <strong>and</strong> the total crosssectionby the integral of this over all solid angles, i.e. 2π ∫ π0 |f(θ)|2 sin θdθ.]4.9 Hints <strong>and</strong> answers4.1 Write as 2( ∑ 1000n=1 n − ∑ 499n=1n) = 751500.4.3 Divergent <strong>for</strong> r ≤ 1; convergent <strong>for</strong> r ≥ 2.4.5 (a) ln(N + 1), divergent; (b) 1 [1 − 3 (−2)n ], oscillates infinitely; (c) Add 1 S 3 N to theS N series; 3 [1 − 16 (−3)−N ]+ 3 4 N(−3)−N−1 ,convergentto 3 . 164.7 Write the nth term as the difference between two consecutive values of a partialfractionfunction of n. Thesumequals 1 (1 − 2 N−2 ).4.9 Sum the geometric series with rth term exp[i(θ + rα)]. Its real part is{cos θ − cos [(n +1)α + θ] − cos(θ − α)+cos(θ + nα)} /4sin 2 (α/2),which can be reduced to the given answer.4.11 (a) −1 ≤ x

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