Simple Nature - Light and Matter

Page 549, problem 29:(a) Conservation of energy givesU A = U B + K BK B = U A − U B12 mv2 = e∆V√2e∆Vv =m(b) Plugging in numbers, we get 5.9 × 10 7 m/s. This is about 20% of the speed of light, so thenonrelativistic assumption was good to at least a rough approximation.Page 550, problem 32: It’s much more practical to measure voltage differences. To measurea current, you have to break the circuit somewhere and insert the meter there, but it’s notpossible to disconnect the circuits sealed inside the board.Solutions for Chapter 10Page 635, problem 16: By symmetry, the field is always directly toward or away from thecenter. We can therefore calculate it along the x axis, where r = x, and the result will be validfor any location at that distance from the center.E = − dVdx= − d (x −1 e −x)dx= −x −2 e −x − x −1 e −xAt small x, near the proton, the first term dominates, and the exponential is essentially 1, so wehave E ∝ x −2 , as we expect from the Coulomb force law. At large x, the second term dominates,and the field approaches zero faster than an exponential.Page 643, problem 56:(sin(a + b) = e i(a+b) − e −i(a+b)) /2i(= e ia e ib − e −ia e −ib) /2i= [(cos a + i sin a)(cos b + i sin b) − (cos a − i sin a)(cos b − i sin b)] /2i= [(cos a + i sin a)(cos b + i sin b) − (cos a − i sin a)(cos b − i sin b)] /2i= cos a sin b + sin a cos bBy a similar computation, we find cos(a + b) = cos a cos b − sin a sin b.Page 643, problem 57: If z 3 = 1, then we know that |z| = 1, since cubing z cubes itsmagnitude. Cubing z triples its argument, so the argument of z must be a number that, whentripled, is equivalent to an angle of zero. There are three possibilities: 0×3 = 0, (2π/3)×3 = 2π,and (4π/3) × 3 = 4π. (Other possibilities, such as (32π/3), are equivalent to one of these.) Thesolutions are:z = 1, e 2πi/3 , e 4πi/3938 Chapter Appendix 4: Hints and Solutions