SOLUTION MANUAL MODERN PHYSICS WITH MODERN COMPUTATIONAL METHODS: FOR SCIENTISTS AND ENGINEERS 3RD EDITION BY MORRISON CHAPTERS 1- 15

1 The Wave-Particle Duality - Solutions 1. The energy of photons in terms of the wavelength of light is given by Eq. (1.5). Following Example 1.1 and substituting λ = 200 eV gives: Ephoton = hc 1240 eV · nm = = 6.2 eV λ 200 nm 2. The energy of the beam each second is: power = Etotal = time 100 W = 100 J 1 s The number of photons comes from the total energy divided by the energy of each photon (see Problem 1). The photon’s energy must be converted to Joules using the constant 1.602 × 10−19 J/eV , see Example 1.5. The result is: Etotal 100 J N = = = 1.01 × 1020 photons Epho ton 9.93 × 10−19 for the number of photons striking the surface each second. 3. We are given the power of the laser in milliwatts, where 1 mW = 10−3 W . The power may be expressed as: 1 W = 1 J/s. Following Example 1.1, the energy of a single photon is: 1240 eV · nm = 1.960 eV hc Ephoton = λ = 632.8 nm We now convert to SI units (see Example 1.5): 1.960 eV × 1.602 × 10−19 J/eV = 3.14 × 10−19 J Following the same procedure as Problem 2: 1 × 10−3 J/s 15 photons Rate of emission = 3.14 × 10−19 J/photon = 3.19 × 10 s 2 4. The maximum kinetic energy of photoelectrons is found using Eq. (1.6) and the work functions, W, of the metals are given in Table 1.1. Following Problem 1, Ephoton = hc/λ = 6.20 eV . For part (a), Na has W = 2.28 eV : (KE)max = 6.20 eV − 2.28 eV = 3.92 eV Similarly, for Al metal in part (b), W = 4.08 eV giving (KE)max = 2.12 eV and for Ag metal in part (c), W = 4.73 eV , giving (KE)max = 1.47 eV . 5. This problem again concerns the photoelectric effect. As in Problem 4, we use Eq. (1.6): (KE)max = hc − W λ where W is the work function of the material and the term hc/λ describes the energy of the incoming photons. Solving for the latter: hc λ = (KE)max + W = 2.3 eV + 0.9 eV = 3.2 eV Solving Eq. (1.5) for the wavelength: 1240 eV · nm λ = 3.2 = 38