SOLUTIONS CHAPTER 1 1.1. Show that Equation (1.6) follows from Equation (1.3). q2 Solution: Equation (1.3) is dEP = dr . Integrating both sides we obtain 4πε 0 r 2 1 q2 ⎡ 1 ⎤ q2 q2 dE = dr = dr = − + const = EP 2 ∫ P ∫ 4πε 0r2 4πε0 ∫ r 4πε 0 ⎢⎣ r ⎥⎦ To find the constant, we employ the boundary condition that at r = ∞ , EP=Evac: −q 2 EP (r = ∞) = Evac = + const = 0 + const 4πε 0 ⋅ ∞ ∴const = Evac and EP = Evac − −q 2 = Evac , Equation (1.6) 4πε 0 ⋅ ∞ 1.2. Consider a lithium nucleus, of charge +3q. Calculate the first three electron energies for an electron in a Li++ ion, using the Bohr model. We repeat the analysis that we used for the hydrogen atom, except that now the charge of the nucleus Q1 is equal to 3q = +3(1.6 × 10−19 )C . The results of the key steps are QQ −3q 2 F = 1 22 = 4πε 0 r 4πε 0 r 2 EP (r ) = Evac − 3q 2 4πε 0 r mv 2 3q 2 − =0 r 4πε 0 r 2 mvn rn = nh vn = rn = ⎛1⎞ ( 4πε 0 ) h ⎜⎝ n ⎟⎠ 3q 2 ( 4πε 0 ) h 2 EK = 3mq 2 (n 2 ) m32 q 4 2 ( 4πε 0 ) 2 ⎛ 1 ⎞ ⎜ 2⎟ h ⎝n ⎠ 2 E n = EPn + EKn = Evac − 9mq 4 2 ( 4πε 0 ) 2 ⎛ 1 ⎞ ⎜ 2⎟ h ⎝n ⎠ 2 Thus E1 = Evac − 9 (13.6eV ) = Evac − 122eV Anderson & Anderson Solutions Chapter 1 1 2/15/04 E2 = Evac − and and E3 = Evac − 9 (13.6eV ) 22 9 (13.6eV ) 32 = Evac − 30.6eV = Evac − 13.6eV 1.3. Show that Equations (1.12) and (1.13) follow from (1.8) and (1.11). Equation (1.8) is: mv ...
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