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chapter01sm

# chapter01sm - SOLUTIONS CHAPTER 1 1.1 Show that...

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SOLUTIONS CHAPTER 1 1.1. Show that Equation (1.6) follows from Equation (1.3). Sol ution: Equation (1.3) is dE P = q 2 4 πε 0 r 2 dr . Integrating both sides we obtain dE P = q 2 4 0 r 2 dr = q 2 4 0 1 r 2 dr = q 2 4 0 1 r + const = E P To find the constant, we employ the boundary condition that at , E r =∞ P =E vac : 2 0 () 0 4 Pv a c q Er E c o n s t c o n s t =∞ = = + = + ⋅∞ const = E vac and 2 0 4 P vac vac q EE E =− = , Equation (1.6) 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 Q 1 is equal to . The results of the key steps are 19 33 ( 1 . 6 1 0 ) qC =+ × 2 12 22 00 3 44 QQ q F rr π επε == 2 0 3 4 a c q Er E r 2 0 3 0 4 mv q −= nn mv r n = = 2 0 31 4 n q v n ⎛⎞ = ⎜⎟ ⎝⎠ = 2 0 2 2 4 3 n rn mq = = 24 2 0 K mq E n = = 4 2 0 91 nP nK nv a c mq EEE E n =+=− = Thus ( ) 1 9 13.6 122 vac vac E Ee V E e V Anderson & Anderson 1 2/15/04 Solutions Chapter 1

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and () 2 2 9 13.6 30.6 2 vac vac eV EE E e V =− and 3 2 9 13.6 13.6 3 vac vac eV E e V 1.3. Show that Equations (1.12) and (1.13) follow from (1.8) and (1.11). Equation (1.8) is: 22 2 0 0 4 mv q rr πε −= Multiply both sides by r 2 and divide by : v 2 0 4 q mvr v = which from Equation (1.11) is 2 0 4 q mvr n v == = Solving the right hand equality for : v 2 0 4 q v n = = (Equation (1.13)) Solving the left hand equality for r substituting in : v 0 0 2 4 4 nn n n r mv mq mq = = = 2 (Equation (1.12)). 1.3. In each of the potential energy distributions in Figure 1P.1, sketch the magnitude and direction of the force on the electron. Anderson & Anderson 2 2/15/04 Solutions Chapter 1
The force is minus the gradient of the potential energy (Equation (1.2)). 1.5. Consider the electron in the energy diagram of Figure 1P.2. Taking the energy the electron has at Point A as E total , at each of the indicated positions, find the total energy, the kinetic energy, the potential energy, and the electron’s velocity. Indicate the direction of force (if any). Recall that total energy is conserved. At point “D” the electron collides inelastically with something (perhaps an atom in the crystal). After the collision, the electron’s energy is equal to its potential energy, and its kinetic energy is zero. Its total energy is much less than before the collision; where did the extra energy go? The electron at Point A is at rest. Its total energy is E total , its kinetic energy and velocity are zero, and its potential energy is equal to its total energy. The force on the electron is to the right, because the slope of the potential energy is negative. At Point B, the total energy of the electron is the same (by conservation of energy), but the potential energy is E P ( B ) = E total 5 eV . The kinetic energy is the difference between the total and the potential energies, so E K ( B ) = E total E P ( B ) = E total E total 5 eV ( ) = 5 eV . The velocity of the electron is ( ) 19 6 31 2( 5 ) 1 .

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chapter01sm - SOLUTIONS CHAPTER 1 1.1 Show that...

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