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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 πε 0 r 2 dr = q 2 4 πε 0 1 r 2 dr = q 2 4 πε 0 - 1 r + const = E P To find the constant, we employ the boundary condition that at r = E P =E vac : 2 0 ( ) 0 4 P vac q E r E const const πε - = = = + = + ץ    const = E vac and 2 0 4 P vac vac q E E E πε - = - = ץ , 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 19 3 3(1.6 10 ) q C - = + . The results of the key steps are 2 1 2 2 2 0 0 3 4 4 Q Q q F r r πε πε - = = 2 0 3 ( ) 4 P vac q E r E r πε = - 2 2 2 0 3 0 4 mv q r r πε - = n n mv r n = h ( 29 2 0 3 1 4 n q v n πε � � = � � � � h ( 29 2 0 2 2 4 ( ) 3 n r n mq πε = h ( 29 2 4 2 2 2 0 3 1 2 4 K m q E n πε = h ( 29 4 2 2 2 0 9 1 2 4 n Pn Kn vac mq E E E E n πε = + = - h Thus ( 29 1 9 13.6 122 vac vac E E eV E eV = - = - Anderson  & Anderson 1 2/15/04 Solutions Chapter  1

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and ( 29 2 2 9 13.6 30.6 2 vac vac eV E E E eV = - = - and ( 29 3 2 9 13.6 13.6 3 vac vac eV E E E eV = - = - 1.3. Show  that Equations (1.12) and (1.13) follow  from (1.8) and (1.11). Equation (1.8) is: 2 2 2 0 0 4 mv q r r πε - = Multiply both sides by r 2 and divide by v : 2 0 4 q mvr v πε = which from Equation (1.11) is 2 0 4 q mvr n v πε = = h Solving the right hand equality for v : 2 0 4 q v n πε = h (Equation (1.13)) Solving the left hand equality for r substituting in v : ( 29 2 2 0 0 2 2 4 4 n n n n r mv mq mq πε πε = = = h h h h (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.

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