Chapter 41

# Chapter 41 - 41.1. Model: Model the electron as a particle...

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41.1. Model: Model the electron as a particle in a rigid one-dimensional box of length L . Solve: Absorption occurs from the ground state n = 1. It’s reasonable to assume that the transition is from n = 1 to n = 2. The energy levels of an electron in a rigid box are 2 2 2 8 n h En mL = The absorbed photons must have just the right energy, so () ( ) ( ) 2 ph elec 2 1 2 34 7 10 31 8 3 8 3 6.63 10 J s 6.00 10 m 3 7.39 10 m 0.739 nm 8 8 9.11 10 kg 3.0 10 m/s hc h Eh f E E E mL h L mc λ −− === Δ =−= ×× ⇒= = = × =

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41.2. Model: Model the electron as a particle in a rigid one-dimensional box of length L . Solve: (a) The wavelength 1484 nm is in the infrared range. (b) The energy levels of an electron in a rigid box are 2 2 2 8 n h En mL = The emitted photons must have just the right energy, so 2 ph elec 3 2 2 5 8 hc h Eh f E E E mL λ === Δ =−= () ( ) ( ) 34 9 9 31 8 5 6.63 10 J s 1484 10 m 5 1.5 10 m 1.5 nm 8 89 .11 10 kg 3 .0 10 m / s h L mc −− ×× ⇒= = = × =
41.3. Model: Model the electron as a particle in a rigid one-dimensional box of length L . Solve: The energy levels for a particle in a rigid box are 22 2 2 2 n En L π = = The wave function shown in Figure Ex41.3 corresponds to n = 3. This is also shown in Figure 41.7. Thus, ( ) () ( ) 34 31 19 33 36 .63 10 J s 0.75 nm 2 2 2 9.11 10 kg 6.0 eV 1.6 10 J/eV h L mE mE −− × == = = ×× × =

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41.4. Model: Model the electron as a particle in a rigid one-dimensional box of length L . Solve: From Equation 41.23, the energies of the stationary states for a particle in a box are E n = n 2 E 1 , where E n is the energy of the stationary state with quantum number n . It can be seen either from Figure 41.7 or from the wave function equation () sin n x An x L ψπ = that the wave function given in Figure Ex41.4 corresponds to n = 4. Thus, 4 41 12.0 eV 16 0.75 eV 16 16 E EE E =⇒ = = =
41.5. Solve: From Equation 41.41, the units of the penetration distance are () ( ) 22 2 0 kg m /s s Js k gm / s k / s m kg m/s kg J 2 kg kg m /s kg m /s mU E η ×× × =⇒ = = = = × × =

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41.6. Solve: (a) (b) For n = 2, the probability of finding the particle at the center of the well is zero. This is because the wave function is zero at that point. (c) This is consistent with standing waves. The n = 2 standing wave on a string has a node at the center of the string.
41.7. Model: The wave function decreases exponentially in the classically forbidden region. Solve: The probability of finding a particle in the small interval x δ at position x is Prob(in x at x ) 2 =| ( )| . x x ψ Thus the ratio 22 P r o b ( i n a t ) | () | | | Prob(in at ) | ( ) | | ( ) | xxL L x L L x L δη η δψ =+ + + == = The wave function in the classically forbidden region x L is / edge xL xe ψψ = At the edge of the forbidden region, at x = L , ( L ) = edge . At x = L + , ( L + ) = edge e –1 . Thus 12 2 edge 2 edge Prob(in at ) | ( ) | 0.135 Prob(in at ) | ( ) | ( ) e L e L + = = =

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41.8. Solve: (a) According to Equation 41.41, the penetration distance is () 34 31 19 0 1.05 10 J s 0.159 nm 2 2 9.11 10 kg 2.0 eV 0.5 eV 1.60 10 J/eV mU E η −− × == = ×− × × = (b) Likewise for E = 1.00 eV, 0.195 nm. = (c) For E = 1.50 eV, = 0.275 nm.
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## This note was uploaded on 01/30/2011 for the course PHYS 131 taught by Professor E.salik during the Winter '10 term at Cal Poly Pomona.

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Chapter 41 - 41.1. Model: Model the electron as a particle...

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