Chapter40

# Chapter40 - Solve: Absorption occurs from the ground state...

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40.1. 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 En h mL n = 2 2 2 8 The absorbed photons must have just the right energy, so Eh f hc EE E h mL L h mc ph elec J s m 8 9.11 10 kg m / s m 0.739 nm == = =−= ⇒= = × () × × × = −− λ 21 2 2 34 7 31 8 10 3 8 3 8 3 6 63 10 6 00 10 30 10 739 10 .. . .

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40.2. 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 En h mL n = 2 2 2 8 The emitted photons must have just the right energy, so Eh f hc EE E h mL ph elec == = =−= λ 32 2 2 5 8 ⇒= = × () × × × = −− L h mc 5 8 5 6 63 10 1484 10 30 10 15 10 34 9 31 8 9 . . . J s m 8 9.11 10 kg m / s m 1.5 nm
40.3. Model: Model the electron as a particle in a rigid one-dimensional box of length L . Visualize: Please refer to Figure Ex40.3. Solve: The energy levels for a particle in a rigid box are En L n = 2 22 2 2 π h The wave function shown in Figure Ex40.3 corresponds to n = 3. This is also shown in Figure 40.7. Thus, L mE h mE == = × () × ×× = −− 3 2 3 3 6 63 10 2291 11 0 60 16 1 0 33 34 31 19 h . .. . J s kg eV J / eV 0.752 nm

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40.4. Model: Model the electron as a particle in a rigid one-dimensional box of length L . Visualize: Please refer to Figure Ex40.4. Solve: From Equation 40.23, 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 40.7 or from the wave function equation ψπ n xAn x L () = () sin that the wave function given in Figure Ex40.4 corresponds to n = 4. Thus, E 4 = 16 E 1 ⇒= = = E E 1 4 16 12.0 eV 16 0.75 eV
40.5. Solve: From Equation 40.41, the units of the penetration distance are η = () × × = × × × = × × = × × = h 2 0 mU E Js kg J kg m / s s kg kg m / s kg m / s kg m / s kg m / s kg m / s m 22 2 2 2

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40.6. Visualize: Please refer to Figure 40.14 (a). 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.
40.7. Model: The wave function decreases exponentially in the classically forbidden region. Solve: The probability of finding a particle in the small interval δ x at position x is Prob(in x at x ) = | ψ ( x )| 2 x . Thus the ratio Prob in at Prob in at () |( ) | |() | ) | | δη ψη ψδ xxL Lx L L =+ = = + = + 2 2 2 2 The wave function in the classically forbidden region x L is ψψ η / xe xL = −− edge At the edge of the forbidden region, at x = L , ( L ) = edge . At x = L + , ( L + ) = edge e –1 . Thus Prob in at Prob in at edge edge ) | | . L L e e = = + == = 2 2 12 2 2 0 135

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40.8.
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## This note was uploaded on 02/27/2010 for the course PHYS 161,260,27 taught by Professor Staff during the Spring '08 term at Maryland.

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Chapter40 - Solve: Absorption occurs from the ground state...

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