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Hwksolns14_Publishers - 41.6. Wave function a is an...

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41.6. Wave function a is an antibonding orbital and wave function b is a bonding orbital. You can tell because in the bottom one the electron has a high probability of being found between adjacent protons indicating that the adjacent protons are sharing the electron.
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41.7. tunnel d tunnel b tunnel a tunnel c () PPPP >>> because 2 tunnel w Pe η = and 0 2( ) mU E = = .
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41.16. Model: The electron is a quantum harmonic oscillator. Solve: (a) The energy levels of a harmonic oscillator are ( ) 35 11 2 222 ,,, En ωω ω =− = == = = The classical angular frequency of a mass on a spring is 15 31 2.0 N/m 1.48 10 rad/s 9.11 10 kg k m × () ( ) 34 15 19 1.05 10 J s 1.48 10 rad/s 1.56 10 J 0.972 eV −− ⇒= × × = × = = The first three energy levels are E 1 = 0.49 eV, E 2 = 1.46 eV, and E 3 = 2.43 eV. (b) The photon energy equals the energy lost by the electron: E photon = Δ E elec = E 3 E 1 = 1.94 eV 19 3.10 10 J. The wavelength is ( )( ) 34 8 19 elec 6.63 10 J s 3.0 10 m/s 640 nm 3.10 10 J ch c fE λ ×× = = Δ×
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41.17. Model: See Example 41.9. Visualize: For a harmonic oscillator the energy difference between adjacent energy levels is e E ω Δ= = . We also know the emitted photon has energy photon photon . E hf E = Combine these to get e 2. c λ π = Solve: We also recall that e . k m = Solve this for k and substitute for e . 2 2 8 23 1 e 9 30 10 ±m/s 2 (911 10 ±kg) 2 225±N/m 1200 10 m c km m ωπ ⎛⎞ == = . × = . ⎜⎟ × ⎝⎠ Assess: This answer is one-fourth of the answer in Example 41.9, which makes sense. An electron would emit a 1200 nm photon in any 1 nn →− jump in this quantum harmonic oscillator; not just the 3 2 jump.
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41.20. Solve: Electrons are bound inside metals by an amount of energy called the work function E 0 . This is the energy that must be supplied to lift an electron out of the metal. In our case, E 0 is the amount of energy ( U 0 E ) appearing in Equation 41.41 for the penetration distance. Thus, () ( ) ( ) 34 11 31 19 0 1.05 10 J s 9.72 10 m 2 2 9.11 10 kg 4.0 eV 1.60 10 J/eV mU E η −− × == = × ×× × = The probability that an electron will tunnel through a w = 4.54 nm gap (from a metal to an STM probe) is ( ) ( ) 91 1 2 0.45 10 m 9.72 10 m 25 tunnel 9.5 10 0.0095% w Pe e −× × = × =
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41.21. Solve: (a) The probability of an electron tunneling through a barrier is 2 tunnel w Pe η = () 0 2 mU E = = [] ( ) 2 2 2 0 2 tunnel tunnel 0 tunnel 22 2 ln 2 ln 4 ln 8 PwP w E U P mw ⇒= = = = = ( ) 2 34 2 tunnel 2 19 31 9 1.05 10 J s 1 eV 5.0 eV ln 1.60 10 J 8 9.11 10 kg 1.0 10 m EP −− × × ×× [ ] 2 tunnel 5.0 eV 0.009455 eV ln P =− For P tunnel = 0.10, E = 5.0 eV – 0.050 eV = 4.95 eV. (b) For P tunnel = 0.010, E = 5.0 eV – 0.20 eV = 4.80 eV. (c) For P tunnel = 0.0010, E = 5.0 eV – 0.45 eV = 4.55 eV.
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41.31. Model: The nucleus can be modeled as a potential well. Visualize: Please refer to Figure 41.17. Solve: The gamma ray wavelength λ = 1.73 × 10 4 nm corresponds to a photon energy of E photon = hc/ = 7.2 MeV. From Fig. 41.17, we can see that a photon of this energy is emitted in a transition from the n = 2 to n = 1 energy level. This can happen after a proton-nucleus collision if the proton’s impact excites the nucleus from the n = 1 ground state to the n = 2 excited state. To cause such an excitation, the proton’s kinetic energy at the instant of impact must be K 7.2 MeV. Let v 1
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This note was uploaded on 12/26/2011 for the course PHYSICS 270 taught by Professor Drake during the Fall '08 term at Maryland.

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Hwksolns14_Publishers - 41.6. Wave function a is an...

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