40_InstructorSolutions

# 40_InstructorSolutions - QUANTUM MECHANICS 40 n2h 2 8mL2...

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40-1 Q UANTUM M ECHANICS 40.1. IDENTIFY and SET UP: The energy levels for a particle in a box are given by 22 2 . 8 n nh E mL = EXECUTE: (a) The lowest level is for 1, n = and 34 2 67 1 2 (1)(6.626 10 J s) 1.2 10 J. 8(0.20 kg)(1.5 m) E ×⋅ == × (b) 2 1 2 Em v = so 67 33 2 2(1.2 10 J) 1.1 10 m/s. 0.20 kg E v m × = × If the ball has this speed the time it would take it to travel from one side of the table to the other is 33 33 1.5 m 1.4 10 s. 1.1 10 m/s t × × (c) 2 12 1 2 , 4 , 8 h EE E mL so 67 67 21 1 3 3(1.2 10 J) 3.6 10 J EE E E −− Δ= − = = × = × (d) EVALUATE: No, quantum mechanical effects are not important for the game of billiards. The discrete, quantized nature of the energy levels is completely unobservable. 40.2. 1 8 h L mE = 34 15 27 6 19 (6.626 10 J s) 6.4 10 m. 8(1.673 10 kg)(5.0 10 eV)(1.602 10 J eV) L × ×× × 40.3. IDENTIFY: An electron in the lowest energy state in this box must have the same energy as it would in the ground state of hydrogen. SET UP: The energy of the n th level of an electron in a box is 2 2 . 8 n nh E mL = EXECUTE: An electron in the ground state of hydrogen has an energy of 13.6 eV, so find the width corresponding to an energy of 1 13.6 eV. E = Solving for L gives 1 8 h L mE = 34 10 31 19 (6.626 10 J s) 1.66 10 m. 8(9.11 10 kg)(13.6 eV)(1.602 10 J eV) × EVALUATE: This width is of the same order of magnitude as the diameter of a Bohr atom with the electron in the K shell. 40.4. (a) The energy of the given photon is 3 34 18 9 (3.00 10 m/s) (6.63 10 J s) 1.63 10 J. (122 10 m) c Eh f h λ × == = × = × × The energy levels of a particle in a box are given by Eq.40.9 2 2 2 2 () . 8 h En n mL 3 4 22 2 10 31 20 ( 6 . 6 3 1 0 J s ) ( 2 1 ) 3.33 10 m. 8 8(9.11 10 kg)(1.63 10 J) hn n L mE −× = × Δ× × (b) The ground state energy for an electron in a box of the calculated dimensions is 23 4 2 19 11 0 2 (6.63 10 J s) 5.43 10 J 3.40 eV 8 8(9.11 10 kg)(3.33 10 m) h E mL = × = (one-third of the original photon energy), which does not correspond to the 13.6 eV ground state energy of the hydrogen atom. Note that the energy levels for a particle in a box are proportional to 2 , n whereas the energy levels for the hydrogen atom are proportional to 2 1 . n 40

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40-2 Chapter 40 40.5. IDENTIFY and SET UP: Eq.(40.9) gives the energy levels. Use this to obtain an expression for 21 EE and use the value given for this energy difference to solve for L. EXECUTE: Ground state energy is 2 1 2 ; 8 h E mL = first excited state energy is 2 2 2 4 . 8 h E mL = The energy separation between these two levels is 2 2 3 . 8 h EE E mL Δ= − = This gives 3 8 Lh mE == Δ 34 10 31 19 3 6.626 10 J s 6.1 10 m 0.61 nm. 8(9.109 10 kg)(3.0 eV)(1.602 10 J/1 eV) L −− = × = ×× EVALUATE: This energy difference is typical for an atom and L is comparable to the size of an atom. 40.6. (a) The wave function for 1 n = vanishes only at 0 x = and xL = in the range 0 . ≤≤ (b) In the range for , x the sine term is a maximum only at the middle of the box, / 2. = (c) The answers to parts (a) and (b) are consistent with the figure.
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## This homework help was uploaded on 03/14/2008 for the course MSE 250 taught by Professor Jabbour during the Spring '08 term at Arizona.

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40_InstructorSolutions - QUANTUM MECHANICS 40 n2h 2 8mL2...

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