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Unformatted text preview: 40.1: a) 2 2 34 2 2 1 2 2 2 m) kg)(1.5 20 . ( 8 s) J 10 63 . 6 ( 8 8 ⋅ × = = ⇒ = mL h E mL h n E n J. 10 22 . 1 67 1 × = ⇒ E b) s m 10 1 . 1 kg 0.20 J) 10 2 . 1 ( 2 2 2 1 33 67 2 × = × = = ⇒ = m E v mv E s. 10 4 . 1 s m 10 1.1 m 5 . 1 33 33 × = × = = ⇒ v d t c) J. 10 3.7 J) 10 22 . 1 ( 3 8 3 ) 1 4 ( 8 67 67 2 2 2 2 1 2 × = × = = = mL h mL h E E d) No. The spacing between energy levels is so small that the energy appears continuous and the balls particlelike (as opposed to wavelike). 40.2: 1 8 mE h L = m. 10 4 . 6 ) eV J 10 eV)(1.602 10 kg)(5.0 10 8(1.673 s) J 10 626 . 6 ( 15 19 6 27 34 × = × × × ⋅ × = 40.3: ) ( 8 3 8 3 ) 1 4 ( 8 1 2 2 2 2 2 1 2 E E m h L mL h mL h E E = ⇒ = = ) eV J 10 eV)(1.60 kg)(3.0 10 8(9.11 3 s) J 10 63 . 6 ( 19 31 34 × × ⋅ × = nm. 0.61 m 10 1 . 6 10 = × = ⇒ L 40.4: a) The energy of the given photon is J. 10 63 . 1 m) 10 (122 m/s) 10 (3.00 s) J 10 63 . 6 ( λ 18 9 3 34 × = × × ⋅ × = = = c h hf E The energy levels of a particle in a box are given by Eq. 40.9 J) 10 kg)(1.63 10 11 . 9 ( 8 ) 1 2 ( s) J 10 63 . 6 ( 8 ) ( ) ( 8 20 31 2 2 2 34 2 2 2 2 2 2 2 × × ⋅ × = ∆ = ⇒ = ∆ E m m n h L m n mL h E m. 10 33 . 3 10 × = b) The ground state energy for an electron in a box of the calculated dimensions is eV 3.40 J 10 43 . 5 19 m) 10 kg)(3.33 10 11 . 9 ( 8 ) s J 10 63 . 6 ( 8 2 10 31 2 34 2 2 = × = = = × × ⋅ × mL h E (onethird of the original photon energy), which does not correspond to the eV 6 . 13 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 . 40.5: J 10 5 . 2 ) s m kg)(0.010 10 00 . 5 ( 7 2 3 2 1 2 2 1 × = × = = mv E b) 1 2 2 1 8 so 8 mE h L mL h E = = m 10 6 . 6 gives J 10 5 . 2 30 7 1 × = × = L E 40.6: a) The wave function for 1 = n vanishes only at = x and L x = in the range . L x ≤ ≤ b) In the range for , x the sine term is a maximum only at the middle of the box, . 2 / L x = c) The answers to parts (a) and (b) are consistent with the figure. 40.7: The first excited state or ) 2 ( = n wave function is . 2 sin 2 ) ( 2 = L πx L x ψ So . 2 sin 2 ) ( 2 2 2 = L πx L x ψ a) If the probability amplitude is zero, then 2 sin 2 = L πx . . . 2 , 1 , , 2 2 = = ⇒ = ⇒ m Lm x mπ L πx So probability is zero for . , 2 , L L x = b) The probability is largest if . 1 2 sin ± = L πx . 4 ) 1 2 ( 2 ) 1 2 ( 2 L m x π m L πx + = ⇒ + = ⇒ So probability is largest for . 4 3 and 4 L L x = c) These answers are consistent with the zeros and maxima of Fig. 40.5....
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This homework help was uploaded on 04/08/2008 for the course PHY 205 taught by Professor Nearing during the Fall '07 term at University of Miami.
 Fall '07
 Nearing
 Physics

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