This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 ( 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 x L x So . 2 sin 2 ) ( 2 2 2 = L x L x a) If the probability amplitude is zero, then 2 sin 2 = L x . . . 2 , 1 , , 2 2 = = = m Lm x m L x So probability is zero for . , 2 , L L x = b) The probability is largest if . 1 2 sin = L x . 4 ) 1 2 ( 2 ) 1 2 ( 2 L m x m L x + = + = 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....
View Full
Document
 Fall '07
 Nearing
 Physics

Click to edit the document details