{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Chapter 40 - 40.1 a E n E1 n h 8 mL 2 2 2 E1 h 2 2 6 63 10...

Info icon This preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
40.1: a) 2 2 34 2 2 1 2 2 2 m) kg)(1.5 20 . 0 ( 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 particle-like (as opposed to wave-like). 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 (one-third 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 - .
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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 0 = x and L x = in the range . 0 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 0 2 sin 2 = L πx . . . 2 , 1 , 0 , 2 2 = = = m Lm x L πx So probability is zero for . , 2 , 0 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. 40.8: a) The third excited state is , 4 = n so 2 2 2 8 ) 1 4 ( mL h E - = eV. 361 J 10 78 . 5 m) 10 kg)(0.125 10 11 . 9 ( 8 s) J 10 626 . 6 ( 15 17 2 9 31 2 34 = × = × × × = - - - - b) J 10 5.78 m/s) 10 s)(3.0 J 10 63 . 6 ( λ 17 8 34 - - × × × = = E hc nm. 44 . 3 λ =
Image of page 2
40.9: Recall . 2 λ mE h p h = = a) m 10 6.0 m) 10 0 . 3 ( 2 2 8 / 2 λ 8 10 10 2 2 1 2 2 1 - - × = × = = = = L mL mh h mL h E m/s kg 10 1 . 1 m 10 6.0 s) J 10 63 . 6 ( λ 24 10 34 1 1 × = × × = = - - - h p b) m 10 0 . 3 λ 8 4 10 2 2 2 2 - × = = = L mL h E m/s. kg 10 2 . 2 2 λ 24 1 2 2 × = = = - p h p c) m 10 0 . 2 3 2 λ 8 9 10 3 2 2 3 - × = = = L mL h E m/s. kg 10 3 . 3 3 24 1 3 × = = - p p 40.10: , 2 2 2 ψ k dx ψ d - = and for ψ to be a solution of Eq.
Image of page 3

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern