PHY
solutions 06

# Modern Physics

• Homework Help
• davidvictor
• 4
• 100% (1) 1 out of 1 people found this document helpful

This preview shows pages 1–2. Sign up to view the full content.

PHY251 Homework Set 6 Reading: Chapter 6 and 7 Homework: Chapter 6, Question 9 Problems 2, 9,18 Chapter 7, Problems 16,20 Hints and Solutions Question VI.9 Suppose you have a box a few centimeters wide and an electron with the energy of a few electron volts (i.e. of the same order of magnitude as an electron in a Bohr orbit). In what range would the n -values for the electron in the box lie? Hints: No hints Solution: The wavelength of the electron with eV energy would be of order the Bohr radius, i.e. 0.1 nm = 10 -10 m. If the box is cm-sized, i.e. 10 -2 m, then such a wavelength would fit 10 8 times in the box, i.e. n 2245 10 8 . Problem VI.2 Consider the normalized wave function of the form ψ ( x ) = C exp{ - x ² / 2 a ² }. Calculate the expectation values (a) x , (b) x ² , (c) p , and (d) p ² for this wave function. Give a physical justification of your results for (a) and (c). Hints: The value of the normalization constant C should not matter. Use the definition of the expectation value, and the operator identity: p = - ih ( / x ) . Solution: The value of C is such that 1 = -∞ | ψ ( x )| 2 dx = | C | 2 -∞ exp{ - x ² / a ² } dx = | C | 2 a π (see Appendix B-2). x = -∞ x | ψ ( x )| 2 dx = -∞ x | C | 2 exp{ - x ² / a ² } dx , which is an integral of an odd function of x , over an interval symmetric with respect to x =0: therefore, the result is 0: x = 0. The result makes sense physically: the wavefunction is perfectly symmetric around x =0, and therefore the average should be zero as well.

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

This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• 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.

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

• 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.

Dana University of Pennsylvania ‘17, Course Hero Intern

• 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.

Jill Tulane University ‘16, Course Hero Intern