This preview shows pages 1–2. 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: Physics 361 Midterm exam 10/26/09 Name: Please indicate your reasoning for your answers. You may consult notes that you yourself have written, but not the book or things that others have written. Test results Total scores: 24 23 22 22 20 20 18 18 16 16 14 14 13 | 12 12 12 11 11 9 9 6 6 Though some students did quite well, almost half the class got less than half the points on the test. People scoring below the vertical line are not really getting material and need to make some change. M.1 Nanocrystal A small cubic crystal has a side length L of only 10 nanometers. The speed of sound in this material is 1000 meters/sec. a) (3 points) Estimate the lowest frequency of vibration ω that can exist in this crystal. b) (4 points) For temperatures below some T the specific heat of the nanocrystal becomes much different than that of a large crystal. Estimate this T in degrees Kelvin. c) (1 point) Qualitatively how do the two specific heats differ for T T ? Solution: Scores: 8 8 8 8 8 7 7 7 7 6 5 | 4 4 3 3 3 2 2 1 0 0 If you were below the vertical line. You don’t understand this adequately. Several people seemed to do this problem by rote rather than looking at what was asked. a) We saw that the lowest frequency vibrations in a crystal are the acoustic modes with wavevectors k given by ω = c k . In a finite crystal the allowed k ’s are discretely spaced. The lowest frequencies correspond to the longest wavelengths 2 π/k . The longest wavelengths that fit in a 10 nm crystal are of the order of 10 nm, i.e., 10- 8 meters. The corresponding frequency ω is thus of order ( c 2 π/ 10- 8 ). Using c = 1000 meters/second, we have ω ’ 6 × 10 11 /sec....
View Full Document