Unformatted text preview: Q7B.4 If we treat the atom in the crystal as a harmonic oscillator, then its energies are given by eq. Q7.16 ( not Q7.12 !): E n = ~ ω ( n + 1 2 ) ; n = 0 , 1 , 2 ,..., (2) where ~ = h/ 2 π = 1 . 055(1034 ) J · s = 6 . 58(1016 ) eV · s. If we measure the lowest energy (corresponding to n = 0) to be E = 0 . 31 eV, then we calculate a classical oscillation frequency ω equal to ω = 2 E ~ = 2 (0 . 31 eV) 6 . 58(1016 ) eV · s = 9 . 4(10 14 ) rad/s . (3) 1...
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This note was uploaded on 03/07/2011 for the course PHYS 274 taught by Professor Ericszarmes during the Spring '11 term at University of Hawaii, Manoa.
 Spring '11
 EricSzarmes
 Physics, Energy, Work

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