Daily_015.SOLN

# Daily_015.SOLN - Q7B.4 If we treat the atom in the crystal...

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Phys 274 – c ± Eric B. Szarmes 25 February 2011 Daily Homework 15 SOLUTIONS Q7B.2 For an electron conﬁned to a wire, which we consider to be a one-dimensional box of length L , the energy of the n th energy eigenstate is E n = h 2 n 2 8 mL 2 = ( hc ) 2 n 2 8 mc 2 L 2 . (1) If we substitute L = 5 . 0 nm, together with hc = 1240 eV and mc 2 = 511 , 000 eV for the electron, we obtain E n = (0 . 015) n 2 eV. Therefore, if E n = 0 . 376 eV we calculate n 2 25 . 07, so that n = 5 . (By the rules of quantum mechanics, after we obtain this value of E n from our energy measurement, the electron collapses into the n = 5 energy eigenstate.)
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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(10-34 ) J · s = 6 . 58(10-16 ) 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(10-16 ) eV · s = 9 . 4(10 14 ) rad/s . (3) 1...
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