Daily_016.SOLN - E n = 0.025 eV = 4.0(10-21 J the nitrogen molecule would be excited to the n th quantum state where n 2 = 8 m N 2 L 2 E n h 2 = 8

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Phys 274 – c ± Eric B. Szarmes 28 February 2011 Daily Homework 16 SOLUTIONS Q7S.1 For a quanton in an “infinite square-well” (i.e. box) potential of length L , we have the fundamental quanti- zation condition that λ n = 2 L n ; n = 1 , 2 , 3 ,... (1) (The perfectly reflecting mirrors in this case ensure that the walls of the box yield a potential energy function that goes to infinity, so the photon cannot penetrate them.) The deBroglie relation, also fundamental, then yields p n = h λ n = hn 2 L ; n = 1 , 2 , 3 ,... (2) But now, for the photon the energy-momentum relation is E = pc , and therefore E n = p n c = hn 2 L c = h ( c 2 L ) n ; n = 1 , 2 , 3 ,... (3) Q7S.8 The mass of a nitrogen molecule is m N 2 = 28 g/mol = 0 . 028 kg 6 . 022(10 23 ) = 4 . 65(10 - 26 ) kg. If this nitrogen molecule is confined to a one-dimensional box of length L = 10 cm = 0.1 m, then it has an energy of E n = h 2 n 2 8 m N 2 L 2 . (4) For thermal motion with
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Unformatted text preview: E n = 0.025 eV = 4.0(10-21 ) J, the nitrogen molecule would be excited to the n th quantum state where n 2 = 8 m N 2 L 2 E n h 2 = 8 ± 4 . 65(10-26 ) kg ²± . 10 m ² 2 ± 4 . 0(10-21 ) J ² ± 6 . 626(10-34 ) J · s ² 2 = 3 . 39(10 19 ) , (5) so that n = 5 . 82(10 9 ) . Also note that eq. (4) can be ‘differentiated’ with respect to n to yield Δ E n Δ n = dE n dn = 2 h 2 n 8 m N 2 L 2 = ³ 2 n ´ h 2 n 2 8 m N 2 L 2 = ³ 2 n ´ E n , (6) so that Δ E n E n = 2Δ n n . (7) If Δ n = 1 for adjacent energy states, we calculate Δ E n E n = 2 n = 2 5 . 82(10 9 ) = 3 . 4(10-10 ) . (8) This fractional difference is so small that E would appear truly continuous and not quantized. 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.

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