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Unformatted text preview: E n = 0.025 eV = 4.0(1021 ) 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(1026 ) kg ²± . 10 m ² 2 ± 4 . 0(1021 ) J ² ± 6 . 626(1034 ) 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(1010 ) . (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.
 Spring '11
 EricSzarmes
 Physics, Work

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