Daily_018.SOLN - L 2 = hc 8( mc 2 ) ( 4 2-3 2 ) = 0 . 9406...

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Phys 274 – © Eric B. Szarmes 4 March 2011 Daily Homework 18 SOLUTIONS Q8B.6 Consider the possible transitions in the box potential, which I have arranged below in order of increasing energy difference (i.e. increasing photon energy ): n i n f E photon = Δ E ( n 2 i - n 2 f ) 2 1 2 2 - 1 2 = 3 3 2 3 2 - 2 2 = 5 4 3 4 2 - 3 2 = 7 3 1 3 2 - 1 2 = 8 5 4 5 2 - 4 2 = 9 The two visible wavelengths emitted by this system must appear sequentially in the above table, or else there would be a third wavelength between them. Now, the ratio of the energy difference for each transition equals the inverse ratio of the photon wavelengths involved, so we have λ long λ short = 620 nm 443 nm = 1 . 39955 1 . 4 = 7 5 = E short E long . (1) Since the photon energies are in the ratio 7 to 5, we see from the above table that the 4 3 transition corresponds to the higher photon energy (443 nm wavelength), and the 3 2 transition corresponds to the lower photon energy (620 nm wavelength). Then, using Δ E i f = hc λ = ( hc ) 2 8( mc 2 ) L 2 ( n 2 i - n 2 f ) (2) with n i = 4, n f = 3, and λ = 443 nm for the 4 3 transition, we calculate
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Unformatted text preview: L 2 = hc 8( mc 2 ) ( 4 2-3 2 ) = 0 . 9406 nm 2 , and L = 0.97 nm . (3) Q8B.8 Electrons are spin- fermions, so according to the Pauli Exclusion Principle only two electrons (with spins anti-aligned) can occupy each level. Thus, if we put 10 electrons into the box, the rst 5 levels are occupied by two electrons each, and the highest level is n = 5 . If we ignore the electrostatic repulsion, then the total energy of all 10 electrons is E total = ( hc ) 2 8( mc 2 ) L 2 ( 2 1 2 + 2 2 2 + 2 3 2 + 2 4 2 + 2 5 2 ) = 110 ( hc ) 2 8( mc 2 ) L 2 . (4) For a box of length L = 3 . 0 nm we calculate E total = 4 . 6 eV . If electrons were bosons, then the total energy would certainly be dierent: all 10 such electrons would settle into the lowest ( n = 1) energy level, and the total energy would be lower by a factor of 10/110. 1...
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