(a) With L= 3.0 ×109nm, the energy difference is EE212392221912408 511 1030 101310−=××−=×−chch..eV.(b) Since (n+ 1)2– n2= 2n+ 1, we have ∆EE EhmLnhcmc Lnnn=−=+=++122288bgbgchbg.Setting this equal to 1.0 eV, we solve for n: ()( )2391944 511 10 eV3.0 10 nm1.0eV111.2 10 .1240eV nmmcL Enhc∆××=−≈×⋅(c) At this value of n, the energy is En=×≈×12408 511 1061002218218ch.eV.Thus, 1813230eV1.2 10 .511 10 eVnEmc×==××(d) Since
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.