hw_7_solu

hw_7_solu - Solution Set 7 ME 141A Intro to Nanotechnology...

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ME 141A Intro to Nanotechnology Due: 2/23/2009 1 Solution Set 7 Problem 6.14 in book We can look at the band structure of an element to get an idea of how many electrons per atom participate in conduction. We can also determine this number based on the free electron density, the atomic mass, and the mass density of the material. (a) Calculate the free electron density of lithium. The Fermi energy is 4.72 eV. (b) Derive a formula for the number of atoms per unit volume. (c) Calculate the number of atoms per unit volume of lithium (535 kg/m 3 , 6.9 g/mol) using the formula you derived in part (b). (d) Determine the number of free electrons per lithium atom. (e) What is the electron configuration for Li? Explain whether your answer to (d) makes sense according to this configuration. The free electron(s) of Li is in which sublevel? ANSWER A) = 4.7 × 10 28 electrons/m 3 B) The number of atoms per unit volume = , where N A is Avogadro’s number, ρ is the density of the material and A is the atomic mass. C) 4.67 × 10 28 atoms/ m 3 D) the number of free electrons per atom is: = 1 free electron per atom E) Li: 1s 2 2s 1 ; yes the answer makes sense because the lower electron band is fully occupied and stable, while the upper valence band is only partially occupied with a single electron in it. This single electron can be energized to become a free, conduction electron; the free electron comes from the 2s sublevel. Problem 6.27 in book
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This note was uploaded on 03/23/2009 for the course ME 141A taught by Professor Pennathur during the Winter '09 term at UCSB.

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hw_7_solu - Solution Set 7 ME 141A Intro to Nanotechnology...

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