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Unformatted text preview: Homework # 3 – Solution Problem #1 (a) The equilibrium interatomic distance R is determined by the minimum of the binding energy as a function of interatomic distance R . The binding energy per atom is given by 2 6 n A q U R R α = , (1) where it is assumed that the repulsive energy is short range (i.e. 1 n ≫ ), so that the repulsive interaction only between 6 nearestneighbor atoms is important. The R can be found from the condition R dU dR = . (2) As follows from Eqs. (1) and (2), 2 1 2 6 n A q n R R α + + = , (3) and therefore 1 1 2 6 n nA R q α = . (4) (b) Substituting Eq.(4) to Eq.(1) we find 2 2 2 2 2 1 6 1 6 1 1 6 1 6 n n A q A q q U q A q R R R R R nA R n α α α α α = = = =  . (5) (c) It follows from Eq.(5) that the constant n is determined by 1 2 1 U R n e α = + , (6) where U 0 =  7.94 eV, R = a /2 = 2.815Å and α...
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This note was uploaded on 09/21/2011 for the course PHYSICS 101 taught by Professor Wormer during the Spring '08 term at Aarhus Universitet.
 Spring '08
 WORMER
 Energy, Work

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