chem5314_hwk2_soln

# chem5314_hwk2_soln - Chem 5314 homework #2 solutions out of...

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Chem 5314 homework #2 solutions – out of 34 marks Problem 1 – 6 marks The heat equation is: ∂u ∂t = α 2 u (1) where α is the thermal diﬀusivity of a substance, and u ( x,y,z,t ) decribes the temperature of a substance as a function of space and time. Show that the separation of variables procedure can be successfully applied to this partial diﬀerential equation to separate the space and time variables. You should obtain, as your ﬁnal answer, a series of ordinary diﬀerential equations. solution: Postulate u ( x,y,z,t ) = S ( x,y,z ) T ( t ). Then LHS = SdT/dt and RHS = αT 2 S . Divide through by u to get LHS = ( dT/dt ) /T and RHS = ( α 2 S ) /S . For LHS = RHS both sides must be independent of x,y,z,t . So to make it work put LHS = RHS = c . This gives dT ( t ) dt = cT ( t ) (2) and 2 S ( x,y,z ) = c α S ( x,y,z ) (3) Now you could also separate the 3 spatial variables in Eq. (3) since this is a linear partial diﬀerential equation to get each variable by itself in a separate equation. Problem 2 – 9 marks Consider a particle in a one-dimensional box of length L in its lowest energy (ground) stationary state. Calculate the probability that the particle is a – 3 marks) in the left half of the box solution: For the lowest energy state, we have ψ ( x ) = r 2 L sin πx L (4) The probability density of this state is thus 2 L sin 2 πx L (5) 1

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You need to integrate this expression. Normally this integral is found in textbooks as
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## This note was uploaded on 02/08/2012 for the course CHEM 5314 taught by Professor Nielsen during the Fall '11 term at University of Texas at Dallas, Richardson.

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chem5314_hwk2_soln - Chem 5314 homework #2 solutions out of...

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