problem 18.7

problem 18.7 - Tanisha Billups = = R= = r0= = U= r 2.65E-10...

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Tanisha Billups problem 18.7 = ε hcD = 1.51E-23 R = 2^(1/6)r = 297 r0= R*2^(-1/6) = 2.65E-10 U= (4 )*((r0/r)^12-(r0/r)^6)) ε r U 2.65E-10 0 2.66E-10 0 2.67E-10 0 2.68E-10 0 2.69E-10 0 2.70E-10 0 2.71E-10 0 2.72E-10 0 2.73E-10 0 2.74E-10 0 2.75E-10 0 2.76E-10 0 2.77E-10 0 2.78E-10 0 2.79E-10 0 2.80E-10 0 2.81E-10 0 2.82E-10 0 2.83E-10 0 2.84E-10 0 2.85E-10 0 2.86E-10 0 2.87E-10 0 2.88E-10 0 2.89E-10 0 2.90E-10 0 2.91E-10 0 2.92E-10 0 2.93E-10
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This note was uploaded on 05/01/2011 for the course CHEM 346 taught by Professor Cardelino during the Spring '11 term at Spelman.

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problem 18.7 - Tanisha Billups = = R= = r0= = U= r 2.65E-10...

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