ph135cS6 - Ph135c Solution set#6 1 We write the...

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Unformatted text preview: Ph135c. Solution set #6, 5/13/10 1. We write the semi-empirical mass formula as: E b = aA- bA 2 / 3- c Z ( Z- 1) A 1 / 3- d ( A- 2 Z ) 2 A + Δ A First, we maximize this with respect to Z : 0 = ∂E b ∂Z =- c 2 Z- 1 A 1 / 3 + d 4( A- 2 Z ) A =- Z 2 c A 1 / 3 + 8 d A + c A 1 / 3 + 4 d ⇒ Z = cA 2 / 3 + 4 dA 2 cA 2 / 3 + 8 d One can now plug this in to the expression for E b above to get the energy of the most stable atom of a given atomic weight A , which is now a function of A only. This function is ugly to write down, but the plot is shown in figure 1. This can be maximized numerically, and, taking the nearest integral values, we find the most stable nucleus has Z = 28 ,A = 62, which is 62 Ni . This is verified experimentally. 2. In the last homework we found: E T = C 1 A 2 / 3 ( Z 5 / 3 + N 5 / 3 ) This is minimized at Z = N = A 2 . The fourth term in the semi-empirical mass formula comes from the extra energy we get when Z is not equal to N . To approximate this, we Taylor expand the above expression around the minimum. The linear term must vanish, and the quadratic term is given by: d 2 E T dZ 2 Z = A 2 = d 2 dZ 2 C 1 A 2 / 3 ( Z 5 / 3 + ( A- Z ) 5 / 3 ) Z = A 2 = C 1 A 2 / 3 ( 5 3 2 3 Z- 1 / 3 + 5 3 2 3 ( A- Z )- 1 / 3...
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ph135cS6 - Ph135c Solution set#6 1 We write the...

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