Question

Using the semi-empirical mass formula for the binding energy of nuclei, calculate the
value of the atomic number (Z) for the most stable nucleus at a given mass number.
Calculate Z0 for A = 60. Take the values of g = 23.7 and d = 0.71 .

Answer 1

Answer on question # 51449, Physics, Solid State Physics

Question Using the semi-empirical mass formula for the binding energy of nuclei, calculate the value of the atomic number (Z) for the most stable nucleus at a given mass number. Calculate Z0 for A = 60. Take the values of g = 23.7 and d = 0.71 .

Solution The most stable nucleus at a given mass number will be the one with the biggest binding energy:

$E_{c}=\alpha A-\beta A^{2/3}-d\frac{Z^{2}}{A^{1/3}}-g\frac{(A/2-Z)^{2}}{A}+\delta,\qquad\delta=\pm\chi A^{3/4}\mbox{ or }0$

A is give and equal to 60, but we can change Z. Let us take derivative with respect to Z, to find local extremum of this function of Z.

$E_{c}^{\prime}=-2d\frac{Z}{A^{1/3}}+2g\frac{A/2-Z}{A}=0$

So we can find Z now:

$-0.71\frac{Z}{60^{1/3}}+23.7\frac{60/2-Z}{60}=0$

$Z\approx 21$

So this formula tells us that nuclei with Z=21 will be most stable. It can be true, however, because most stable with A=60 is ${}^{60}_{28}Ni$.

http://www.AssignmentExpert.com/