Question

Consider nuclei with small mass number A such that .

N = Z =A/2. Neglecting
pairing term (∈) show that semi-empirical formula is given by

BE/A = α −βA(to the power -1/3)− δA(to the power 2/3) / 4.
Obtain the value of A and Z for which binding energy per nucleon (BE / A) is
maximum. Take β = 17 ,8. δ = 71.0 .

Answer 1

Answer on Question #75411-Physics-Quantum Mechanics

Consider nuclei with small mass number A such that $N = Z = A/2$. Neglecting pairing term (€) show that semi-empirical formula is given by

$$BE/A = \alpha - \beta A \text{ (to the power } -1/3) - \delta A \text{ (to the power } 2/3) / 4.$$

Obtain the value of A and Z for which binding energy per nucleon (BE / A) is maximum. Take $\beta = 17, 8, \delta = 71.0$.

Solution

Binding energy per nucleon (BE / A) is maximum when

$$\frac{d}{dA} \left(\frac{BE}{A}\right) = 0$$ $$\frac{d}{dA} \left(\alpha - \beta A^{-\frac{1}{3}} - \frac{\delta A^{\frac{2}{3}}}{4}\right) = - \left(-\frac{1}{3}\right) \beta A^{-\frac{1}{3} - 1} - \left(\frac{2}{3}\right) \frac{\delta A^{\frac{2}{3} - 1}}{4} = 0$$ $$\left(\frac{1}{3}\right) \beta - \left(\frac{2}{3}\right) \frac{\delta A}{4} = 0$$ $$A = \frac{2\beta}{\delta} = 2 \left(\frac{17.8}{71.0}\right) = 0.5$$ $$Z = \frac{0.5}{2} = 0.25.$$

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