Question

according to liquid model Binding energy is given by B= a1-a2A^2/3-a3 z(z-1)/A^1/3-a4(A-2z)^2/A+p where a1=15.5Mev a2=16.8Mev a3=0.7Mev a4=23Mev p - constant. find z for which binding energy is minimum for A=64

Answer 1

29

Answer 2

For $A = 64$, the binding energy (ignoring constants independent of $z$) is

$$B(z) = -\frac{a_3}{A^{1/3}}\,z(z-1) -\frac{a_4}{A}\,(A-2z)^2.$$

Differentiate with respect to $z$ and set the derivative to zero.

1. The derivative of the first term: $$\frac{d}{dz}\left[-\frac{a_3}{A^{1/3}}(z^2-z)\right] = -\frac{a_3}{A^{1/3}}(2z-1).$$
2. The derivative of the second term: $$\frac{d}{dz}\left[-\frac{a_4}{A}(A-2z)^2\right] = -\frac{a_4}{A}\cdot 2(A-2z)(-2) = \frac{4a_4}{A}(A-2z).$$

Set the sum to zero:

$$-\frac{a_3}{A^{1/3}}(2z-1) + \frac{4a_4}{A}(A-2z)=0.$$

For $A=64$, note that $A^{1/3} = 4$:

$$-\frac{a_3}{4}(2z-1) + \frac{4a_4}{64}(64-2z)=0.$$

Since $\frac{4}{64}=\frac{1}{16}$, the equation becomes:

$$-\frac{a_3}{4}(2z-1) + \frac{a_4}{16}(64-2z)=0.$$

Multiplying through by 16 to clear denominators:

$$-4a_3(2z-1) + a_4(64-2z)=0.$$

Rearrange:

$$a_4(64-2z) = 4a_3(2z-1).$$

Substitute $a_3 = 0.7\,\text{MeV}$ and $a_4 = 23\,\text{MeV}$:

$$23(64-2z) = 4(0.7)(2z-1) \quad \Longrightarrow \quad 23(64-2z) = 2.8(2z-1).$$

Expanding:

$$1472 - 46z = 5.6z - 2.8,$$ $$1472 + 2.8 = 51.6z,$$ $$51.6z = 1474.8,$$ $$z \approx \frac{1474.8}{51.6} \approx 28.6.$$

Since $z$ must be an integer, the optimum $z$ is approximately 29.