This question requires us to calculate the energy released during the spontaneous symmetric fission of a heavy nucleus, utilizing the principles of the Bethe-Weizsäcker mass formula and neglecting the pairing energy term. We need to identify the correct expression representing this energy release.

Understanding Fission Energy Release using Bethe-Weizsäcker Formula

Nuclear fission involves the splitting of a heavy nucleus into lighter nuclei. In spontaneous symmetric fission, the nucleus divides into two fragments of approximately equal mass and charge.

The Bethe-Weizsäcker mass formula approximates the binding energy ($B$) of a nucleus. The key terms relevant to fission energy release are:

• Surface Energy Term: Represented as $-\alpha_s A^{2/3}$, this term arises from the surface tension of the nuclear matter, penalizing smaller surface area-to-volume ratios.
• Coulomb Energy Term: Represented as $-\alpha_c \frac{Z^2}{A^{1/3}}$, this term accounts for the electrostatic repulsion between protons within the nucleus, which destabilizes it.

Energy is released in fission ($Q$) because the total binding energy of the resulting fragments is greater than the binding energy of the original heavy nucleus. Mathematically, the energy released is $Q = B_{final} - B_{initial}$.

Deriving the Energy Release for Symmetric Fission

Consider a heavy nucleus with mass number A and atomic number Z. Upon symmetric fission, it splits into two identical nuclei, each having mass number $A/2$ and atomic number $Z/2$.

Surface Energy Change Contribution

Let's calculate the change in the surface energy component that contributes to the energy release:

• Initial Surface Energy ($SE_{initial}$): $-\alpha_s A^{2/3}$
• Final Surface Energy ($SE_{final}$): For two fragments, each with mass $A/2$, the total surface energy is the sum of the surface energies of the two fragments: $2 \times (-\alpha_s (A/2)^{2/3})$. $$ SE_{final} = -2 \alpha_s \frac{A^{2/3}}{2^{2/3}} = -\alpha_s A^{2/3} \frac{2}{2^{2/3}} = -\alpha_s A^{2/3} 2^{(1 - 2/3)} = -\alpha_s A^{2/3} 2^{1/3} $$
• Change in Surface Energy Contribution to Binding Energy: This change drives energy release. We calculate it as $\Delta SE = SE_{final} - SE_{initial}$. $$ \Delta SE = (-\alpha_s A^{2/3} 2^{1/3}) - (-\alpha_s A^{2/3}) $$ $$ \Delta SE = \alpha_s A^{2/3} (1 - 2^{1/3}) $$

Since $2^{1/3} \approx 1.26$, the term $(1 - 2^{1/3})$ is negative. This negative change in the surface energy contribution means the binding energy increases due to reduced surface effects relative to volume, thus releasing energy.

Coulomb Energy Change Contribution

Now, let's calculate the change in the Coulomb energy component:

• Initial Coulomb Energy ($CE_{initial}$): $-\alpha_c \frac{Z^2}{A^{1/3}}$
• Final Coulomb Energy ($CE_{final}$): For two fragments, each with charge $Z/2$ and mass $A/2$, the total Coulomb energy is $2 \times (-\alpha_c \frac{(Z/2)^2}{(A/2)^{1/3}})$. $$ CE_{final} = -2 \alpha_c \frac{Z^2/4}{A^{1/3}/2^{1/3}} = -2 \alpha_c \frac{Z^2}{A^{1/3}} \frac{2^{1/3}}{4} $$ $$ CE_{final} = -\alpha_c \frac{Z^2}{A^{1/3}} \frac{2^{1/3}}{2} = -\alpha_c \frac{Z^2}{A^{1/3}} 2^{-2/3} $$
• Change in Coulomb Energy Contribution to Binding Energy: $\Delta CE = CE_{final} - CE_{initial}$. $$ \Delta CE = (-\alpha_c \frac{Z^2}{A^{1/3}} 2^{-2/3}) - (-\alpha_c \frac{Z^2}{A^{1/3}}) $$ $$ \Delta CE = \alpha_c \frac{Z^2}{A^{1/3}} (1 - 2^{-2/3}) $$

Since $2^{-2/3} < 1$, the term $(1 - 2^{-2/3})$ is positive. This indicates that the Coulomb repulsion effect becomes less dominant relative to the volume after splitting, contributing positively to the energy release.

Total Energy Released

The total energy released ($Q$) in symmetric fission is the sum of the changes in the surface and Coulomb energy contributions:

Note: The question and options use the coefficients $\alpha_s$ and $a_c$. We match our derived formula using these notations.

Matching the Result with Options

Comparing our derived expression for the energy released with the given options:

• Option 1: $\alpha_s A^{2/3}(1-2^{1/3})+a_c \frac{Z^2}{A^{1/3}}(1-2^{-2/3})$
• Option 2: $\alpha_s A^{2/3}(1+2^{1/3})+a_c \frac{Z^2}{A^{1/3}}(1-2^{2/3})$
• Option 3: $\alpha_s A^{2/3}(1+2^{1/3})+a_c \frac{Z^2}{A^{1/3}}(1+2^{-2/3})$
• Option 4: $\alpha_s A^{2/3}(1-2^{1/3}) + a_c \frac{Z^2}{A^{1/3}}(1-2^{2/3})$

Our derived formula matches Option 1 exactly.