In a nuclear fission process, a high mass nuclide (A ≈ 236)$(A \approx 236)$ with binding energy 7.6 MeV/Nucleon$7.6\text{ MeV/Nucleon}$ dissociated into middle mass nuclides (A ≈ 118)$(A \approx 118)$, having binding energy of 8.6 MeV/Nucleon$8.6\text{ MeV/Nucleon}$. The energy released in the process would be ______ MeV.

Related Formula

Nuclear radius is empirically related to mass number by: R = R₀ A1/3$R = R_0 A^{1/3}$

Core Logic

Given R₁ = (R₂)/(2)$R_1 = \frac{R_2}{2}$ where A₂ = 192$A_2 = 192$. We need to find A₁$A_1$.

Step 1: Forming the Ratio

Step 2: Cubing Both Sides

Pattern Recognition

Since R ∝ A1/3$R \propto A^{1/3}$, scaling R$R$ by k$k$ means scaling A$A$ by k³$k^3$. Half the radius (k = 1/2$k = 1/2$) means 1/8$1/8$th the mass number.

Chapter Mix

Class 12 Physics: Nuclei

Related Formula

R = R₀ A1/3$R = R_0 A^{1/3}$

Core Logic

Since radius R$R$ is proportional to A1/3$A^{1/3}$, the volume V$V$ (which depends on R³$R^3$) will be directly proportional to the mass number A$A$.

Step 1: Show Proportionality

This proves that V ∝ A$V \propto A$.

Step 2: Calculate Ratio

Given that A₂ = 4A₁$A_2 = 4A_1$:

Pattern Recognition

Nuclear density is constant for all nuclei. Therefore, Mass ∝$\propto$ Volume. Since Mass number (A$A$) represents mass, Volume is strictly linearly proportional to Mass number.

Chapter Mix

Class 12 Physics: Nuclei

Related Formula

Core Logic

Given the mass defect:

The total energy liberated in Joules is:

Step 2: Conversion to kWh

We need the answer in kWh$\mathrm{kWh}$. Since 1 kWh = 1000 W × 3600 s = 3.6 × 10⁶ J$1\mathrm{\,kWh} = 1000\mathrm{\,W} \times 3600\mathrm{\,s} = 3.6 \times 10^6\mathrm{\,J}$:

Comparing this to n × 10⁷ kWh$n \times 10^7\mathrm{\,kWh}$, we get: n = 1$n = 1$

Chapter Mix

Class 12 Physics: Nuclei