Given below are two statements: Statement I : For all elements, greater the mass of the nucleus, greater is the binding energy per nucleon. Statement II : For all elements, nuclei with less binding energy per nucleon transforms to nuclei with greater binding energy per nucleon. In the light of the above statements, choose the correct answer from the options given below:

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