Why is nuclear density $\rho$ independent of mass number $A$ across all elements?

The nuclear radius follows $R = R_0 A^{1/3}$, making volume $V = \frac{4}{3}\pi R^3 = \frac{4}{3}\pi R_0^3 A$. Since mass $M \approx A \cdot m_{\text{nucleon}}$, density $\rho = \frac{M}{V} = \frac{A \cdot m_{\text{nucleon}}}{\frac{4}{3}\pi R_0^3 A} = \frac{3 m_{\text{nucleon}}}{4 \pi R_0^3}$. The variable $A$ cancels out completely, giving a constant density of $\approx 2.3 \times 10^{17} \text{ kg/m}^3$.

How do I determine whether a nuclear fission or fusion reaction releases energy ($Q > 0$)?

Check the binding energy per nucleon ($E_{bn}$) of reactants versus products. If product nuclei lie closer to the peak of the binding energy curve (around Fe-56 with $E_{bn} \approx 8.75 \text{ MeV}$), total product binding energy is higher. Since $Q = E_{b,\text{products}} - E_{b,\text{reactants}}$, a net gain in stability releases positive energy ($Q > 0$).

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Solved High-Yield Sample Questions

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Question #1NEET 2024

A radioactive nucleus $X$ decays into a stable nucleus $Y$ . Initially at time $t = 0$ , a sample contains pure $X$ with $N_0$ nuclei. The half-life of $X$ is $T$ . The time $t$ at which the ratio of the number of nuclei $Y$ to that of $X$ becomes $7:1$ is:

Solution Locked

Step-by-Step Solution & Concept

Step 1

Apply conservation of total nuclei $N_0 = N_X + N_Y$ . The target ratio gives $\frac{N_Y}{N_X} = 7$ , which means $N_Y = 7 N_X$ .

Step 2

Substitute $N_Y$ into the total conservation equation: $N_0 = N_X + 7 N_X = 8 N_X$ . Thus, remaining nuclei $N_X = \frac{N_0}{8} = N_0 \left(\frac{1}{2}\right)^3$ .

Step 3

Compare with the standard decay expression $N_X = N_0 \left(\frac{1}{2}\right)^{t/T}$ . Equating powers gives $\frac{t}{T} = 3$ , leading to final decay time $t = 3T$ .

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Question #2JEE Main 2023

The binding energy per nucleon for a parent nucleus $A^{100}$ is $7.6 \text{ MeV}$ and for two daughter nuclei $B^{50}$ formed after fission of $A^{100}$ is $8.5 \text{ MeV}$ each. The total energy released ( $Q$ -value) in this fission process is:

Solution Locked

Step-by-Step Solution & Concept

Step 1

Calculate total initial binding energy of parent nucleus $A^{100}$ using $E_{b,\text{initial}} = A \times (E_{bn})_{\text{initial}} = 100 \times 7.6 \text{ MeV} = 760 \text{ MeV}$ .

Step 2

Calculate total final binding energy of the two daughter nuclei $2 \times B^{50}$ using $E_{b,\text{final}} = 2 \times [50 \times 8.5 \text{ MeV}] = 100 \times 8.5 \text{ MeV} = 850 \text{ MeV}$ .

Step 3

Evaluate released $Q$ -value as $Q = E_{b,\text{final}} - E_{b,\text{initial}} = 850 \text{ MeV} - 760 \text{ MeV} = 90 \text{ MeV}$ .

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Mastering Nuclei: Formulas, Traps & Exam Strategy | ExamiFly

Chapter Overview

Core Subtopics & Syllabus Breakdown

Key Formulas & High-Yield Equations

Nuclear Radius Formula

Mass Defect & Binding Energy

Law of Radioactive Decay

Half-Life and Decay Constant Relationship

Q-Value of Nuclear Reaction

Common Mistakes & Traps to Avoid

Past Year Question (PYQ) Trends

Solved High-Yield Sample Questions

Frequently Asked Questions

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