Q.7 Physics ⭐ 360° Option Elimination Active

NEET 2021 Physics Question 7 Solution

Topic: Nuclei · Official NTA Booklet Code M1

Question

A nucleus with mass number 240 breaks into two fragments each of mass number 120, the binding energy per nucleon of unfragmented nuclei is 7.6 MeV while that of fragments is 8.5 MeV. The total gain in the Binding Energy in the process is:

✓

Verified Against NTA Final Answer Key

Option (4) : 216 MeV

📚 Nuclei Easy

📋 Given Parameters & Problem Setup

Parent Mass Number (A) 240

Parent BE per nucleon 7.6 MeV/nucleon

Daughter Fragments Two fragments, each A = 120

Daughter BE per nucleon 8.5 MeV/nucleon

⚡

The 30-Second Exam Hall Shortcut: ⚡ Fast Delta-BE: ΔE = A × Δ(BE/A)

⏱️ Save ~20 seconds

Total nucleons $A = 240$ is conserved. Increase in binding energy per nucleon $\Delta(\text{BE}/A) = 8.5 - 7.6 = 0.9\text{ MeV}$ . Total gain in binding energy $= 240 \times 0.9 = 216\text{ MeV}$ ! Match Option (4)!

1 Conserved total nucleon count:

A = 240

2 Difference in binding energy per nucleon:

\Delta b = 8.5 - 7.6 = 0.9\text{ MeV/nucleon}

3 Total gain in Binding Energy

= A \times \Delta b = 240 \times 0.9\text{ MeV}

240 \times 0.9 = 216\text{ MeV}

5 Match Option (4).

🛡️

Why You Can 100% Trust This Shortcut: Canonical nuclear conservation of baryon/nucleon number and definition of total binding energy.

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Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Calculating the Initial Binding Energy of Parent Nucleus

The parent nucleus has mass number $A = 240$ and binding energy per nucleon $\text{BE}_i/A = 7.6\text{ MeV}$ :

\text{BE}_i = 240 \times 7.6\text{ MeV} = 1824\text{ MeV}

Step 2

Calculating the Final Binding Energy of Daughter Fragments

The parent splits into two equal fragments of mass number $A = 120$ each, with $\text{BE}_f/A = 8.5\text{ MeV}$ :

\text{BE}_f = 2 \times (120 \times 8.5\text{ MeV}) = 240 \times 8.5\text{ MeV} = 2040\text{ MeV}

Step 3

Computing the Net Gain in Binding Energy

The total gain in binding energy (which is also the Q-value energy released in the fission reaction) is:

\Delta\text{BE} = \text{BE}_f - \text{BE}_i = 2040\text{ MeV} - 1824\text{ MeV} = 216\text{ MeV}

⭐

360° Option Elimination Engine Key Feature

Diagnose every choice: Know exactly why the right option works and why distractors fail.

4-Option Deep Scan

1 1

✗ Incorrect

Verdict & Scientific Reason:

This is only the gain per single nucleon ( $8.5 - 7.6 = 0.9\text{ MeV}$ ), not the total gain for the entire nucleus of 240 nucleons.

⚠️

Student Fallacy (Why candidates pick this trap): Forgetting to multiply by the total number of nucleons

A = 240

2 2

✗ Incorrect

Verdict & Scientific Reason:

Arbitrary sum/difference error ( $8.5 + 0.9 = 9.4$ ).

⚠️

Student Fallacy (Why candidates pick this trap): Misinterpreting binding energy difference formulas.

3 3

✗ Incorrect

Verdict & Scientific Reason:

Computational arithmetic error.

⚠️

Student Fallacy (Why candidates pick this trap): Incorrect multi-step multiplication.

4 4

✓ 100% Correct

Verdict & Scientific Reason:

Exact total gain in binding energy: $240 \times (8.5 - 7.6) = 216\text{ MeV}$ .

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Why this is airtight: None; precise nuclear physics computation.

⚠️

NTA Trap Alert: BE per Nucleon vs. Total Binding Energy Trap

NTA intentionally sets Option (1) as $0.9\text{ MeV}$ . Students who subtract $8.5 - 7.6 = 0.9$ see $0.9\text{ MeV}$ and select it immediately, failing to notice the prompt asks for "total gain in Binding Energy", which requires multiplying by all 240 nucleons.

🛡️ How to avoid the -1 mark: Total Binding Energy = (BE per nucleon)

\times

(Total mass number A).

📖

Core NEET Syllabus Standard

NCERT Physics · Nuclei Standard Benchmark

Core Scientific Principle Tested:

"Energy released in nuclear fission equals the gain in total binding energy: $Q = \text{BE}_{\text{products}} - \text{BE}_{\text{reactants}} = A \times \Delta(\text{BE}/A)$ ."

Syllabus Focus: Binding Energy per Nucleon and Nuclear Fission

✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

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NEET 2021 Physics Question 7 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ Fast Delta-BE: ΔE = A × Δ(BE/A)

Step-by-Step Rigorous Workflow & Mathematical Proof

Calculating the Initial Binding Energy of Parent Nucleus

Calculating the Final Binding Energy of Daughter Fragments

Computing the Net Gain in Binding Energy

360° Option Elimination Engine Key Feature

NTA Trap Alert: BE per Nucleon vs. Total Binding Energy Trap

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

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