Quick Summary
Concept Tested: Nuclear mass defect and neutron separation energy via Einstein’s mass-energy equivalence.
Chapter – Subtopic: Atoms and Nuclei – Nuclear Binding Energy
Difficulty: ★★☆☆☆ (2/5)
Estimated Time: 3 minutes
Key Formula: $$E=\Delta m\times931.5\text{ MeV},\quad \Delta m=[M(^{12}\text{C})+m_n]-M(^{13}\text{C})$$
Answer: C
One-line Reason: The mass defect on separating one neutron from $^{13}$C is $0.005311\,u$, which converts to $4.95$ MeV.
The Question
The atomic mass of $^{12}\text{C}$ is $12.000000\,u$ and that of $^{13}\text{C}$ is $13.003354\,u$. The required energy to remove a neutron from $^{13}\text{C}$, if the mass of a neutron is $1.008665\,u$, will be:
(A) $62.5$ MeV
(B) $6.25$ MeV
(C) $4.95$ MeV
(D) $49.5$ MeV
Quick Answer
Answer: C. Removing a neutron from $^{13}\text{C}$ leaves $^{12}\text{C}+n$. The combined mass of the products exceeds the mass of $^{13}\text{C}$ by $\Delta m=0.005311\,u$; converting this mass defect to energy via $E=\Delta mc^2$ ($1\,u=931.5$ MeV) gives $4.95$ MeV.
Why Other Options Are Incorrect
Option A (62.5 MeV) and Option B (6.25 MeV): These are exactly $10\times$ apart from each other, indicating a shared decimal-placement error unrelated to the correct mass-defect value — neither matches $\Delta m\times931.5$ using the given masses.
Option D (49.5 MeV): Exactly $10\times$ the correct answer (C) — this is the classic result of misplacing a decimal point in $\Delta m$ (using $0.05311\,u$ instead of $0.005311\,u$).
Video Solution
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Video Solution
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Understanding the Concept
A nucleus is slightly lighter than the sum of its free, separated constituent particles — this “missing mass” (mass defect) is the energy that binds the nucleus together, per Einstein’s $E=mc^2$. To remove a nucleon from a bound nucleus, you must supply energy exactly equal to this mass defect. Here, removing one neutron from $^{13}\text{C}$ converts it into $^{12}\text{C}+n$; the required energy equals the mass increase (products heavier than reactant) converted via $1\,u=931.5$ MeV.
Detailed Step-by-Step Solution
Step 1: Write the neutron-removal reaction
$$^{13}_{6}\text{C}\ +\ \text{Energy}\ \longrightarrow\ ^{12}_{6}\text{C}\ +\ ^{1}_{0}n$$
Step 2: Compute the mass defect
Mass of products $=M(^{12}\text{C})+m_n=12.000000+1.008665=13.008665\,u$
$$\Delta m=13.008665-13.003354=0.005311\,u$$
Step 3: Convert mass defect to energy
$$E=\Delta m\times931.5\text{ MeV}/u=0.005311\times931.5=4.947\approx4.95\text{ MeV}$$
Final Answer
✔ The correct answer is (C) — the required energy is $4.95$ MeV.
Essential Formulas for This Topic
$$E=\Delta mc^2,\qquad 1\,u=931.5\text{ MeV}/c^2$$
$$\Delta m_{\text{separation}}=[M(A-1,Z)+m_n]-M(A,Z)\quad\text{(neutron separation energy)}$$
Note: always add the mass of the ejected particle (here, the neutron) to the lighter product before subtracting — a common source of sign/setup errors.
Common Mistakes to Avoid
Mistake 1: Forgetting to add the neutron mass to $^{12}$C before subtracting
Computing $\Delta m$ as simply $M(^{13}\text{C})-M(^{12}\text{C})=1.003354\,u$ (omitting $m_n$ entirely) gives an absurdly large mass defect and an energy of nearly $935$ MeV — a clear sign that the neutron’s own mass must be added to the $^{12}$C side before subtracting.
Mistake 2: Decimal misplacement in the mass defect
Writing $\Delta m$ as $0.05311\,u$ instead of $0.005311\,u$ inflates the answer by $10\times$, producing option D instead of C.
Mistake 3: Using $1\,u=931$ MeV instead of $931.5$ MeV
Small rounding of the conversion constant shifts the final answer away from the precise listed value — always use $931.5$ MeV/u unless the question specifies otherwise.
Key Concept Summary
- A bound nucleus is always lighter than the sum of its free constituent particles.
- This mass deficit, converted via $E=mc^2$, equals the nucleus’s binding energy.
- Removing any one nucleon requires energy exactly equal to that nucleon’s separation energy.
- Always include the mass of the ejected particle on the “products” side before subtracting.
- $1\,u=931.5$ MeV is the standard JEE conversion factor — memorise it exactly.
Golden Rule: Mass defect = (mass of separated products) − (mass of original nucleus); energy required = mass defect × 931.5 MeV.
Frequently Asked Questions
Q: Why is the mass of $^{13}$C greater than $^{12}$C by less than a full neutron mass?
A: Because part of the neutron’s mass-energy is “used up” as binding energy holding it inside the nucleus — only the leftover $0.005311\,u$ needs to be supplied to free it again.
Q: Is this the same as the binding energy of $^{13}$C itself?
A: Not exactly — this is the one-neutron separation energy, i.e., the marginal binding energy of the last neutron, not the total binding energy of the whole nucleus (which would involve separating every nucleon into individual protons and neutrons).
Q: What if the mass defect had come out negative?
A: It would mean $^{13}$C is less stable than $^{12}\text{C}+n$ combined, so the neutron would separate spontaneously (unbound) — not the case here, since $\Delta m>0$.
Q: Does this method work for removing a proton instead of a neutron?
A: Yes, with two changes: use atomic (not nuclear) masses carefully since removing a proton also removes an electron in neutral-atom bookkeeping, and replace $m_n$ with $m_{^1\text{H}}$ (hydrogen atomic mass) to keep electron counts consistent.
Prerequisites to Solve This Question
- Einstein’s mass-energy equivalence, $E=mc^2$.
- The concept of nuclear mass defect and binding energy.
- Correctly balancing a nuclear reaction (mass and atomic numbers).
- Careful decimal arithmetic when subtracting near-equal masses.
After Solving This, You Can:
- ✔ Compute mass defects for any nuclear reaction given isotope masses.
- ✔ Distinguish total binding energy from single-nucleon separation energy.
- ✔ Convert atomic mass units to MeV confidently using $931.5$.
- ✔ Spot decimal-placement traps common in JEE nuclear physics numericals.
Study Tips for This Topic
1. Always write the full balanced reaction first — it prevents sign/setup errors.
2. Keep at least 6 decimal places throughout mass-defect subtraction — rounding early causes visible answer drift.
3. Memorise $1\,u=931.5$ MeV cold; it appears in nearly every nuclear-physics numerical.
4. After computing, sanity-check the order of magnitude — nucleon separation energies for light nuclei are typically a few MeV, not tens or hundreds.
Difficulty Rating & Exam Frequency
Difficulty: ★★☆☆☆ (straightforward, arithmetic-heavy)
JEE Main Frequency: Common — mass-defect/binding-energy numericals appear almost every year.
JEE Advanced Frequency: Occasional, often combined with per-nucleon binding energy comparisons.
Overall Importance: High — foundational to all of nuclear physics.
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Last Updated: August 2026
Question Source: JEE Main 2024 PYQ
Topic: Atoms And Nuclei