E = m c² (Einstein, 1905). Mass and energy are interconvertible. 1 u = 931.5 MeV/c².
-- NCERT Class 12 Physics, Ch. 13, p. 310Atomic Masses
8 MCQs1 revision card9-step worked example
Source: NCERT Atoms and NucleiOfficial key: NTA-verifiedLast updated: 26 Sep 2026
Atomic Masses, explained for NEET
A stem hands you "atomic mass of ¹⁶O = 15.9949 u" and asks for the mass of the nucleus. Write 15.9949 u and you have silently carried eight electrons into the nucleus. Tabulated atomic masses are masses of neutral atoms — nucleus plus Z electrons. To get the nuclear mass, subtract Z × 0.000548 u. Most of the time the electrons cancel on both sides and you may ignore them; in a question that names the nucleus specifically, they do not.
The unit itself is fixed by definition: 1 u is one-twelfth the mass of a neutral carbon-12 atom, equal to 1.660539 × 10⁻²⁷ kg (NCERT Class 12 Physics, Chapter 13 "Nuclei", printed page 307). Two consequences follow. Carbon-12 has an atomic mass of exactly 12 u — no experimental uncertainty, it is the anchor. And the definition uses the atom, not the bare nucleus, which is why the electron correction above is built into every tabulated value.
The energy equivalent is 1 u ≡ 931.5 MeV/c². The /c² is not decoration. 931.5 MeV is an energy; 931.5 MeV/c² is a mass. Both appear in options, and only one of them is a mass unit. (The relation behind the conversion, E = mc², is developed in the mass-energy relation topic.)
Finally, the atomic mass printed for an element in a periodic table is an abundance-weighted average over its isotopes — chlorine is 35.45 u because natural chlorine is roughly three parts ³⁵Cl to one part ³⁷Cl. No chlorine atom weighs 35.45 u. Individual isotope masses stay close to integers; element averages need not.
Watch out for three swaps: atomic mass read as nuclear mass, element average read as isotope mass, and MeV written where MeV/c² was required.
Can you answer these Atomic Masses MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
The unified atomic mass unit (1 u) is defined as
Show answer and why every option is right or wrong
Answer: C. The modern definition fixes 1 u as one-twelfth the mass of a neutral ¹²C atom, which makes the atomic mass of carbon-12 exactly 12 u (NCERT Class 12 Physics, Chapter 13 "Nuclei", printed page 307).
Why A is wrong: A is wrong because the reference is the neutral atom, not the bare nucleus; the ¹²C nucleus is lighter by six electron masses, so this definition would shift the value of u.
Why B is wrong: B is wrong because the proton mass is 1.007276 u, not 1 u — the two are close but not equal, and defining u from the proton would make carbon-12 a non-integer.
Why D is wrong: D is wrong because the oxygen-16 scale was abandoned; the current standard is carbon-12, and ¹⁶O is 15.9949 u on it, not exactly 16 u.
MCQ 2Easy RecallPractice
The value of 1 u in SI units is closest to
Show answer and why every option is right or wrong
Answer: A. 1 u = 1.660539 × 10⁻²⁷ kg, the mass scale of a single nucleon (NCERT Class 12 Physics, Chapter 13 "Nuclei", printed page 307).
Why B is wrong: B is wrong because 1.66 × 10⁻²⁴ is the value in grams, not kilograms — a factor-of-10³ unit slip.
Why C is wrong: C is wrong because 9.11 × 10⁻³¹ kg is the electron mass, about 1/1836 of a nucleon mass.
Why D is wrong: D is wrong because 1.60 × 10⁻¹⁹ is the elementary charge in coulombs, a charge and not a mass.
MCQ 3Easy RecallPractice
Expressed in energy-equivalent units, the mass 1 u is correctly written as
Show answer and why every option is right or wrong
Answer: D. Dividing the energy equivalent by c² returns a mass, so 1 u = 931.5 MeV/c²; the energy released if that mass were fully converted is 931.5 MeV (NCERT Class 12 Physics, Chapter 13 "Nuclei", printed page 311).
Why A is wrong: A is wrong because 931.5 MeV is an energy, not a mass; it is the energy equivalent of 1 u, which is a different physical quantity from 1 u itself.
Why B is wrong: B is wrong because the magnitude is off by 10⁶ — the figure 931.5 belongs with MeV, so this is 931.5 eV/c², about a millionth of the correct mass.
Why C is wrong: C is wrong on both counts: the joule is an energy unit, and the energy equivalent of 1 u is 1.49 × 10⁻¹⁰ J, not 931.5 J.
MCQ 4Direct ApplicationPractice
The atomic mass of ¹⁶O is 15.9949 u. Taking the electron mass as 0.000548 u and neglecting electronic binding energies, the mass of the ¹⁶O nucleus is
Show answer and why every option is right or wrong
Answer: B. A tabulated atomic mass is that of the neutral atom, so remove Z = 8 electrons: 15.9949 − 8 × 0.000548 = 15.9949 − 0.004384 = 15.9905 u (NCERT Class 12 Physics, Chapter 13 "Nuclei", printed page 311).
Why A is wrong: A is wrong because it treats the tabulated atomic mass as the nuclear mass and leaves eight electrons inside the nucleus.
Why C is wrong: C is wrong because the electron masses were added instead of subtracted; the nucleus must be lighter than the neutral atom, not heavier.
Why D is wrong: D is wrong because 16 u is the mass number, an integer count of nucleons, and only carbon-12 is exactly an integer number of u by definition.
MCQ 5Direct ApplicationPractice
Natural chlorine consists of ³⁵Cl (mass 34.9689 u, abundance 75.77%) and ³⁷Cl (mass 36.9659 u, abundance 24.23%). The atomic mass of the element chlorine is
Show answer and why every option is right or wrong
Answer: C. The element's atomic mass is the abundance-weighted mean: 0.7577 × 34.9689 + 0.2423 × 36.9659 = 26.494 + 8.957 = 35.45 u (NCERT Class 12 Physics, Chapter 13 "Nuclei", printed page 307).
Why A is wrong: A is wrong because it quotes only the lighter isotope; the element value must lie between the two isotope masses, weighted towards the more abundant one.
Why B is wrong: B is wrong because it is the unweighted mean (34.9689 + 36.9659)/2; equal weighting ignores that ³⁵Cl is roughly three times as abundant.
Why D is wrong: D is wrong because it quotes only the heavier, minority isotope.
MCQ 6Direct ApplicationPractice
A mass of 2.00 × 10⁻³ u is fully converted into energy. The energy released is
Show answer and why every option is right or wrong
Answer: A. Using 1 u ≡ 931.5 MeV, energy = 2.00 × 10⁻³ × 931.5 = 1.863 MeV, which is 1.86 MeV to three significant figures (NCERT Class 12 Physics, Chapter 13 "Nuclei", printed page 311).
Why B is wrong: B is wrong by a factor of 10³ — the conversion factor is 931.5 MeV per u, so the product lands in MeV, not keV.
Why C is wrong: C is wrong because it converts 1 u rather than 2.00 × 10⁻³ u; the 10⁻³ factor in the given mass has been dropped.
Why D is wrong: D is wrong because it reports the mass number with the unit changed to MeV, i.e. it treats 1 u as 1 MeV and never applies the 931.5 conversion at all.
MCQ 7Concept TrapPractice
The atomic mass of naturally occurring neon is listed as 20.18 u, although every neon nucleus contains a whole number of nucleons. The principal reason for the non-integer value is that
Show answer and why every option is right or wrong
Answer: D. Neon occurs as ²⁰Ne, ²¹Ne and ²²Ne, and the listed element mass is the abundance-weighted average of the three isotope masses, so it need not be near an integer (NCERT Class 12 Physics, Chapter 13 "Nuclei", printed page 307).
Why A is wrong: A is wrong because ten electrons contribute only 10 × 0.000548 = 0.00548 u, roughly thirty times too small to explain a 0.18 u offset.
Why B is wrong: B is wrong in direction as well as magnitude: binding lowers the mass of a bound nucleus below the sum of its free constituents, so a mass defect cannot push the value above 20 u.
Why C is wrong: C is wrong because the oxygen-16 reference was replaced by carbon-12; in any case the choice of reference does not make a single isotope's mass non-integer by 0.18 u.
MCQ 8CalculationPractice
An element has exactly two stable isotopes, of atomic masses 62.93 u and 64.93 u. The atomic mass of the element is 63.55 u. The percentage abundance of the lighter isotope is
Show answer and why every option is right or wrong
Answer: B. Let the lighter fraction be x: 62.93x + 64.93(1 − x) = 63.55, giving 64.93 − 2.00x = 63.55, so x = 1.38/2.00 = 0.690, i.e. 69.0% (NCERT Class 12 Physics, Chapter 13 "Nuclei", printed page 307).
Why A is wrong: A is wrong because 31.0% is the complementary fraction, the abundance of the heavier isotope; the element mass 63.55 u sits nearer 62.93 u, so the lighter isotope must be the majority one.
Why C is wrong: C is wrong because equal abundance would give the unweighted mean (62.93 + 64.93)/2 = 63.93 u, not 63.55 u.
Why D is wrong: D is wrong because it imports the chlorine abundance figure instead of solving the weighted-average equation for this element.
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Atomic Masses: quick recall before you leave
How do you solve a Atomic Masses question? A worked example
- 1
Given
Mass m = 1.6605 × 10⁻²⁷ kg (this is 1 u).
Speed of light c = 2.99792458 × 10⁸ m/s (exact, SI definition).
1 eV = 1.602177 × 10⁻¹⁹ J (exact, SI definition of the electronvolt). - 2
Required
The energy equivalent of 1 u, expressed in MeV.
- 3
Concept
A mass has an energy equivalent given by Einstein's mass-energy relation. Converting that energy from joules to electronvolts turns the atomic mass unit into the MeV scale used throughout nuclear physics (NCERT Class 12 Physics, Chapter 13 "Nuclei", printed page 311).
- 4
Formula
E = mc²
- 5
Substitution
E = (1.6605 × 10⁻²⁷ kg) × (2.99792458 × 10⁸ m/s)²
- 6
Calculation
c² = 8.98755 × 10¹⁶ m²/s²
E = 1.6605 × 10⁻²⁷ × 8.98755 × 10¹⁶ = 1.49238 × 10⁻¹⁰ J
Convert to eV: E = (1.49238 × 10⁻¹⁰) / (1.602177 × 10⁻¹⁹) = 9.3148 × 10⁸ eV
E = 9.3148 × 10² MeV
Both c and the eV-to-joule factor are exact defined SI constants, so they place no limit on the significant figures; the five significant figures in the answer come entirely from the measured mass 1.6605 × 10⁻²⁷ kg. - 7
Final answer
E = 931.5 MeV (to four significant figures). Equivalently, 1 u = 931.5 MeV/c².
- 8
Common trap
Two slips recur. First, squaring is skipped — computing mc instead of mc² gives 4.98 × 10⁻¹⁹ J, out by eight orders of magnitude, and the resulting option is always on the paper. Second, the answer is reported as "1 u = 931.5 MeV". That equates a mass with an energy. The energy released by annihilating 1 u is 931.5 MeV; the mass itself is 931.5 MeV/c². Read the stem for which of the two it asked for.
- 9
Similar NEET-style question
The atomic mass of ⁴He is 4.0026 u. Express the mass of the neutral ⁴He atom in MeV/c², and then state its mass in kilograms. (Answer: 4.0026 × 931.5 = 3.728 × 10³ MeV/c²; 4.0026 × 1.6605 × 10⁻²⁷ = 6.6465 × 10⁻²⁷ kg.)
What to remember before solving Atomic Masses questions
Which Atomic Masses formulas do you need for NEET?
1 formula — click to collapse
Mass-energy equivalence
Einstein's mass-energy relation. 1 atomic mass unit = 931.5 MeV in energy units.
| Symbol | Quantity | SI Unit |
|---|---|---|
| E | energy | J or MeV |
| m | mass | kg or u |
| c | speed of light | m/s |
Valid when
- Conversion of mass to energy (or vice versa)
More in Atoms and Nuclei: 3 exam traps and mistakes · 5 formulas · 5 question patterns from its other lessons.
Atomic Masses questions from past NEET papers
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Sources
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