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Atomic Mass Nuclear Composition

🎓 Class 12 Physics CBSE Theory Ch 13 – Nuclei ⏱ ~14 min
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Atomic Mass Nuclear Composition

13.1 Introduction — Inside the Atomic Nucleus

In Chapter 12 we learnt that almost the entire mass of an atom and all of its positive charge sit inside a tiny central body — the nucleus. Rutherford's α-scattering experiment showed that the nuclear radius is roughly 10⁴ times smaller than the atom itself. That means the volume of a nucleus is a tiny 10⁻¹² times the volume of the atom. Picture the atom as a classroom; the nucleus would be smaller than a pinhead at its centre — yet it still contains more than 99.9% of the atom's mass.

Does this densely packed nucleus itself have an internal structure? What holds it together? Why are some nuclei stable while others spontaneously decay? In this chapter we systematically study nuclear masses, sizes, the strong force that binds nucleons, radioactivity, and the energy released in fission and fusion.

Big questions for this chapter: What are the constituents of the nucleus? How is its mass measured? Why is its mass less than the sum of the masses of its parts? How does energy get released in nuclear reactions?

13.2 Atomic Masses and the Atomic Mass Unit (u)

The mass of a single atom is far too small for kilograms to be a comfortable unit. The mass of one carbon-12 atom is only \(1.992647 \times 10^{-26}\) kg. To make the numbers manageable, physicists use the atomic mass unit (u), defined as exactly one-twelfth of the mass of one neutral \(^{12}_{6}\mathrm{C}\) atom:

1 u = (1/12) × mass of one ¹²C atom = (1/12) × 1.992647 × 10⁻²⁶ kg = 1.660539 × 10⁻²⁷ kg

Atomic masses expressed in u are close to whole numbers for most elements — but not exactly so. Chlorine, for example, has an atomic mass of 35.46 u, far from a whole number. The explanation lies in the existence of isotopes.

Isotopes — same chemistry, different mass

A mass spectrometer reveals that almost every element is actually a mixture of atomic species that differ in mass but share the same chemical behaviour. These species are isotopes (Greek: iso-topos, "same place" — they sit in the same slot of the periodic table). The atomic mass listed in the periodic table is the weighted average over the natural abundances of all isotopes.

Worked example — chlorine: Chlorine has two stable isotopes, masses 34.98 u and 36.98 u, with abundances 75.4% and 24.6%. The average mass is

m̄ = (75.4 × 34.98 + 24.6 × 36.98) / 100 = 35.47 u ✓

which matches the periodic-table entry.

Hydrogen — the lightest element — has three isotopes:

NameSymbolMass (u)AbundanceStability
Protium (proton)\(^{1}_{1}\mathrm{H}\)1.0078399.985%stable
Deuterium\(^{2}_{1}\mathrm{H}\) (D)2.014100.015%stable
Tritium\(^{3}_{1}\mathrm{H}\) (T)3.01605radioactive (T½ ≈ 12.3 y)

Proton, Neutron and Discovery of the Neutron

The nucleus of the lightest hydrogen atom is the proton, with mass

m_p = 1.00727 u = 1.67262 × 10⁻²⁷ kg, charge = +e

The proton mass equals the hydrogen-atom mass minus one electron mass (\(m_e = 0.00055\) u).

But deuterium and tritium have masses ≈ 2 u and ≈ 3 u while still containing only one proton. Therefore the nucleus must also contain neutral matter — particles whose mass is close to that of a proton. James Chadwick proved this in 1932 by bombarding beryllium with α-particles and observing emission of a penetrating, neutral, energetic radiation. Conservation of energy and momentum showed it could not be photons; it had to be a new particle, the neutron:

m_n = 1.00866 u = 1.6749 × 10⁻²⁷ kg, charge = 0

Chadwick won the 1935 Nobel Prize for this discovery. A free neutron is unstable (mean life ≈ 1000 s, decaying into a proton, an electron and an antineutrino), but inside a nucleus it is stable.

Nuclear notation: A nuclide is written \(^{A}_{Z}\mathrm{X}\) where
  • Z = atomic number = number of protons
  • N = neutron number = number of neutrons
  • A = mass number = Z + N = number of nucleons (protons + neutrons)
Example: \(^{197}_{79}\mathrm{Au}\) contains 197 nucleons — 79 protons and 118 neutrons.

Isotopes, Isobars and Isotones

FamilySameDifferentExamples
IsotopesZN (and A)\(^{1}_{1}\mathrm{H}, ^{2}_{1}\mathrm{H}, ^{3}_{1}\mathrm{H}\); \(^{12}_{6}\mathrm{C}, ^{14}_{6}\mathrm{C}\)
IsobarsAZ and N\(^{3}_{1}\mathrm{H}\) and \(^{3}_{2}\mathrm{He}\); \(^{40}_{18}\mathrm{Ar}\) and \(^{40}_{20}\mathrm{Ca}\)
IsotonesNZ (and A)\(^{198}_{80}\mathrm{Hg}\) and \(^{197}_{79}\mathrm{Au}\) (both N = 118)
Three families of nuclei Isotopes (same Z) p¹₁H pn²₁H ³₁H Isobars (same A) ³₁H (A=3) ³₂He (A=3) Isotones (same N) ¹³₆C (N=7) ¹⁴₇N (N=7) proton (p) neutron (n)
Fig 13.A: Three nuclear families. Isotopes share Z, isobars share A, isotones share N.
Activity 13.1 — Counting nucleons

For the nuclide \(^{40}_{19}\mathrm{K}\):

  1. How many protons does it have?
  2. How many neutrons?
  3. Is \(^{40}_{20}\mathrm{Ca}\) its isotope, isobar or isotone?
Predict before revealing.
(1) Z = 19 protons. (2) N = A − Z = 40 − 19 = 21 neutrons. (3) \(^{40}_{20}\mathrm{Ca}\) has the same mass number A = 40 but a different Z, so it is an isobar of potassium-40.

13.3 Size of the Nucleus

From Geiger–Marsden experiments, Rutherford deduced that the distance of closest approach of a 5.5-MeV α-particle to a gold nucleus is about \(4.0 \times 10^{-14}\) m. The nucleus must therefore be smaller than this. Higher-energy α-particles probe still more closely, until the short-range nuclear force kicks in and Coulomb-only calculations break down.

Modern measurements use fast electrons as projectiles (electrons feel only the electromagnetic force, so they probe the charge distribution cleanly). A vast set of such experiments yields the simple empirical rule:

R = R₀ A1/3, R₀ = 1.2 × 10⁻¹⁵ m = 1.2 fm

Some immediate consequences:

  • Volume \(V = \tfrac{4}{3}\pi R^3 \propto A\) — i.e. nuclear volume is proportional to the number of nucleons.
  • Therefore nuclear density is independent of A: every nucleus has approximately the same density, like droplets of an incompressible liquid.
  • Numerically, ρ_nuc ≈ 2.3 × 10¹⁷ kg m⁻³ — roughly 10¹⁴ times the density of water.
Astronomy connection: Neutron stars are essentially "giant nuclei" — gravity has compressed ordinary matter to nuclear density. A teaspoon of neutron-star matter weighs about a billion tonnes.

Worked Example 13.1 — Nuclear density of iron

Given mass of \(^{56}_{26}\mathrm{Fe}\) nucleus = 55.85 u (so m = 9.27 × 10⁻²⁶ kg) and A = 56, find the nuclear density.

Radius: \(R = R_0 A^{1/3} = 1.2 \times 10^{-15} \times 56^{1/3}\) m. Since \(56^{1/3} \approx 3.83\), \(R \approx 4.6 \times 10^{-15}\) m.

Volume: \(V = \tfrac{4}{3}\pi R^3 = \tfrac{4}{3}\pi (1.2\times 10^{-15})^3 \times 56\) m³.

Density:

\[ \rho = \frac{m}{V} = \frac{9.27 \times 10^{-26}}{\tfrac{4}{3}\pi (1.2\times 10^{-15})^3 \times 56} \approx 2.29 \times 10^{17}\ \text{kg m}^{-3} \]

The same answer would emerge for any nucleus, confirming the constant-density rule.

Worked Example 13.2 — Ratio of radii

Find the ratio of nuclear radii of \(^{197}_{79}\mathrm{Au}\) and \(^{107}_{47}\mathrm{Ag}\).
\[ \frac{R_{\text{Au}}}{R_{\text{Ag}}} = \left( \frac{197}{107} \right)^{1/3} = (1.841)^{1/3} \approx 1.226 \]

So a gold nucleus is about 22.6% larger in radius than a silver nucleus.

Interactive — Nuclear Size & Density Calculator

Nuclear-radius simulator

Pick an isotope; the simulator computes its radius from \(R = R_0 A^{1/3}\), draws a scale-correct sphere, and displays the constant nuclear density.

Mass number A: 16 Radius R: 3.0 fm Density ρ: 2.3×10¹⁷ kg/m³
0 scale: 1 px ≈ 0.04 fm ¹⁶O R

As A increases, R grows only as A1/3. Density stays virtually constant.

Competency-Based Questions

Q1 (MCQ). The atomic mass unit (1 u) equals:

  • (a) 1.673 × 10⁻²⁷ kg
  • (b) 1.661 × 10⁻²⁷ kg
  • (c) 9.11 × 10⁻³¹ kg
  • (d) 1.602 × 10⁻¹⁹ kg
(b) 1.660539 × 10⁻²⁷ kg by definition (1/12 the mass of a ¹²C atom).

Q2 (MCQ). Two nuclei \(^{40}_{18}\mathrm{Ar}\) and \(^{40}_{20}\mathrm{Ca}\) are best described as:

  • (a) Isotopes
  • (b) Isobars
  • (c) Isotones
  • (d) Identical nuclei
(b) Same A = 40, different Z — they are isobars.

Q3 (Short Answer). State two reasons why electron scattering gives a more accurate value of nuclear radius than α-scattering.

(i) Electrons interact only via the well-understood electromagnetic force; α-particles also feel the strong nuclear force at small distances, which complicates analysis. (ii) Electrons can be accelerated to much higher energies, giving smaller de Broglie wavelengths and hence finer spatial resolution.

Q4 (Numerical). Estimate the radius of \(^{27}_{13}\mathrm{Al}\) given R₀ = 1.2 fm.

\(R = 1.2 \times 27^{1/3} = 1.2 \times 3 = 3.6\) fm.

Q5 (HOTS). Show that the nuclear matter density is independent of A and estimate it numerically.

Mass of nucleus ≈ A·m_n where m_n ≈ 1.67×10⁻²⁷ kg. Volume = (4/3)πR³ = (4/3)π(R₀A^(1/3))³ = (4/3)πR₀³·A. Therefore density ρ = mass/volume = m_n / ((4/3)πR₀³), which is independent of A. Plugging in: ρ ≈ 1.67×10⁻²⁷ / ((4/3)π(1.2×10⁻¹⁵)³) ≈ 2.3 × 10¹⁷ kg/m³.

Assertion–Reason Questions

Options: (A) Both true, R is the correct explanation. (B) Both true, R is not the correct explanation. (C) A true, R false. (D) A false, R true.

Assertion: Two atoms of the same element with different mass numbers can have identical chemical properties.

Reason: Chemistry is determined by the electron configuration, which depends on Z, not on N.

(A) Both correct — and the reason explains the assertion. Isotopes occupy the same place in the periodic table.

Assertion: The density of nuclear matter is the same for the lightest and heaviest stable nuclei.

Reason: The nuclear radius is exactly proportional to the mass number A.

(C) Assertion is true but the reason is false. The radius scales as A^(1/3), not A. Volume scales as A, which is what makes density independent of A.

Assertion: A free neutron decays spontaneously, but a neutron inside a stable nucleus does not.

Reason: Inside a nucleus, the energetics of the strong-force binding can forbid the decay channel that is open to a free neutron.

(A) Both correct and the reason explains the assertion.

Frequently Asked Questions - Atomic Mass Nuclear Composition

What is the main concept covered in Atomic Mass Nuclear Composition?
In NCERT Class 12 Physics Chapter 13 (Nuclei), "Atomic Mass Nuclear Composition" covers the core principles and equations students need for board exam success. The MyAiSchool lesson explains the topic with definitions, derivations, worked examples, and interactive simulations. Key formulas and dimensional analysis are included to build conceptual depth and problem-solving skills aligned with the CBSE 2025-26 syllabus.
How is Atomic Mass Nuclear Composition useful in real-life applications?
Real-life applications of "Atomic Mass Nuclear Composition" from NCERT Class 12 Physics Chapter 13 include electronics, communication systems, medical imaging, solar energy, semiconductor devices, and modern technology. The MyAiSchool lesson links every concept to a tangible example so students see physics as a problem-solving framework for the physical world, not as abstract formulas.
What are the key formulas in Atomic Mass Nuclear Composition?
Key formulas in "Atomic Mass Nuclear Composition" (NCERT Class 12 Physics Chapter 13 Nuclei) are derived step-by-step in the MyAiSchool lesson. Students should memorize the final formula AND understand its derivation for full board marks. Each formula is listed with its dimensional formula, SI unit, applicability range, and common pitfalls. The Summary section at the end of each part includes a quick-reference formula card.
How does this part connect to other parts of Chapter 13?
NCERT Class 12 Physics Chapter 13 (Nuclei) is structured so each part builds on the previous one. "Atomic Mass Nuclear Composition" connects directly to neighbouring parts via shared definitions, units, and methodology. The MyAiSchool lesson cross-references related concepts with internal links so students can navigate the whole chapter as one connected story rather than disconnected fragments.
What types of CBSE board questions come from Atomic Mass Nuclear Composition?
CBSE board questions from "Atomic Mass Nuclear Composition" typically include: (1) 1-mark MCQs on definitions and formulas, (2) 2-mark short-answer derivations or applications, (3) 3-mark numerical problems with units, (4) 5-mark long-answer derivations followed by application. The MyAiSchool lesson tags each Competency-Based Question (CBQ) with Bloom level (L1-L6) so students know how to study for each weight.
How can students use the interactive simulation effectively?
The interactive simulation in the "Atomic Mass Nuclear Composition" lesson allows students to adjust input parameters (sliders or selectors) and see physical quantities update in real time. To use it effectively: (1) try extreme values to understand limiting cases, (2) compare with the analytical formula, (3) check unit consistency, (4) test special configurations from worked examples. The simulation reinforces conceptual intuition that pure formula manipulation cannot.
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