આ MCQ મોડ્યુલ આના પર આધારિત છે: Atomic Mass Nuclear Composition
Atomic Mass Nuclear Composition
આ મૂલ્યાંકન આના પર આધારિત હશે: Atomic Mass Nuclear Composition
મૂલ્યાંકન બનાવવામાં તેમની સામગ્રી સામેલ કરવા ચિત્રો, PDF અથવા Word દસ્તાવેજ અપલોડ કરો.
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.
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:
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
which matches the periodic-table entry.
Hydrogen — the lightest element — has three isotopes:
| Name | Symbol | Mass (u) | Abundance | Stability |
|---|---|---|---|---|
| Protium (proton) | \(^{1}_{1}\mathrm{H}\) | 1.00783 | 99.985% | stable |
| Deuterium | \(^{2}_{1}\mathrm{H}\) (D) | 2.01410 | 0.015% | stable |
| Tritium | \(^{3}_{1}\mathrm{H}\) (T) | 3.01605 | — | radioactive (T½ ≈ 12.3 y) |
Proton, Neutron and Discovery of the Neutron
The nucleus of the lightest hydrogen atom is the proton, with mass
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:
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.
- Z = atomic number = number of protons
- N = neutron number = number of neutrons
- A = mass number = Z + N = number of nucleons (protons + neutrons)
Isotopes, Isobars and Isotones
| Family | Same | Different | Examples |
|---|---|---|---|
| Isotopes | Z | N (and A) | \(^{1}_{1}\mathrm{H}, ^{2}_{1}\mathrm{H}, ^{3}_{1}\mathrm{H}\); \(^{12}_{6}\mathrm{C}, ^{14}_{6}\mathrm{C}\) |
| Isobars | A | Z and N | \(^{3}_{1}\mathrm{H}\) and \(^{3}_{2}\mathrm{He}\); \(^{40}_{18}\mathrm{Ar}\) and \(^{40}_{20}\mathrm{Ca}\) |
| Isotones | N | Z (and A) | \(^{198}_{80}\mathrm{Hg}\) and \(^{197}_{79}\mathrm{Au}\) (both N = 118) |
For the nuclide \(^{40}_{19}\mathrm{K}\):
- How many protons does it have?
- How many neutrons?
- Is \(^{40}_{20}\mathrm{Ca}\) its isotope, isobar or isotone?
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:
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.
Worked Example 13.1 — Nuclear density of iron
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
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.
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:
Q2 (MCQ). Two nuclei \(^{40}_{18}\mathrm{Ar}\) and \(^{40}_{20}\mathrm{Ca}\) are best described as:
Q3 (Short Answer). State two reasons why electron scattering gives a more accurate value of nuclear radius than α-scattering.
Q4 (Numerical). Estimate the radius of \(^{27}_{13}\mathrm{Al}\) given R₀ = 1.2 fm.
Q5 (HOTS). Show that the nuclear matter density is independent of A and estimate it numerically.
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.
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.
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.
Frequently Asked Questions - Atomic Mass Nuclear Composition
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Physics — CBSE Class XII Sample Paper 1 (2025-26)
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