Atoms & Nuclei – JIPMER Study Notes
Audience: 12th‑grade physics students preparing for the JIPMER entrance exam.
Goal: Build a strong conceptual foundation and problem‑solving practice for the atoms & nuclei chapter.
1. Simple Explanation
| Concept | What It Means | Key Take‑away |
|---|---|---|
| Atom | Small, electrically neutral particle made of a nucleus (protons + neutrons) surrounded by electrons. | Electrons carry negative charge; protons positive; neutrons neutral. |
| Nucleus | Dense core containing protons and neutrons (collectively called nucleons). | Holds most of the atom’s mass; governed by strong nuclear force. |
| Isotope | Elements with same proton number (Z) but different neutron number (N). | Same chemical behaviour, different nuclear stability. |
| Mass Number (A) | Total number of nucleons: (A = Z + N). | Used to label isotopes. |
| Binding Energy | Energy required to separate all nucleons in a nucleus. | More binding energy → more stable nucleus. |
| Radioactive Decay | Spontaneous transformation of an unstable nucleus into a more stable one, emitting radiation. | Types: α, β, γ, fission, fusion. |
| Half‑Life ((t_{1/2})) | Time taken for half the nuclei in a sample to decay. | Short (t_{1/2}): rapid decay; long (t_{1/2}): very stable. |
| Nuclear Fusion | Light nuclei combine to form a heavier nucleus (energy released). | Sun’s energy source. |
| Nuclear Fission | Heavy nucleus splits into lighter nuclei (energy released). | Nuclear reactors, atomic bombs. |
2. Key Points (Bullet‑style for quick recall)
-
Atomic Structure
- Proton (p⁺) mass ≈ 1.007 u, charge +e.
- Neutron (n⁰) mass ≈ 1.009 u, charge 0.
- Electron (e⁻) mass ≈ 0.0005 u, charge –e.
- Atomic mass ≈ mass of nucleus (electrons negligible).
-
Isotopes & Mass Defect
- Mass defect ( \Delta m = (Z,m_p + N,m_n) - m_{\text{nucleus}}).
- Binding energy (E_b = \Delta m,c^2).
- Binding energy per nucleon peaks for Fe‑56 (~8.8 MeV).
-
Types of Decay
Decay Equation Change in Z & A Energy α‑decay ((A,Z) \rightarrow (A-4, Z-2) + \alpha) (\Delta Z=-2,; \Delta A=-4) ~5–10 MeV β⁻‑decay ((A,Z) \rightarrow (A, Z+1) + e^- + \bar\nu_e) (\Delta Z=+1) ≤3 MeV (continuous spectrum) β⁺‑decay / electron capture ((A,Z) \rightarrow (A, Z-1) + e^+ + \nu_e) (\Delta Z=-1) ~2 MeV γ‑decay ((A,Z)^* \rightarrow (A,Z) + \gamma) No change in A,Z 0.1–10 MeV -
Fission & Fusion
- Fission: (^{235}\mathrm{U} + n \rightarrow) 2 fragments + 2–3 n + ~200 MeV.
- Fusion: (^{1}\mathrm{H} + {}^{1}\mathrm{H} \rightarrow {}^{2}\mathrm{D} + e^+ + \nu_e + 0.42;\text{MeV}).
(^{4}\mathrm{He} + 2,^{1}\mathrm{H} \rightarrow {}^{4}\mathrm{He} + 2,\gamma + 28.3;\text{MeV}).
-
Half‑Life Formula [ N(t) = N_0 e^{-\lambda t},\quad \lambda = \frac{\ln 2}{t_{1/2}} ] where (N(t)) is the remaining number of nuclei after time (t).
-
Radioactivity Units
- Becquerel (Bq): 1 decay/s = (1,\text{s}^{-1}).
- Curie (Ci): (3.7\times10^{10}) decays/s.
3. Worked Example
Problem:
A sample of (^{226}\mathrm{Ra}) (half‑life (t_{1/2}=1600) years) contains (0.1) g initially.
(a) How many atoms are present initially?
(b) After 2400 years, what mass of (^{226}\mathrm{Ra}) remains?
Given:
- Molar mass of (^{226}\mathrm{Ra}) ≈ 226 g/mol.
- Avogadro’s number (N_A = 6.022\times10^{23},\text{mol}^{-1}).
(a) Initial number of atoms
[ N_0 = \frac{m}{M} \times N_A = \frac{0.1,\text{g}}{226,\text{g/mol}} \times 6.022\times10^{23} \approx 2.66\times10^{20},\text{atoms} ]
(b) Remaining mass after 2400 years
- Compute decay constant (\lambda).
[ \lambda = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{1600,\text{yr}} = 4.33\times10^{-4},\text{yr}^{-1