Nuclei
1. Mass-Energy and Binding Energy
Einstein's mass-energy equivalence: E = mc2.
Mass Defect (Δm): Difference between mass of nucleons and nucleus mass.
Δm = [Z mp + (A-Z) mn] - M.
Binding Energy (Eb): Energy equivalent to mass defect which holds nucleus together.
Eb = Δm × 931.5 MeV (if Δm in amu).
Δm = [Z mp + (A-Z) mn] - M.
Binding Energy (Eb): Energy equivalent to mass defect which holds nucleus together.
Eb = Δm × 931.5 MeV (if Δm in amu).
Stability: Higher B.E./nucleon → More Stable.
- Max stability at A ≈ 56 (Fe).
- Light nuclei fuse (Fusion) and Heavy nuclei split (Fission) to increase B.E./nucleon.
- Max stability at A ≈ 56 (Fe).
- Light nuclei fuse (Fusion) and Heavy nuclei split (Fission) to increase B.E./nucleon.
2. Radioactivity
Spontaneous emission of radiation (α, β, γ) from unstable nuclei.
Law of Radioactive Decay: N = N0 e-λt.
Rate of Decay (Activity): R = -dN/dt = λN.
Half Life (T1/2): Time for N to become N0/2.
T1/2 = ln(2)/λ = 0.693/λ.
Mean Life (τ): τ = 1/λ = 1.44 T1/2.
Rate of Decay (Activity): R = -dN/dt = λN.
Half Life (T1/2): Time for N to become N0/2.
T1/2 = ln(2)/λ = 0.693/λ.
Mean Life (τ): τ = 1/λ = 1.44 T1/2.
(Asked in NEET 2018, 2019, 2022)
3. Nuclear Energy
Fission: Heavy nucleus splits into lighter nuclei + Energy.
92U235 + 0n1 → 56Ba144 + 36Kr89 + 3 0n1 + Energy.
Fusion: Light nuclei combine to form heavy nucleus + Energy (Source of Sun's energy).
4 1H1 → 2He4 + 2 +1e0 + Energy (26.7 MeV).
92U235 + 0n1 → 56Ba144 + 36Kr89 + 3 0n1 + Energy.
Fusion: Light nuclei combine to form heavy nucleus + Energy (Source of Sun's energy).
4 1H1 → 2He4 + 2 +1e0 + Energy (26.7 MeV).
(Asked in NEET 2016, 2020)
Numericals: Nuclei
Q1. Half life of a radioactive substance is 30 days. How much remains after 90 days?
Solution:
Number of half lives n = t / T1/2 = 90 / 30 = 3.
Remaining fraction = (1/2)n = (1/2)3 = 1/8.
Number of half lives n = t / T1/2 = 90 / 30 = 3.
Remaining fraction = (1/2)n = (1/2)3 = 1/8.
Q2. Calculate energy equivalent of 1 gram of substance.
Solution:
E = mc2.
m = 10-3 kg, c = 3 × 108 m/s.
E = 10-3 × (3 × 108)2 = 10-3 × 9 × 1016.
E = 9 × 1013 J.
E = mc2.
m = 10-3 kg, c = 3 × 108 m/s.
E = 10-3 × (3 × 108)2 = 10-3 × 9 × 1016.
E = 9 × 1013 J.
Q3. Mass numbers of two nuclei are 1 and 27. Find ratio of their nuclear radii.
Solution:
R ∝ A1/3.
R1 / R2 = (A1 / A2)1/3.
Ratio = (1 / 27)1/3 = (1 / 33)1/3 = 1:3.
R ∝ A1/3.
R1 / R2 = (A1 / A2)1/3.
Ratio = (1 / 27)1/3 = (1 / 33)1/3 = 1:3.
Q4. Mass defect of He-4 nucleus is 0.0303 u. Find Binding Energy per nucleon.
Solution:
Total B.E. = Δm × 931.5 MeV.
B.E. = 0.0303 × 931.5 = 28.22 MeV.
B.E. per nucleon = B.E. / A = 28.22 / 4 = 7.05 MeV.
Total B.E. = Δm × 931.5 MeV.
B.E. = 0.0303 × 931.5 = 28.22 MeV.
B.E. per nucleon = B.E. / A = 28.22 / 4 = 7.05 MeV.
Q5. Half life is 20 mins. Time interval between 33% decay and 67% decay?
Solution:
33% decay → 67% remains ≈ 2/3 N0.
67% decay → 33% remains ≈ 1/3 N0.
From 2/3 to 1/3 is halving the amount.
Time taken is one half life = 20 mins.
33% decay → 67% remains ≈ 2/3 N0.
67% decay → 33% remains ≈ 1/3 N0.
From 2/3 to 1/3 is halving the amount.
Time taken is one half life = 20 mins.
Q6. Mean life is 100 sec. Find decay constant λ.
Solution:
Mean life τ = 1/λ.
λ = 1/τ = 1/100 = 0.01 s-1.
Mean life τ = 1/λ.
λ = 1/τ = 1/100 = 0.01 s-1.
Q7. Nucleus A (mass 200) at rest splits into B (mass 100) and C (mass 100). Q-value is 200 MeV. KE of B?
Solution:
By symmetry and conservation of momentum, Energy divides equally if masses equal.
KE of B = KE of C = Q/2.
KE of B = 200 / 2 = 100 MeV.
By symmetry and conservation of momentum, Energy divides equally if masses equal.
KE of B = KE of C = Q/2.
KE of B = 200 / 2 = 100 MeV.
Q8. Count rate falls from 1600 per min to 100 per min in 8 hours. Find half life.
Solution:
1600 → 800 → 400 → 200 → 100.
Total 4 half lives passed.
4 T1/2 = 8 hours.
T1/2 = 2 hours.
1600 → 800 → 400 → 200 → 100.
Total 4 half lives passed.
4 T1/2 = 8 hours.
T1/2 = 2 hours.
Q9. Four protons fuse to form He-4. Mass defect is 0.02866 u. Energy released?
Solution:
E = Δm × 931.5 MeV.
E = 0.02866 × 931.5.
E ≈ 26.7 MeV.
E = Δm × 931.5 MeV.
E = 0.02866 × 931.5.
E ≈ 26.7 MeV.
Q10. Ratio of nuclear density of O-16 and Ca-40?
Solution:
Nuclear density is constant for all nuclei and independent of Mass number A.
Ratio = 1:1.
Nuclear density is constant for all nuclei and independent of Mass number A.
Ratio = 1:1.
Important Formulae
20 NEET Golden Facts
- 1. Nuclear Force: Strongest force in nature. Short range (~2-3 fm). Charge independent.
- 2. Isotopes: Same Z, different A (Chemical properties same).
- 3. Isobars: Same A, different Z (Physical properties differ).
- 4. Isotones: Same number of neutrons (A-Z).
- 5. 1 amu (u): 1/12th mass of C-12 atom = 1.66 × 10-27 kg.
- 6. Energy of 1 amu: 931.5 MeV.
- 7. Nuclear Density: Constant for all nuclei (Independent of size A). ρ ~ 1017 kg/m3.
- 8. Nuclear Radius: R = R0 A1/3. (R0 = 1.2 fm).
- 9. Alpha Decay: Z → Z-2, A → A-4. He nucleus emitted.
- 10. Beta Minus Decay: Z → Z+1. Neutron converts to Proton + Electron + Antineutrino.
- 11. Beta Plus Decay: Z → Z-1. Proton converts to Neutron + Positron + Neutrino.
- 12. Gamma Decay: No change in Z or A. Excited nucleus returns to ground state.
- 13. Activity Unit: 1 Becquerel (Bq) = 1 decay/s. 1 Curie (Ci) = 3.7 × 1010 Bq.
- 14. Fusion Condition: High Temperature (109K) and High Pressure.
- 15. Iron (Fe-56): Most stable nucleus (Max B.E./nucleon ~8.75 MeV).
- 16. Control Rods: Cadmium or Boron. Absorb neutrons to control fission rate.
- 17. Moderator: Heavy water (D2O) or Graphite. Slows down fast neutrons.
- 18. Mean Life Relation: After time τ, 37% of original nuclei remain undecayed.
- 19. Pair Production: γ-ray photon converts into Electron-Positron pair (Energy → Mass).
- 20. Solar Energy: Result of p-p cycle (Nuclear Fusion) in Sun's core.
📱 Practice MCQs for this topic inside our App
📱 Practice MCQs for this topic inside our App
📱 Practice MCQs for this topic inside our App
📱 Practice MCQs for this topic inside our App