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Radioactive Decay, Nuclear Forces and Stability

This category covers the fundamental concepts of radioactive decay, the various nuclear forces that bind protons and neutrons, and the factors that determine nuclear stability. Topics include alpha, beta and gamma decay, binding energy, and how stability influences element behavior.

30 questions

The binding energy per nucleon of a nucleus is \( 8.5 \, \text{MeV} \). What is the total binding energy for a nucleus w

**Nuclear force** strong, short-range ~1 fm, attractive, charge independent, saturated in large nuclei because each nucleon interacts only with neighbors, not all others, so BE/A saturates ~8 MeV, primary factor limiting stable nuclei size is Coulomb repulsion between protons growing as Z² vs strong force saturating, beyond Z≈83 no stable nuclei, competition between Coulomb and strong. Total binding energy = Ebₙ × A . Ebₙ = 8.5 MeV , A = 20 . E_b = 8.5 × 20 = 170 MeV . Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀ A^¹/³, BE = Δm c² and 1

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How much energy is equivalent to a mass defect of \( 0.1 \, \text{u} \)? (Given \( 1 \, \text{u} = 931.5 \, \text{MeV/c}

**Stability** belt of stability N≈Z for light, N>Z for heavy due to Coulomb, beyond leads to alpha decay, fission, stability requires balance, nuclear force saturated explains constant density and BE/A. Energy = Δ M · c² . Δ M = 0.1 u . Energy = 0.1 × 931.5 = 93.15 MeV . Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀ A^¹/³, BE = Δm c² and 1 u = 931.5 MeV, evaluation yields 93.15 MeV, consistent with Bohr model and nuclear binding energy systematics.

Ref: NCERT > Physics Book > Atoms and Nuclei > Radioactive Decay, Nuclear Forces and Stability

What is the primary factor limiting the size of stable nuclei?

**Radioactive decay** occurs when nucleus unstable, alpha decay emits He-4, beta decay neutron→proton+electron+antineutrino, gamma decay photon emission, decay law N=N₀ e^{-λt}, half-life T½=ln2/λ, nuclear density ~10¹⁷ kg/m³, nuclear force saturated means BE/A constant for A>20. In large nuclei, the Coulomb repulsion between protons increases with atomic number, counteracting the nuclear force and reducing stability, limiting the size of stable nuclei. Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀ A^¹/³, BE = Δm c² and 1 u = 931.5 MeV, evaluation yields Coulomb repulsion, consisten

Ref: NCERT > Physics Book > Atoms and Nuclei > Radioactive Decay, Nuclear Forces and Stability

A nucleus with mass number 18 has a binding energy of \( 144 \, \text{MeV} \). What is its binding energy per nucleon?

**Nuclear force** strong, short-range ~1 fm, attractive, charge independent, saturated in large nuclei because each nucleon interacts only with neighbors, not all others, so BE/A saturates ~8 MeV, primary factor limiting stable nuclei size is Coulomb repulsion between protons growing as Z² vs strong force saturating, beyond Z≈83 no stable nuclei, competition between Coulomb and strong. Ebₙ = (E_b/A) . E_b = 144 MeV , A = 18 . Ebₙ = (144/18) = 8.0 MeV . Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀ A^¹/³, BE = Δm c² and 1 u = 931.5 MeV, evaluation yields

Ref: NCERT > Physics Book > Atoms and Nuclei > Radioactive Decay, Nuclear Forces and Stability

A nucleus with mass number 28 has a binding energy of \( 224 \, \text{MeV} \). What is its binding energy per nucleon?

**Stability** belt of stability N≈Z for light, N>Z for heavy due to Coulomb, beyond leads to alpha decay, fission, stability requires balance, nuclear force saturated explains constant density and BE/A. Ebₙ = (E_b/A) . E_b = 224 MeV , A = 28 . Ebₙ = (224/28) = 8.0 MeV . Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀ A^¹/³, BE = Δm c² and 1 u = 931.5 MeV, evaluation yields 8.0 MeV, consistent with Bohr model and nuclear binding energy systematics.

Ref: NCERT > Physics Book > Atoms and Nuclei > Radioactive Decay, Nuclear Forces and Stability

A nucleus has a binding energy of \( 127.5 \, \text{MeV} \) and mass number 16. What is its binding energy per nucleon?

**Radioactive decay** occurs when nucleus unstable, alpha decay emits He-4, beta decay neutron→proton+electron+antineutrino, gamma decay photon emission, decay law N=N₀ e^{-λt}, half-life T½=ln2/λ, nuclear density ~10¹⁷ kg/m³, nuclear force saturated means BE/A constant for A>20. Ebₙ = (E_b/A) . E_b = 127.5 MeV , A = 16 . Ebₙ = (127.5/16) ≈ 7.97 MeV ≈ 8 MeV . Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀ A^¹/³, BE = Δm c² and 1 u = 931.5 MeV, evaluation yields 8.0 MeV, consistent with Bohr model and nuclear binding energy systematics.

Ref: NCERT > Physics Book > Atoms and Nuclei > Radioactive Decay, Nuclear Forces and Stability

What is the radius of a nucleus with mass number 16? (Given \( R_0 = 1.2 \times 10^{-15} \, \text{m} \))

**Nuclear force** strong, short-range ~1 fm, attractive, charge independent, saturated in large nuclei because each nucleon interacts only with neighbors, not all others, so BE/A saturates ~8 MeV, primary factor limiting stable nuclei size is Coulomb repulsion between protons growing as Z² vs strong force saturating, beyond Z≈83 no stable nuclei, competition between Coulomb and strong. The radius of a nucleus is given by R = R₀ A¹/³ , where A = 16 . A¹/³ = 16¹/³ = (2⁴)¹/³ = 2⁴/³ ≈ 2.52 . R = 1.2 × 10⁻¹⁵ × 2.52 ≈ 3.0 × 10⁻¹⁵ m . Using E_n = -13.6/n² eV, r_n = n²

Ref: NCERT > Physics Book > Atoms and Nuclei > Radioactive Decay, Nuclear Forces and Stability

What is the nuclear density of a nucleus with mass \( 3.67 \times 10^{-27} \, \text{kg} \) and radius \( 2.0 \times 10^{

**Stability** belt of stability N≈Z for light, N>Z for heavy due to Coulomb, beyond leads to alpha decay, fission, stability requires balance, nuclear force saturated explains constant density and BE/A. Density = (mass/volume) , Volume = (4/3) π R³ . R³ = (2.0 × 10⁻¹⁵)³ = 8.0 × 10⁻⁴⁵ m³ . Volume = (4/3) × 3.14 × 8.0 × 10⁻⁴⁵ ≈ 3.35 × 10⁻⁴⁴ m³ . Density = (3.67 × 10⁻²⁷/3.35 × 10⁻⁴⁴) ≈ 1.10 × 10¹⁷ kg/m³ . Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀ A^¹/³, BE = Δm c² and 1 u = 931.5

Ref: NCERT > Physics Book > Atoms and Nuclei > Radioactive Decay, Nuclear Forces and Stability

A nucleus with mass number 12 has a binding energy of \( 96 \, \text{MeV} \). What is its binding energy per nucleon?

**Nuclear force** strong, short-range ~1 fm, attractive, charge independent, saturated in large nuclei because each nucleon interacts only with neighbors, not all others, so BE/A saturates ~8 MeV, primary factor limiting stable nuclei size is Coulomb repulsion between protons growing as Z² vs strong force saturating, beyond Z≈83 no stable nuclei, competition between Coulomb and strong. Ebₙ = (E_b/A) . E_b = 96 MeV , A = 12 . Ebₙ = (96/12) = 8.0 MeV . Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀ A^¹/³, BE = Δm c² and 1 u = 931.5 MeV, evaluation yields

Ref: NCERT > Physics Book > Atoms and Nuclei > Radioactive Decay, Nuclear Forces and Stability

Why is the nuclear force considered saturated in large nuclei?

**Stability** belt of stability N≈Z for light, N>Z for heavy due to Coulomb, beyond leads to alpha decay, fission, stability requires balance, nuclear force saturated explains constant density and BE/A. The nuclear force is short-ranged, affecting only a fixed number of neighboring nucleons, so adding more nucleons in large nuclei does not proportionally increase the binding energy, leading to saturation. Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀ A^¹/³, BE = Δm c² and 1 u = 931.5 MeV, evaluation yields Limited range of interaction, consistent with Bohr model and nuclear binding

Ref: NCERT > Physics Book > Atoms and Nuclei > Radioactive Decay, Nuclear Forces and Stability

A nucleus with mass number 36 has a binding energy of \( 288 \, \text{MeV} \). What is its binding energy per nucleon?

**Nuclear force** strong, short-range ~1 fm, attractive, charge independent, saturated in large nuclei because each nucleon interacts only with neighbors, not all others, so BE/A saturates ~8 MeV, primary factor limiting stable nuclei size is Coulomb repulsion between protons growing as Z² vs strong force saturating, beyond Z≈83 no stable nuclei, competition between Coulomb and strong. Ebₙ = (E_b/A) . E_b = 288 MeV , A = 36 . Ebₙ = (288/36) = 8.0 MeV . Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀ A^¹/³, BE = Δm c² and 1 u = 931.5 MeV, evaluation yields

Ref: NCERT > Physics Book > Atoms and Nuclei > Radioactive Decay, Nuclear Forces and Stability

A nucleus with mass number 100 has a binding energy of \( 850 \, \text{MeV} \). What is its binding energy per nucleon?

**Stability** belt of stability N≈Z for light, N>Z for heavy due to Coulomb, beyond leads to alpha decay, fission, stability requires balance, nuclear force saturated explains constant density and BE/A. Ebₙ = (E_b/A) . E_b = 850 MeV , A = 100 . Ebₙ = (850/100) = 8.5 MeV . Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀ A^¹/³, BE = Δm c² and 1 u = 931.5 MeV, evaluation yields 8.5 MeV, consistent with Bohr model and nuclear binding energy systematics.

Ref: NCERT > Physics Book > Atoms and Nuclei > Radioactive Decay, Nuclear Forces and Stability