#nucleon

Radioactive Decay, Nuclear Forces and Stability

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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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

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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

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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

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Nuclear Reactions - Fission, Fusion and Energy Release

What is the significance of the binding energy per nucleon curve?

**Nuclear fusion** source of energy in Sun, proton-proton cycle 4p→He+2e⁺+2ν+26.7 MeV, high temperature ~10⁷ K needed to give kinetic energy to overcome repulsion, thermal motion at high T allows tunneling, energy release because He BE/A higher than H. Fission releases energy because heavy nucleus BE/A ~7.6 MeV splits to intermediate ~8.5 MeV. The binding energy per nucleon curve shows how tightly nucleons are bound, indicating that energy is released in fission (heavy nuclei) or fusion (light nuclei) when transitioning to higher binding energy states. Using E_n = -13.6/n² eV, r_n = n² a₀, L =

Ref: NCERT > Physics Book > Atoms and Nuclei > Nuclear Reactions - Fission, Fusion and Energy Release

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Radioactive Decay, Nuclear Forces and Stability

Why is the binding energy per nucleon a useful measure of nuclear stability?

**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. It represents the average energy needed to remove a nucleon, with higher values indicating a more stable nucleus due to stronger binding against disruptive forces like Coulomb repulsion. Using E_n = -13.6/n² eV,

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

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Mass Defect, Binding Energy and Binding Energy per Nucleon

What is the binding energy per nucleon if a nucleus with mass number 56 has a total binding energy of \( 490 \, \text{Me

**Mass defect** Δm = Z m_p + N m_n - M_nucleus, binding energy BE = Δm c², 1 u =931.5 MeV/c², BE per nucleon = BE/A, measures stability, peak ~8.8 MeV at Fe-56, for A=36 BE=288 MeV BE/A=8 MeV, for A=12 BE=96 MeV BE/A=8 MeV, for A=16 BE=127.5 MeV BE/A≈7.97 MeV, higher BE/A more stable. Ebₙ = (E_b/A) . E_b = 490 MeV , A = 56 . Ebₙ = (490/56) ≈ 8.75 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 8.75

Ref: NCERT > Physics Book > Atoms and Nuclei > Mass Defect, Binding Energy and Binding Energy per Nucleon

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Mass Defect, Binding Energy and Binding Energy per Nucleon

Why does the binding energy per nucleon remain nearly constant for nuclei with mass numbers between 30 and 170?

**Mass defect** Δm = Z m_p + N m_n - M_nucleus, binding energy BE = Δm c², 1 u =931.5 MeV/c², BE per nucleon = BE/A, measures stability, peak ~8.8 MeV at Fe-56, for A=36 BE=288 MeV BE/A=8 MeV, for A=12 BE=96 MeV BE/A=8 MeV, for A=16 BE=127.5 MeV BE/A≈7.97 MeV, higher BE/A more stable. The nuclear force is short-ranged, affecting only neighboring nucleons. In medium-sized nuclei (A = 30 to 170), most nucleons are inside the nucleus, surrounded by a constant number of neighbors, leading to a saturation effect and nearly constant binding energy per nucleon. Using E_n = -13.6/n² eV, r_n = n²

Ref: NCERT > Physics Book > Atoms and Nuclei > Mass Defect, Binding Energy and Binding Energy per Nucleon

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