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#nuclear stability

5 public questions tagged with this topic.

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

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

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

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

What role does the binding energy play in nuclear stability?

**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. Higher binding energy indicates a more tightly bound nucleus, making it more stable as it requires more energy to break apart, reflecting the strength of the nuclear force. 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 Indicates tighter binding, consi

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