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Question

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

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Explanation

**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

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