For an isotope the mass defect per nucleus is

PHYSICS · Single correct · AP · Physics 2: Algebra-based · Modern Physics · Modern Physics

Problem

For an isotope the mass defect per nucleus is Δm=0.0024 u\Delta m = 0.0024\,\text{u}. Given 1 u=931 MeV/c21\,\text{u}=931\,\text{MeV}/c^{2}, 1 MeV=1.602×10−13 J1\,\text{MeV}=1.602\times10^{-13}\,\text{J} and Avogadro number NA=6.022×1023 mol−1N_A = 6.022\times10^{23}\,\text{mol}^{-1}, the energy released when 1 mol1\,\text{mol} of the nuclei is completely disassembled into free nucleons is approximately (a) 2.2×1010 J2.2\times10^{10}\,\text{J} (b) 2.2×1013 J2.2\times10^{13}\,\text{J} (c) 2.2×1011 J2.2\times10^{11}\,\text{J} (d) 3.6×109 J3.6\times10^{9}\,\text{J}

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