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InterviewSolution
| 1. |
If the mass defect of 4X9 is 0⋅09 a.m.u, then the binding energy per nucleon is : (a) 8.38 MeV(b) 83⋅8 MeV (c) 931⋅5 MeV (d) 9⋅315 MeV |
| Answer» <p style="text-align: justify;"><span style="font-family:Arial,Helvetica,sans-serif"><span style="color:#000000"><strong>Correct option (d) 9⋅315 MeV</strong></span></span></p><p style="text-align: justify;"><span style="font-family:Arial,Helvetica,sans-serif"><span style="color:#000000"><strong>Explanation:</strong></span></span></p><p style="text-align: justify;"><span style="font-family:Arial,Helvetica,sans-serif"><span style="color:#000000">Binding energy of nucleus = Mass defect in amu × 931.48 MeV</span></span></p><p style="text-align: justify;"><span style="font-family:Arial,Helvetica,sans-serif"><span style="color:#000000">= 0.09 × 931.48</span></span></p><p style="text-align: justify;"><span style="font-family:Arial,Helvetica,sans-serif"><span style="color:#000000">= 83.83 MeV</span></span></p><p style="text-align: justify;"><span style="font-family:Arial,Helvetica,sans-serif"><span style="color:#000000">Binding energy per nucleon</span></span></p><p style="text-align: justify;"><span style="font-family:Arial,Helvetica,sans-serif"><span style="color:#000000">= binding energy of the nucleus/number of nucleon (mass number)</span></span></p><p style="text-align: justify;"><span style="font-family:Arial,Helvetica,sans-serif"><span style="color:#000000">= 83.83/9 = 9.315 MeV</span></span></p> | |