$$ \\\\ $$ Find the energy required for each nucleon.","comment":{"@type":"Comment","text":" Calculate the mass defect and from that, find the total energy and divide it by the mass number to find binding energy per nucleon."},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" 6.06","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" 5.56","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" 8.85","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" 7.07","position":1,"answerExplanation":{"@type":"Comment","text":" The mass defect shown by the particle is $$[2.01565+2.01733-4.002604]$$ which is $$0.030376 amu.$$ So, the energy required per nucleon will be $$\\dfrac{0.030376\\times 931}{4}=7.07MeV$$.","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Nuclear fission and fusion Quiz 1","text":" Find the nuclei at any time instant ‘t’ in a fission reaction, with successive production:","comment":{"@type":"Comment","text":" Find the limiting case, for the maximum number of particles produced, then integrate the equations from 0 to t."},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$\\dfrac{\\lambda }{a}(1-{{e}^{-\\lambda t}})$$","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" $$\\dfrac{a}{\\lambda }(1-{{e}^{\\lambda t}})$$","position":2},{"@type":"Answer","encodingFormat":"text/html","text":"$$\\dfrac{\\lambda }{a}(1-{{e}^{\\lambda t}})$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$\\dfrac{a}{\\lambda }(1-{{e}^{-\\lambda t}})$$","position":0,"answerExplanation":{"@type":"Comment","text":"$$\\xrightarrow{\\alpha }A\\xrightarrow{\\lambda }B,\\,\\,\\,\\alpha \\text{ is rate of production} \\\\ \\text{we have}\\\\ \\dfrac{d{{N}_{A}}}{dt}=\\alpha -\\lambda {{N}_{a}}. \\text{When }{{N}_{a}} \\text{is maximum,}\\\\ \\dfrac{d{{N}_{a}}}{dt}=0=\\alpha -\\lambda {{N}_{a}} \\\\ \\Rightarrow {{N}_{a}}=\\dfrac{rate\\,of\\,production}{\\lambda } \\\\ \\text{So, on integration, we have:}\\\\ \\int\\limits_{0}^{t}{\\dfrac{d{{N}_{a}}}{\\alpha -\\lambda {{N}_{a}}}=\\int\\limits_{0}^{t}{dt}} \\\\ \\text{Number of nuclei obtained at any time ‘t’ ss } \\\\ {{N}_{a}}=\\dfrac{\\alpha }{\\lambda }(1-{{e}^{-\\lambda t}})$$.","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Nuclear fission and fusion Quiz 1","text":" A goes parallel radioactive disintegration to form B and C, the constants are $${{\\lambda }_{1}},{{\\lambda }_{2}}$$ respectively. Find the half-life.","comment":{"@type":"Comment","text":" In a parallel radioactive disintegration, apply conservation of the number of nuclei particles and differentiate to obtain an expression for half-life."},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$\\dfrac{{{t}_{1}}+{{t}_{2}}}{{{t}_{1}}{{t}_{2}}}$$","position":1},{"@type":"Answer","encodingFormat":"text/html","text":"$$\\dfrac{1}{{{t}_{1}}{{t}_{2}}}$$","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" $${{t}_{1}}\\times {{t}_{2}}$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$\\dfrac{{{t}_{1}}{{t}_{2}}}{{{t}_{1}}+{{t}_{2}}}$$","position":0,"answerExplanation":{"@type":"Comment","text":"$$ {{N}_{o}}={{N}_{a}}+{{N}_{b}}+{{N}_{c}} \\text{ and } \\\\ \\dfrac{d{{N}_{a}}}{dt}=-\\dfrac{d({{N}_{b}}+{{N}_{c}})}{dt}.\\\\ \\text{We also have:}\\\\ \\dfrac{d({{N}_{b}}+{{N}_{c}})}{dt}=({{\\lambda }_{1}}+{{\\lambda }_{2}}){{N}_{a}} \\\\ \\Rightarrow \\dfrac{d{{N}_{a}}}{dt}=-({{\\lambda }_{1}}+{{\\lambda }_{2}}){{N}_{a}} \\\\ \\Rightarrow {{\\lambda }_{eff}}={{\\lambda }_{1}}+{{\\lambda }_{2}} \\\\ \\Rightarrow {{t}_{eff}}=\\dfrac{{{t}_{1}}{{t}_{2}}}{{{t}_{1}}+{{t}_{2}}} $$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]}]}