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CBSE Class 12 Physics 2016 Delhi Set 1 Paper

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Question : 8 of 26
Marks: +1, -0
A nucleus with mass number A=240A=240 and BEA=7.6 MeV\frac{BE}{A}=7.6\,\mathrm{MeV} breaks into two fragments each of A=120A=120 with BEA=8.5 MeV\frac{BE}{A}=8.5\,\mathrm{MeV} . Calculate the released energy.
OR
Calculate the energy in fusion reaction :
12H+12H→23He+n_{1}^{2}\mathrm{H}+_{1}^{2}\mathrm{H}\rightarrow_{2}^{3}\mathrm{He}+\mathrm{n} , where BE of 12H=2.23 MeV_{1}^{2}\mathrm{H}=2.23\,\mathrm{MeV}
and of 23He=7.73 MeV_{2}^{3}\mathrm{He}=7.73\,\mathrm{MeV} .
Solution:  
Calculation of energy released
Binding energy of nucleus with mass number 240 ,
Ebn=240×7.6 MeVE_{bn}=240\times7.6\,\mathrm{MeV}
Binding energy of two fragments
  =2×120×8.5 MeV\;=2\times120\times8.5\,\mathrm{MeV}
  Energy released  =  240(8.5−7.6) MeV\;\text{Energy released}\;=\;240(8.5-7.6)\,\mathrm{MeV}
  =240×0.9\;=240\times0.9
  =216 MeV\;=216\,\mathrm{MeV}
OR
Calculation of Energy in the fusion Reaction
Total Binding energy of Initial System i.e.
12H+12H  =(2.23+2.23) MeV_{1}^{2}\mathrm{H}+_{1}^{2}\mathrm{H}\;=(2.23+2.23)\,\mathrm{MeV}
  =4.46 MeV\;=4.46\,\mathrm{MeV}
Binding energy of Final System i.e. 23He_{2}^{3}\mathrm{He}
=7.73 MeV=7.73\,\mathrm{MeV}
Hence energy released
=7.73 MeV−4.46 MeV=7.73\,\mathrm{MeV}-4.46\,\mathrm{MeV}
=3.27 MeV=3.27\,\mathrm{MeV}
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