A) 108 MeV
B) 200 MeV
C) 1000 MeV
D) more than 100 MeV
Correct Answer: B
Solution :
In the process of nuclear fission, huge amount of energy is released, called nuclear energy. \[_{92}{{U}^{235}}{{+}_{0}}{{n}^{1}}{{\to }_{92}}{{U}^{236}}{{\to }_{56}}B{{a}^{144}}{{+}_{36}}K{{r}^{89}}\] \[+3{{\,}_{0}}{{n}^{1}}+\] energy The mass of \[_{92}{{U}^{235}}{{+}_{0}}{{n}^{1}}=234.99\,\,amu+1.01\,\,amu\] = 236 amu Similarly, mass of \[_{55}B{{a}^{144}}{{+}_{36}}K{{r}^{89}}+{{3}_{0}}{{n}^{1}}\] =143.87 + 88.9 + 3 \[\times \] 1.01 = 235.8 amu \[\therefore \] \[\Delta m=236-235.8=0.20\,amu\] From Einsteins mass energy equivalence 1 amu mass is equivalent to 931 MeV energy, \[\therefore \] \[\Delta E=\Delta m{{c}^{2}}\] \[=0.2\times 931\approx 190\,\,MeV\] Which can be approximated to 200 MeV.
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