Question

Energy released when one atom of uranium undergoes nuclear fission according to the following reaction is (atomic mass of U= 235.060; n= 1.009; Ba = 143.881 and Kr= 89.947) about $ _{92}{{U}^{235}}{{+}_{0}}{{n}^{1}}{{\to }_{56}}B{{a}^{144}}{{+}_{36}}K{{r}^{90}}+{{2}_{0}}{{n}^{1}} $

• A. 235 MeV
• B. 208 MeV
• C. 931.5 MeV
• D. $ 5.33\times {{10}^{23}}MeV $
• E. 20.8 MeV

Answer 1

Answer: (B)

Mass defect, $ \Delta m=Z $ mass of reactants $ -\Sigma $ mass of products $ \Delta m=235.060-(143.881+89.947+1.009) $
$ =0.223 $
Binding energy or the energy released
$ =0.223\times 931\,MeV $
$ =207.6MeV $
$ =208\text{ }MeV $