Explanation
$${}_{92}^{235}U + {}_0^1n \to {}_{56}^{141}Ba + {}_{36}^{92}Kr + 3({}_0^1n) + Q$$
Q is Energy released Q = ( Total mass of reactants – total mass ) $$ \times {C^2}$$
$$ = \left[ {Mass\left( {{}_{92}^{235}U} \right) + mass\left( {{}_0^1n} \right)} \right] - \left[ {mass{}_{56}^{141}Ba + mass{}_{36}^{92}Kr + mass3{}_0^1n} \right]{c^2}$$
$$ = \left( {235 \cdot 043933 - 140.917700 - 91.8954 - 2 \times 1.008665} \right)u \times {c^2}$$
$$ = 198.9MeV$$
It is an example of nuclear fusion reaction. In fusion reaction, two smaller nuclei combine to form a heavy nuclei with release of some energy.
Number of mole of radium, $$n=1$$
The number of nuclei of radium, $$N={{N}_{a}}=avagdaro\,\,number=6.023\times {{10}^{23}}$$
Activity, $$A=\lambda N$$
$$ \because 1\,\,curie=3.7\times {{10}^{10}}\,decay/\sec $$
$$ \dfrac{1}{3.7}\,kilocurie=0.27\,kilocurie=9.99\times {{10}^{12}}\,dacay/\sec $$
So, $$\lambda =\dfrac{A}{N}=\dfrac{9.99\times {{10}^{12}}}{6.023\times {{10}^{23}}}=1.65\times {{10}^{-11}}\,$$
Given,
Energy released per atom, $$E=200\,MeV=200\times {{10}^{6}}\times 1.6\times {{10}^{-19}}=3.2\times {{10}^{-11}}\,J$$
Number of atom in $$1\,kg$$ uranium, $$n=\dfrac{m{{N}_{A}}}{M}=\dfrac{1000\times 6.023\times {{10}^{23}}}{235}=2.56\times {{10}^{24}}$$
Total Energy released = $$nE=2.56\times {{10}^{24}}\times 3.2\times {{10}^{-11}}=8.19\times {{10}^{13}}J$$
Hence, Total Energy released = $$8.19\times {{10}^{13}}J$$.
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