A) 0.1%
B) 0.25%
C) 0.3%
D) 2%
Correct Answer: A
Solution :
The fission reaction of \[\text{U}{{}^{\text{235}}}\]is as follows : \[_{92}{{\text{U}}^{235}}{{+}_{0}}{{n}^{1}}{{\xrightarrow{{}}}_{92}}{{\text{U}}^{236}}\xrightarrow{{}}\] \[_{56}B{{a}^{144}}{{+}_{36}}K{{r}^{89}}+{{3}_{0}}{{n}^{1}}+energy\] The mass of \[_{92}{{U}^{235}}{{+}_{0}}n{{\,}^{1}}=234.99\,\text{amu +1}\text{.01}\,\text{amu}\] = 236.00 amu Similarly, the mass of \[_{56}B{{a}^{144}}{{+}_{36}}K{{r}^{89}}+{{3}_{0}}{{n}^{1}}\] \[=143.87\text{ amu }+\text{ }88.90\text{ amu }+\text{ }3\text{ }\times \text{ }1.01\text{ amu}\] \[=235.8\text{ amu}\] \[\therefore \] \[\Delta m=\] mass defect \[=\text{ }236-\text{235}\text{.8 = 0}\text{.20 amu}\] Hence, % of mass converted to energy \[=\frac{0.20}{236}\times 100\approx 0.085\approx 0.1%\]
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