A) 0.001 g
B) 0.1 g
C) 10.0g
D) 100g
Correct Answer: D
Solution :
\[U=Pt\] \[={{10}^{6}}\times 24\times 36\times {{10}^{12}}\] \[=24\times 36\times {{10}^{8}}J\] Energy released per fusion reaction \[=20MeV\] \[=20\times {{10}^{6}}\times 1.6\times {{10}^{-19}}\] \[=32\times {{10}^{-13}}J\] Energy released per atom of\[_{1}{{H}^{2}}=32\times {{10}^{-13}}J\] Number of\[_{1}{{H}^{2}}\]atom of used \[=\frac{24\times 36\times {{10}^{8}}}{32\times {{10}^{-13}}}\] \[=27\times {{10}^{21}}\] Mass of\[6\times {{10}^{23}}\]atom\[=2g\] \[=\frac{2}{6\times {{10}^{23}}}\times 27\times {{10}^{21}}\] \[90\approx 100g\]
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