A) 0.001 g
B) 0.1 g
C) 10.0 g
D) 1000 g
Correct Answer: B
Solution :
Energy produced, \[U=Pt\] \[={{10}^{6}}\times 24\times 36\times {{10}^{2}}\] \[=24\times 36\times {{10}^{8}}J\] Energy released per fusion reaction \[=20\text{ }MeV\] \[=20\times {{10}^{6}}\times 1.6\times {{10}^{-19}}\] \[=32\times {{10}^{-13}}J\] Energy released per atom of\[_{1}{{H}^{2}}\] \[=32\times {{10}^{-13}}J\] Number of\[_{1}{{H}^{2}}\]atoms used \[=\frac{24\times 36\times {{10}^{8}}}{32\times {{10}^{-13}}}\] \[=27\times {{10}^{21}}\] Mass of\[6\times {{10}^{23}}\]atoms\[=2g\] \[\therefore \] Mass of\[27\times {{10}^{21}}\]atoms \[=\frac{2}{6\times {{10}^{23}}}\times 27\times {{10}^{21}}=0.1g\]
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