A) 19.2 MeV
B) 23.6 MeV
C) 26.9 MeV
D) 13.9 MeV
Correct Answer: B
Solution :
\[_{1}{{H}^{2}}{{+}_{1}}{{H}^{2}}{{\to }_{2}}H{{e}^{4}}+\]energy Binding energy of a \[{{(}_{1}}{{H}^{2}})\]deuterium nuclei \[=2\times 1.1=2.2\ MeV\] Total binding energy of two deuterium nuclei \[=2.2\times 2=4.4\ MeV\] Binding energy of a \[{{(}_{2}}H{{e}^{4}})\]nuclei \[=4\times 7=28\ MeV\] So, energy released in fusion \[=28-4.4=23.6\ MeV\]
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