A) 23.6 MeV
B) 2.2 MeV
C) 28.0 MeV
D) 30.2 MeV
Correct Answer: A
Solution :
Mass of \[_{1}{{H}^{2}}=2.01478\,amu\] Mass of \[_{2}H{{e}^{4}}=4.00388\,amu\] Mass of two deuterium = 2 x 2.01478 = 4.02956 amu Energy equivalent to \[_{2}{{H}^{2}}\] \[=4.02956\times 1.2\] = 4.4 MeV Energy equivalent to \[_{2}{{H}^{4}}\] \[=4.00388\times 7=28MeV\] Energy released = (28 - 4.4) = 23.6 MeVYou need to login to perform this action.
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