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If the binding energy of the deuterium is `2.23 MeV`. The mass defect given in a.m.u. is.

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If the binding energy of the deuterium is `2.23 MeV`. The mass defect given in a.m.u. is.
A. `-0.0024`
B. `-0.0012`
C. `0.0012`
D. `0.0024`
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Correct Answer - D
(d) Mass defect `Delta m = (1.23)/(931) = 0.0024`.
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