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Q.

Binding energy of deuterium is 2.23MeV.
Mass defect in amu is

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a

0.0012

b

0.0024

c

0.0036

d

–0.0012

answer is B.

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Detailed Solution

$\begin{array}{l} B . E = Δm \times 931 .5 MeV; 2 .23 = Δm \times 931 .5 \\ Δm = \frac{2 .23}{931 .5} = 0 .0024 amu \end{array}$

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