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Physics-

General

Easy

Question

If the binding energy of the deutriumis $2.23 blank M e V$ . The mass defect given in $a. m. u.$ is

$- 0.0024$
$- 0.0012$
0.0012
0.0024

The correct answer is: 0.0024

Mass defect $increment m equals fraction numerator 2.23 over denominator 931 end fraction equals 0.0024 blank a m u$

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