JEE PHYSICS MCQ: NUCLEI

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Application +4 | -1

If the binding energy per nucleon of deuteron is

1.115 \,{MeV}

, then its mass defect (in amu) is

View answer and explanation

Correct option: C

0.0024

Option C (0.0024) is correct because the mass defect is calculated by dividing the binding energy by 931.5 MeV/amu.

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