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ड्यूटीरियम `(""_(1)^(2)"H")` और हीलियम `(""_(2)^(4)"He")` नाभिकों प्रति न्यूक्लिऑन बंधन ऊर्जा क्रमश: 1.1 MeV तथा 7 MeV यदि दो ड्यूटीरियम नाभिक संयोग कर एक हीलियम नाभिक बनाते हों, तो विमुक्त ऊर्जा होगी -
A. a.13.9 MeV
B. b.26.9 MeV
C. c.23.6 MeV
D. d.19.2 MeV.

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Correct Answer - C
`""_(1)H^(1)+""_(1)H^(2)to""_(2)"He"^(4)`
मुक्त ऊर्जा = उत्पाद की बंधन ऊर्जा - क्रिया कारकों की बंधन ऊर्जा
`=4xx7-2xx11-2xx11=23.6" MeV."`

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