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Two deuterons undergo nuclear fusion to form a Helium nucleus. Energy released in this process is : (given binding energy per nucleon for deuteron=1.1 MeV and for helium=7.0 MeV) a) 30.2 MeV b) 32.4 MeV c) 23.6 MeV d) 25.8 MeV

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 _{1}^{2}\textrm{D} + _{1}^{2}\textrm{D} \rightarrow _{2}^{4}\textrm{He} + Q

Q=E(_{2}^{4}\textrm{He})-E(2\ \times\ _{1}^{2}\textrm{D})

Q=7*4-2(2*1.1) = 23.6\ eV

 

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Safeer PP

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it is mentioned in the question that 2 deuteron nuclei react to form single He nucleus. that means 

 _{1}^{2}\textrm{D} + _{1}^{2}\textrm{D} \rightarrow _{2}^{4}\textrm{He} + Q

Q represents the energy realesed in the reaction

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Safeer PP

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