1.

Calculate the energy released in the given reaction: ""_(6)C^(12) + ""_(6)C^(12) to ""_(10)Ne^(20) + ""_(2)He^(4) Given atomic masses are as follows: M_(He) = 4.002603u , M_(Ne) = 19.992439 u, M_( c)=12.0000u

Answer»


ANSWER :4.6183 MEV


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