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Calculate the energy released when three alpha-particle (""_2He^4) fuse to form a carbon (""_6C^(12)) nucleus. Given m(""_2He^4)=4.002603 amu. |
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Answer» Solution :`3 ""_2He^4 to ""_6C^(12)+Q` `therefore Q=m (3 ""_2He^4)-m(""_6C^(12))=12.0078-12=0.007809` `therefore` Energy released `=0.007809 xx931 MeV =7.27` MeV |
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