1.

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.

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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