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The mass defect for the nucleus of helium is $0.0303 a m u$ . The binding energy per nucleon in $M e v$ is nearly

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Correct option is B

Δm=0.0303amu BE=Δm×931.5 =0.0303×931.5=28.22445Mev Helium has 4 nucleons∴ BE/nucleon=28.224454≃7Mev

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The mass defect for the nucleus of helium is 0.0303 amu . The binding energy per nucleon in Mev is nearly

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The mass defect for the nucleus of helium is $0.0303 a m u$ . The binding energy per nucleon in $M e v$ is nearly