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A nucleuswithmass numberA =240and BE//A= 7.6 Me V breaksintotwofragmentseach ofA =120with BE//A=8.5 Me V .Calculatethe releasedenergy . |
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Answer» Solution :Binding energy of nucleus with MASS NUMBER 240, `BE_1 = 240 xx 7.6 =1824 Me V` Binding energy of two fragments `BE_2 = 2 xx 120 × 85 = 2040 MeV` `:.` Energy released = `BE_2 – BE_1` = (2040 – 1824) MeV = 216 MeV |
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