Vijay Dalakoti Asked a Question
January 14, 2022 1:36 pmpts 30 pts
Consider the following decay process: 228 Th224 Ra +a 224 Ra 224 Ra + y (217 ke) The kinetic energy of the alpha particle is (Your answer should be upto two decimal places in the units MeV) (Given: m Th)=228.028726 u, mRa)=224.020196 u, m(He) =4.00260 u) n= m=mx 2t -M(h)-M(2Ra')-M (jHe)|a =228.028726-224.020196 4.002601c = 0.00593x 931 MeV/u = 5.52 MeV Knetic energy of a-particle KE- MeV-5.4231 MeV But kinetic energy of the a-particle wil be less as some part of energy is taken by the gamma particle KE = 5.423 1 MeV-217 10 MeV=5.20 MeV
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  • Chandra dhawan thankyou
    we can not subtract from q directly because q value just difference of mass of reactant and mass of product. as observed, the kinetic energy is always less than the Q value of any ...
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