CBSE Class 12 Physics 2022 Term 2 Delhi Set 1 Solved Paper

Question : 6

Total: 12

In a fission event of ‌92238U by fast moving neutrons, no neutrons are emitted and final products, after the beta decay of the primary fragments, are ‌58140‌Ce and ‌4499‌Ru. Calculate Q for this process. Neglect the masses of electrons/ positrons emitted during the intermediate steps. 3
Given:
m(‌92238U)=238.05079u;
m(‌58140‌Ce)=139.90543u
m(‌4499‌Ru)=98.90594u;m(‌01n)=1.008665u

Solution:

‌92238U+‌01n→‌92239U→‌58140Ce+‌4499Ru+10‌−10e
Δm=(238.05079+1.008665−139.90543−98.90549)u
[neglecting mass of electron]
Or. Δm=0.24835u
∴Q=0.24835×931=231.386085=231.386MeV

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