CBSE Class 12 Physics 2020 Delhi Set 1 Paper

Question : 24

Total: 37

A heavy nucleus P of mass number 240 and binding energy 7.6‌MeV per nucleon splits in to two nuclei Q and R of mass numbers 110 and 130 and binding energy per nucleon 8.5‌MeV and 8.4 MeV, respectively. Calculate the energy released in the fission.

Solution:

Total BE of P=240×7.6=1824‌MeV
BE of Q=110×8.5=935‌MeV
BE of R=130×8.4=1092‌MeV
Total BE of Q and R=(935+1092)=2027‌MeV
Total energy released in the fission
‌=2027−1824
‌=203‌MeV

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