CBSE Class 12 Physics 2023 Delhi Set 3 Paper

Question : 14

Total: 14

(a) Calculate the binding energy of an alpha particle in MeV. Given
mass of a proton =1.007825u
mass of a neutron =1.008665u
mas of He nucleus =4.002800u
1u=931‌MeV∕c2
OR
(b) A heavy nucleus P of mass number 240 and binding energy 7.6 MeV per nucleon splits into two nuclei Q and R of mass number 110 and 130 and binding energy per nucleon 8.5‌MeV and 8.4‌MeV respectively. Calculate the energy released in the fission.

Solution: 👈: Video Solution

(a) Mass defect = Mass of protons + mass of neutrons - Mass of Helium nucleus
‌ Or, ‌‌‌‌ Mass defect ‌=‌(2×1.007825+2 ‌×1.008665−4.002800)‌u
∴‌‌‌ Mass defect ‌=‌0.03018u
=‌0.03018×931
=‌28.09785‌MeV
OR
(b) According to the question
‌240P→‌110Q+‌130R+Q
Q=‌110×8.5+130×8.4 ‌−240×7.6
Or, Q‌=935+1092−1824
∴ Q‌=202‌MeV

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