Test Index

CBSE Class 12 Physics 2023 Delhi Set 3 Paper

© examsnet.com
Question : 14 of 14
Marks: +1, -0
(a) Calculate the binding energy of an alpha particle in MeV. Given
mass of a proton =1.007825 u=1.007825\,\mathrm{u}
mass of a neutron =1.008665 u=1.008665\,\mathrm{u}
mas of He nucleus =4.002800 u=4.002800\,\mathrm{u}
1 u=931 MeV/c21\,\mathrm{u}=931\,\mathrm{MeV}/c^{2}
OR
(b) A heavy nucleus PP of mass number 240 and binding energy 7.6 MeV\mathrm{MeV} per nucleon splits into two nuclei QQ and RR of mass number 110 and 130 and binding energy per nucleon 8.5 MeV8.5\,\mathrm{MeV} and 8.4 MeV8.4\,\mathrm{MeV} 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\;\text{Or,}\;\; \;\; \;\text{ Mass defect }\;= (2 \times 1.007825+2   ×1.008665−4.002800) u\; \times 1.008665-4.002800) \,\mathrm{u}
∴       Mass defect   =  0.03018 u\therefore \;\; \;\text{ Mass defect }\; = \; 0.03018 \,\mathrm{u}
=  0.03018×931= \; 0.03018 \times 931
=  28.09785 MeV= \; 28.09785 \,\mathrm{MeV}
OR
(b) According to the question
240P→110Q+130R+Q{}^{240}\mathrm{P} \rightarrow {}^{110}\mathrm{Q} + {}^{130}R + Q
Q=  110×8.5+130×8.4Q = \; 110 \times 8.5 + 130 \times 8.4   −240×7.6\; -240 \times 7.6
Or, Q  =935+1092−1824Q \; = 935+1092-1824
∴ Q  =202 MeVQ \; = 202 \,\mathrm{MeV}
© examsnet.com
Go to Question:
Prev Question