Atoms & Nuclei

JEE Main level, numerical answer, medium

204 of 239

Question 204 of 239

From the given data, the amount of energy required to break the nucleus of aluminium Al1327Al1327 is __________x×10−3Jx\times {10}^{-3}J
Mass of neutron =1.00866u=1.00866u
Mass of proton =1.00726u=1.00726u
Mass of Aluminium nucleus =27.18846u=27.18846u
(Assume 1u1u corresponds to xJxJ of energy)
(Round off to the nearest integer)

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Answer: 27

Δm=(ZmP+(A−Z)mn)−MAl\Delta m=\left(Z{m}_{P}+(A-Z){m}_{n}\right)-{M}_{Al}
=(13×1.00726+14×1.00866)−27.18846=(13\times 1.00726+14\times 1.00866)-27.18846
=27.21562−27.18846=27.21562-27.18846
=0.02716u=0.02716u
E=27.16x×10−3JE=27.16x\times {10}^{-3}J

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