Δm=(ZmP+(A−Z)mn)−MAl
=(13×1.00726+14×1.00866)−27.18846
=27.21562−27.18846
=0.02716u
E=27.16x×10−3J
JEE Main 2021 — Physics Modern Physics
From the given data, the amount of energy required to break the nucleus of aluminium Al1327 is __________x×10−3J
Mass of neutron =1.00866u
Mass of proton =1.00726u
Mass of Aluminium nucleus =27.18846u
(Assume 1u corresponds to xJ of energy)
(Round off to the nearest integer)
Held on 25 Jul 2021 · Verified 6 Jul 2026.
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