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From the given data, the amount of energy required to break the nucleus of aluminium ²⁷₁₃Al is ______ x×10⁻³ J. Mass of neutron =1.00866 u, Mass of proton =1.00726 u, Mass of aluminium nucleus =27.18846 u (Assume 1 u corresponds to x J of energy) (Round off to the nearest integer)

Asked in JEE Main 25th July 2nd Shift 2021 · Mass defect and binding energy

Answer: 27

Step-by-step solution

13×1.00726+14×1.00866=27.21562 u

Δ m=27.21562-27.18846=0.02716 u

E=0.02716x J=27.16x×10⁻³ J≈27x×10⁻³

→ 27

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