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)
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 by Shiksha User
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1 Answer
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Binding Energy = (Δm)c²
= [Zmp + (A-Z)mn - MAl]c²
= [ (13×1.00726 + 14×1.00866) - 27.18846] u
= [ (13.09438 + 14.12124) - 27.18846] u
= [27.21562 - 27.18846] u
= 0.02716 u
= 0.02716 x = 27.16x × 10? ³
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