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From the given data, the amount of energy required to break the nucleus of aluminium Al 13 27 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)

Correct answer

0

Step-by-step solution

Δ m = Z m P + ( A - Z ) m n - M Al = ( 13 × 1 . 00726 + 14 × 1 . 00866 ) - 27 . 18846 = 27 . 21562 - 27 . 18846 = 0 . 02716   u E = 27 . 16 x × 10 - 3   J

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