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QJEEM21Ph88da: JEE Main 2021 Physics - Class 12 JEEM21Ph88da Solution: From the given data, the amount of energy required to break the …

From the given data, the amount of energy required to break the nucleus of aluminium 1327_{13}^{27} Al is __________ x ×\times 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)

Step-by-Step Verified Solution

Δ\Delta m = (Zmp + (A −- Z)mn) −- MAl

= (13 ×\times 1.00726 + 14 ×\times 1.00866) −- 27.18846

= 27.21562 −- 27.18846

= 0.02716 u

E = 27.16 x ×\times 10 −- 3 J

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