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JEE Main 2021 (25 Jul, Shift 2) · Mass Defect & Binding Energy

Mass Defect & Binding Energy question from JEE Main 2021 (25 Jul, Shift 2)

Q16JEE Main 2021 · 25 JUL · SHIFT 2Medium

Mass Defect & Binding Energy

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)

Numerical answer type. Work it out, then open the answer.

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Answer: 27

  1. 1327{}_{13}^{27}Al: 13 protons, 14 neutrons.
  2. 13(1.00726)+14(1.00866)=13.09438+14.12124=27.2156213(1.00726) + 14(1.00866) = 13.09438 + 14.12124 = 27.21562 u
  3. Δm=27.21562−27.18846=0.02716\Delta m = 27.21562 - 27.18846 = 0.02716 u
  4. Energy =0.02716 x= 0.02716\,x J =27.16×10−3 x= 27.16 \times 10^{-3}\,x J, so the answer is 27
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