Question:
From the given data, the amount of energy required to break the nucleus of aluminium \(^{27}_{13}\mathrm{Al}\) is __________ \(x \times 10^{-3}\ \mathrm{J}\).
Mass of neutron \(= 1.00866\ \mathrm{u}\)
Mass of proton \(= 1.00726\ \mathrm{u}\)
Mass of Aluminium nucleus \(= 27.18846\ \mathrm{u}\)
(Assume \(1\ \mathrm{u}\) corresponds to \(x\ \mathrm{J}\) of energy)
(Round off to the nearest integer)
From the given data, the amount of energy required to break the nucleus of aluminium \(^{27}_{13}\mathrm{Al}\) is __________ \(x \times 10^{-3}\ \mathrm{J}\).
Mass of neutron \(= 1.00866\ \mathrm{u}\)
Mass of proton \(= 1.00726\ \mathrm{u}\)
Mass of Aluminium nucleus \(= 27.18846\ \mathrm{u}\)
(Assume \(1\ \mathrm{u}\) corresponds to \(x\ \mathrm{J}\) of energy)
(Round off to the nearest integer)
Mass of neutron \(= 1.00866\ \mathrm{u}\)
Mass of proton \(= 1.00726\ \mathrm{u}\)
Mass of Aluminium nucleus \(= 27.18846\ \mathrm{u}\)
(Assume \(1\ \mathrm{u}\) corresponds to \(x\ \mathrm{J}\) of energy)
(Round off to the nearest integer)
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Energy needed to break the nucleus = binding energy = (mass defect in u) x (x joules per u).
Updated On: Nov 4, 2026
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Correct Answer: 27
Solution and Explanation
Aluminium-27 has \(Z=13\) protons and \(N=27-13=14\) neutrons.
Mass of separated nucleons \(= 13(1.00726) + 14(1.00866) = 13.09438 + 14.12124 = 27.21562\ \mathrm{u}\).
Mass defect \(\Delta m = 27.21562 - 27.18846 = 0.02716\ \mathrm{u}\).
Since \(1\ \mathrm{u} \equiv x\ \mathrm{J}\), \(E = 0.02716\,x\ \mathrm{J} = 27.16\times10^{-3}\,x\ \mathrm{J}\).
Comparing with "__ \(x\times10^{-3}\) J" and rounding off: \(27\).
Answer: 27
Mass of separated nucleons \(= 13(1.00726) + 14(1.00866) = 13.09438 + 14.12124 = 27.21562\ \mathrm{u}\).
Mass defect \(\Delta m = 27.21562 - 27.18846 = 0.02716\ \mathrm{u}\).
Since \(1\ \mathrm{u} \equiv x\ \mathrm{J}\), \(E = 0.02716\,x\ \mathrm{J} = 27.16\times10^{-3}\,x\ \mathrm{J}\).
Comparing with "__ \(x\times10^{-3}\) J" and rounding off: \(27\).
Answer: 27
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