From the given data, the amount of energy required to break the nucleus of aluminium ₁₃²⁷Al 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)

Solution

Physics sir · 4 steps

Energy to break the nucleus is its binding energy, so find the mass defect in u and multiply by x.

1. Count the nucleons. ²⁷₁₃Al has Z = 13 protons and A - Z = 27 - 13 = 14 neutrons.
2. Mass of free nucleons. Protons give 13 × 1.00726 = 13.09438 u. Neutrons give 14 × 1.00866 = 14.12124 u. Together that is 27.21562 u.

   Δ m = 27.21562 - 27.18846 = 0.02716 u

3. Convert to energy. Each u is worth x J, so the binding energy is 0.02716 x J = 27.16 x × 10⁻³ J.
4. So. To the nearest integer, the energy needed is 27 x × 10⁻³ J, so the answer is 27.

The common slip: 26. Swapping the counts to 14 protons and 13 neutrons gives a mass defect of 0.02576 u and an answer of 26. The number of protons is Z = 13.

The idea to keep. Breaking a nucleus costs its binding energy. Add up the masses of the free protons and neutrons, subtract the nucleus mass, and convert that mass defect to energy.

Answer from the NTA final answer key. Solution by PrepWiser.