How many J of energy would it take to break a single aluminum-27 nucleus (26.9815386 amu) into its component neutrons (1.008664916 amu) and protons (1.007276467 amu)?

Question

How many J of energy would it take to break a single aluminum-27 nucleus (26.9815386 amu) into its component neutrons (1.008664916 amu) and protons (1.007276467 amu)? (Express your answer to the correct number of significant figures.)

Answer 1

To solve this problem, I will first compare the mass of an Al-27 nucleus to the mass of its constituents:

Al-27 has a known mass of 26.9815386 amu and contains 13 protons and 14 neutrons.

The mass of 13 individual protons is 13 x 1.007276467amu = 13.09459407 amu

The mass of 14 individual neutrons is 14 x 1.008664916amu = 14.12130882 amu

The change in mass in the decomposition of an Al-27 nucleus into 13 protons and 14 neutrons is then:

13.09459407 amu + 14.12130882 amu - 26.9815386 amu = 0.2343643 amu.

0.2343643 amu/(6.02214076 × 10²³amu/g) = 3.891711 x 10^-25g = 3.891711 x 10^-28kg

This change in mass can be related to the change in energy by Einstein's equation ΔE = Δmc².

ΔE = (3.891711 x 10^-28kg)(2.99792458 x 10⁸m/s)² = 3.497695 x 10^-11J

Thus, to the correct number of significant figures (7), it takes 3.497695 x 10^-11J of energy to break an Al-27 nucleus into its component neutrons and protons. The decomposition of a nucleus does indeed require energy because a nucleus is more energetically stable than its component nucleons.