Worked Example

The radioisotope thorium 228 is an alpha emitter. Calculate the energy Q released when one such nucleus decays.

Mass of Th ₂₂₈ = 228.02875u

Mass of Ra ₂₂₄ = 224.02022u

Mass of alpha particle = 4.00260u

Mass defect (Dm) = mass of Th₂₂₈ – mass of Ra₂₂₄ – mass of alpha particle

= 228.02875u – 224.02022u – 4.00260u = 0.005926u

But 1u = 931.3 MeV (from data sheet)

Therefore Mass defect is equivalent to 0.005926 x 931.3 = 5.5188838 MeV

However we can only quote to 4 significant figures (as this is the least number given) therefore the answer is

Q = 5.519 MeV