Nuclear physics question
A single 212Po nucleus, initially at rest, emits an alpha particle as it decays into a 208Pb nucleus. There are no other decay products, and the kinetic energy of the alpha particle is measured to be $1.407 × 10-12 J$. What is the total amount of nuclear potential energy, in joules, that was released in this process?
$PE_{nuclear} = ?J$
What is the difference between the original mass of the 212Po nucleus and the total mass of the fragments after the decay?
$Delta M = ?kg$
Answer
$PE = Ek +m(\alpha)*c^2$
$m (4,2He) =6.644*10^-27 kg (=4.001 amu)$
$PE = 1.407*10^{-12} +6.644*10^-{27}*9*10^{16} =5.9937*10^{-10} J$
Mass defect is
$m(208Pb) =207.9766 amu$ (http://en.wikipedia.org/wiki/Isotopes_of_lead)
$m(212Po) =211.989 amu$ (http://en.wikipedia.org/wiki/Isotopes_of_polonium)
$\Delta(M) = 211.989 -207.9766-4.001 =0.0114 amu =0.0114*1.66*10^-{27} =1.892*10^{-29} kg$