Question

Find the energy liberated in the reaction
²²³Ra → ²⁰⁹Pb + ¹⁴C.
The atomic masses needed are as follows.
²²³Ra         ²⁰⁹Pb        ¹⁴C
22..018 u  208.981 u  14.003 u

Answer 1

Given :
Atomic mass of ²²³Ra, m(²²³Ra) = 223.018 u
Atomic mass of ²⁰⁹Pb, m(²⁰⁹Pb) = 208.981 u 
Atomic mass of ¹⁴C, m(¹⁴C) = 14.003 u

Reaction : 

`""^223"Ra" → ""^209"Pb" + ""^14"C"`

Energy , `E = [m(""^223Ra) - (m(""^209Pb) + m(""^14C))]c^2`

= `[223.018  "u" - (208.981 + 14.003)  "u"] c^2`

= `0.034 xx 931  "MeV"`

= `31.65  "MeV"`