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

In the above reaction, the liberated energy is 31.65 MeV.

Explanation:

223 radium can be split into 209 lead and 14 carbon.

As per the question,  

223Ra has the atomic mass, m(223Ra) = 223.018 u

209Pb has the atomic mass, m(209Pb) = 208.981 u

14C has the atomic mass, m(14C) = 14.003 u

$\mathrm{Ra} 223 \rightarrow \mathrm{Cl} 4+209 \mathrm{Pb}$

Energy,$\mathrm{E}=\mathrm{m} \mathrm{R} 223 \mathrm{a}+\mathrm{mC} 14 \mathrm{c} 2-\mathrm{m} \mathrm{P} 209 \mathrm{b}$

$=223.018 \mathrm{u}+14.003 \mathrm{u} \mathrm{c} 2-208.981$

$=0.034 \times 931 \mathrm{MeV}=31.65 \mathrm{MeV}$