#### Question

Find the energy liberated in the reaction^{223}Ra → ^{209}Pb + ^{14}C.

The atomic masses needed are as follows.^{223}Ra ^{209}Pb ^{14}C

22..018 u 208.981 u 14.003 u

#### Solution

Given :

Atomic mass of ^{223}Ra, m(^{223}Ra) = 223.018 u

Atomic mass of ^{209}Pb, m(^{209}Pb) = 208.981 u

Atomic mass of ^{14}C, m(^{14}C) = 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"`

Is there an error in this question or solution?

Solution 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 Concept: Mass-Energy Relation and Mass Defect.