Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 12
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# 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 - Physics

ConceptMass-Energy Relation and Mass Defect

#### 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

#### Solution

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"

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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.
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