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Find the energy liberated in the reactio...

Find the energy liberated in the reaction
`Ra^223 rarr Pb^209 + C^14 `
The atomic masses needed are as follows .
`Ra^223`, `Pb^209`, `C^14`
223.18u, 208.981u, 14.003u`.

Text Solution

Verified by Experts

The correct Answer is:
A, C
`^(234)R_a rarr `^(209)Pb + `^14C`
`Deltam = mass `^(223)Ra - mass `
`(209)Pb + ^14C`
`[(Data Mass), (223_Rararr 223.018mu),(209_Pb rarr 208.981mu),(14_C rarr 208.981mu),(14_C rarr 14.003mu)]`
`= 223.018 - (208.981 + 14.003)`
`= 0.034`
`Energy = Deltam xx u = 0.034 xx 931`
`=31.65 MeV` .
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