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

Answer

A, C

Explanation

`^(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` .