Find the Q-value and the kinetic energy of the emitted `alpha`-particle in the `alpha`-decya of (a) `""_(88)^(226)Ra` and (b) `""_(86)^(220)Rn`.
Given `m (""_(88)^(226)Ra) = 226.02540 u, " " m(""_(86)^(222)Rn) = 222.01750 u`.
`m(""_(86)^(220)Rn) = 220.01137 u, " " m (""_(84)^(216)Po) = 216.00189u`.

Explanation

`Q = E_(alpha)((A)/(A - 4))` where `E_(alpha)` is the KE of `alpha` particle.
A - mass number parent
a. `""_(88)^(226)Ra to ""_(86)^(222)Ra + alpha + Q`
`226.02540 - (222.01750 + 4.002603) = 0.0055297`
`:. Q = 0.005297 xx 931.5 = 4.934 MeV`
`:. E_(alpha) = 4.934 xx (A - 4)/A = 4.934(222/226) = 4.85 MeV` .