Find the Q value and the kinetic energy of emitted `alpha` particle in the `alpha` decay of
(a) `._88Ra^(226)` (b) `._86Rn^(220)`.
Given `m(._88Ra^(226))=226.02540u, m(._86Rn^(222))=222.01750u`
(b) `m(._86Rn^(220))=220.01137u , m(._84Po^(216))=216.00189u` and `m._(alpha)=4.00260u.`

Explanation

(a) The decay takes place according to the equation
`""_(88)Ra^(226) to ""_(2)He^(4) + ""_(86)Rn^(222) + Q`
Q-value of this relation
`=[m(""_(88)Ra^(226))-[m(""_(2)He^(4)) + m(""_(86)Rn^(222))]c^(2)`
`=0.005297 xx 931.5` MeV
`=4.9342 MeV`
Kinetic energy of emitted `alpha`-particles
`=(A-4)/A xx Q`-value`=(226-4)/226 xx 4.93 = 4.85` MeV
(b) The decay takes place according to the equation
`""_(86)Rn^(220) to ""_(2)He^(4) + ""_(84)Po^(216) + Q`
Q-value of this reaction.
`=[m(""_(86)Rn^(220))-[m(""_(2)He^(4))+m(""_(84)Po^(216))]c^(2)`
`=(220.01137 - 220.00449)u xx c^(2)`
`=0.006877 u xx c^(2) = 0.006877 xx 931.5 = 6.41` MeV
Kinetic energy of emitted `alpha`-particles.
`=(A-4)/A xx Q -"value" = (220-4)/220 xx 6.41`
=6.29 MeV