Question

The nucleus of `._(90)^(230)Th` is unstable against `alpha`-decay with a half-life of `7.6xx10^(3)"years"`. Write down the equation of the decay and estimate the kinetic energy of the emitted `alpha`-particle from the following data: `m(._(90)^(230)Th)=230.0381 "amu", m(._(68)^(226)Ra)=226.0254 "amu" m(._(2)^(4)He)=4.0026"amu"`.

Answer 1

The equation of the decay is
`._(90)^(230)Th rarr_(88)^(225)Ra+_(2)^(4)He+Q`
The energy `Q` is given by
`Q=[m(Th)-m(Ra)-m(He)]c^(2)`
Using the given data and `c^(2)=931.5 MeV//"amu"`, we get `Q=9.41 MeV`. This energy is shared by `Ra` and `He`. If the original nucleus `Th` is at rest, i.e. if the momentum of the system before `alpha`-decay is zero, the total momentum after the decay will also be zero. Thus `Ra` and `He` will have equal and opposite linear momenta.
`:. m_(He)v_(He)= -m_(Ra)V_(Ra)`
or `m_(He)v_(He)=m_(Ra)^(2)v_(ra)^(2)` ......(i)
Now `(K.E.(HE))/(K.E.(Ra))=(1//2m_(He)v_(He)^(2))/(1//2m_(Ra)v_(Ra)^(2))=((m_(He)v_(He))/(m_(ra)v_(Ra)))xx((m_(Ra))/(m_(He)))`
`=(m_(Ra))/(m_(He))`" " [use Eq.(i)]
`~~(226)/(4)~~56.6` ......(ii) i.e. the kinetic energy of `He` is `56.5` times that of `Ra`, the total energy being `K.E.(He)+K.E.(Ra)=9.41 MeV` " "(ii)
From eqs (ii) and (iii) we have `K.E. (Ra)=0.16 MeV` and `K.E.(He)~~9.25 MeV`.