(a) Calculate the disintegration energy when stationary `""_(92)^(232)"U"` nucleus decays to thorium `""_(90)^(228)"Th"` with the emission of `alpha` particle. The atomic masses are of `""_(92)^(232)"U" = 232.037156 u` , `""_(90)^(228)"Th" = 228.028741 u` and ` ""_(2)^(4)"He" = 4.002603 u`
(b) Calculate kinetic energies of `""_(90)^(228)"Th"` and `alpha`- particle and their ratio.

Explanation

(a) The difference in masses `Delta m = (m_(U) - m_(Th) - m_(alpha))`
` = (232.037156 - 228.028741 -4.002603)u`
The mass lost in this decay = `0.005812 u ` Since `1u = 931`MeV, the energy Q released is `Q = (0.005812 u) xx (931 MeV/u) = 5.14 MeV`
This disintegration energy Q appears as the kinetic energy of `alpha` particle and the daughter nucleus.In any decay, the total linear momentum must be conserved. Total linear momentum of the parent nucleus = total linear momentum of the daughter nucelus + `alpha` particle.
Since before decay, the uranium nucleus is at rest,its momentum is zero. By applying conservation of momentum, we get
`0 = m_(Th) vec(v)_(Th) + m_(a) vec(v)a`
`m_(a) vec(v)_(a) = - m_(Th) vec (v)_(Th)`
It implies that the alpha particle and daughter nucleus move in opposite directions.In magnitude `m_(alpha)v_(alpha) = m_(Th) v_(Th)`
The velocity of `alpha` particle `v_(a) = (m_(Th))/(m_(a)) V_(Th)`
Now that `(m_(Th))/(m_(a)) gt 1`, so `v_(alpha) gt v_(Th)`. The ratio of the kinetic energy of `alpha` particle to the daughter nucleus.
`(K.E_(alpha))/(K.E_(Th)) = (1/2 m_(alpha)v_(alpha)^(2))/(1/2 m_(Th)v_(Th)^(2)`
By substituting, the value of `v_(a)` into the above equation, we get
`(K.E_(alpha))/(K.E_(Th)) = (m_(Th))/(m_(alpha)) = (228.02871)/(4.002603) = 57 `
The kinetic energy of `alpha` particle is 57 times greater than the kinetic energy of the daughter nucleus `(""_(90)^(228)"Th")`.
The disintegration energy Q = total kinetyic energy of products
`K.E_(alpha) + K.E_(Th) = 5.41 MeV`
`57 K.E_(Th) + K.E_(Th) = 5.41 MeV`
`K.E_(Th) = (5.41)/(58) MeV = 0.093 meV`
`K.E_(alpha) = 57 K.E_(Th) = 57 xx 0.093 = 5.301 MeV`
In fact, `98 %` of total kinetic energy is taken by the `alpha` particle.