Stationery nucleus `.^(238)U` decays by a emission generaring a total kinetic energy T:
`._(92)^(238) rarr ._(90)^(234)Th +._2^4 alpha`
What is the kinetic energy of the `alpha`-particle?

Options

A. Slightly less than `T//2`
B. `T//2`
C. Slightly less than T
D. Slightly greater than T

Answer

c

Explanation

Let the kinetic energy of the `alpha`-particle be `E_(alpha)` and that of the thorium Th be `E_(th)`. The ratio of kinetic energies is
`(E_(alpha))/(E_(th))=((1)/(2)m_(alpha)v_(alpha)^(2))/((1)/(2) m_(th)v_(th)^(2))=((m_(alpha))/(m_(th)))((v_(alpha))/(v_(th)))^(2)`
By conservation of momentum, the momentum of `alpha`-particle and that of the recoiling thorium must be equal. Thus,
`m_(alpha) v_(alpha)=m_(th) v_(th)`
or `(v_(alpha))/(v_(th)) =(m_(th))/(m_(alpha))`
Substitutuing Eq, (ii) in Eq. (i), we have
`E_(alpha)/(E_(th)) =((m_(alpha))/(m_(th)))((m_(th))/(m_(alpha)))^(2) =(m_(th))/(m_(a))=(234)/(4)=58.5`
Thus, the kinetic energy of the `alpha`-particle expressed as the fraction of the total kinetic energy `T` is given by
`E_(alpha)=(58.5)/(1+58.5)T=(58.5)/(59.5)T=0.98T`
Which is slightly less than `T`.