Correct Answer - c

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`.