A nucleus of `._84Po^210` originally at rest emit `alpha` particle with speed v . Recoil speed of the daughter nucleus is

Explanation

The momentum of the `alpha `-particle is given by ,`p_(alpha)=sqrt (2m _(alpha)K)` ……(1), Let the recoiled momentum of the daughter nucleus be `p_(d) =m_(d)v_(d)` where `m_(d)` and `v_(d)` are the mass and velocity of daughter nucleus respectively.Using the principle of conservation of momentum we get, `P_(d)=P_(alpha)=sqrt(2m_(alpha)k)` `implies V_(d)=sqrt (2m _(alpha)k)/(m_(d))`.....(2), `implies V _(d)=(1)/(196) sqrt ((2xx4xxk)/(m_(p))`=`(2)/(196) sqrt ((2K) /(m_(p))` where m_(p) is the mass of the proton.`implies v_(d) = 3.39 xx10^(5) m//s`.Let the K.E of the daughter nucleus be K' , then, `(K')/(K)=m_(alpha)/m_(d)` as the momenta are same.`therefore (K')/(K+K')=(m_(alpha))/(m_(alpha)+m_(d))`But K+K' is the total energy ,K_(t) say. `therefore (K')/(K_(1))=(m_(alpha))/(m_(alpha)+m_(d))`, `implies K'= (m_(alpha))/(m_(alpha)+m_(d)) K _(t) =(4)/(196+4)k_(t)` `implies =0.02 K_(t)`,`implies (K')/(K_(t)=0.02`