# 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 ki

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

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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.