# 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

21 views
in Physics
closed
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
B. T//2
C. slightly less than T
D. slightly greater than T

by (117k points)
selected

( c) Let then 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 ....(1)
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))/(_(alpha)) ...(2)
Subst. (2) into (1), we have
(E_(alpha))/(E_(th)) = ((m_alpha)/(m_(th)))((m_(th))/(m_(alpha)))^2 = (m_(th))/(m_(alpha)) = (234)/(4) = 58.5
Thus, the kinetic energy is 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 = 0.98 T
which is slightly less than T.