a stationary nucleus with mass numbers a=220 decays by alpha emission. the energy released is q=5.5 MeG .A good estimate for the kinetic energy of the alpha partial will be.
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1 amu = 931.4 MeV
zX220 → z-2Y216 + 2He4
From energy and momentum conservation we can derive the equation for K.E. of alpha particle as
(K.E.)α = [(A−4)/A] Q
Q = (K.E.)α A/(A−4) = 5.4 x 220/216 = 5.5 MeV
K.E. of alpha particle = 5.4 MeV
K.E. of daughter nucleus = Q − K.E. of alpha particle
= 5.5 − 5.4 = 0.1 MeV
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