Question

A nucleus of mass 220 amu in free state decays to amit an alpha particle kinetic energy of the alpha particle diameter is 5.4 mega electron volt the recoil energy of the daughter nucleus is

Answer 1

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

Answer 2

Answer:

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