Question

A polonium isotope with an atomic mass of 218.008973 u undergoes alpha decay, resulting in a daughter isotope with an atomic mass of 213.999805 u. lgnoring any recoil of the daughter, find the kinetic energy of the emitted alpha particle in megaelectronvolts (MeV) Number MeV

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Answer #1

?m = mass defect = 218.008973 u - 213.999805 u - 4.002603 u = 0.006565 u

So kinetic energy of the emitted particle = ?m×931 MeV

= 0.006565×931 = 6.112 MeV

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