The nucleus of 214Po decays radioactively by emitting an alp

The nucleus of 214Po decays radioactively by emitting an alpha particle (mass 6.65×1027 kg ) with kinetic energy 1.20×1012 J , as measured in the laboratory reference frame.

Assuming that the Po was initially at rest in this frame, find the magnitude of the recoil velocity of the nucleus that remains after the decay.

Solution

Apply conservation of momentum to the nucleus and its fragments. The initial momentum is zero

mass of 214 Po = 214*1.67*10^-27 = 3.57*10^-25 kg

Let +x be the direction in which the alpha particle is emitted. The nucleus that is left after the decay has mass

m2 = m1 - malpha = 3.57*10^-25 - 6.65*10^-27

ma = 3.5*10^-25 kg

Pi = Pf

0 = m2*v2 + ma*va

v2 = ma*va/m2

given

KEa = 0.5*m*va^2 = 1.2*10^-12

va = sqrt(2*1.2*10^-12/(6.65*10^-27)) = 1.89*10^7 m/sec

v2 = 6.65*10^-27*1.89*10^7/(3.5*10^-25)

v2 = 3.59*10^5 m/sec

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