Suppose some nucleus undergoes alpha decay, releasing 5.6 MeV of energy. If the

Question

Suppose some nucleus undergoes alpha decay, releasing 5.6 MeV of energy. If the daughter product of this reaction is 92 X 236 (atomic mass = 236.04556) , it will recoil away as the alpha particle leaves. Determine (a) the energy of the daughter product and (b) the energy of the alpha particle (atomic mass = 4.002603 u). Assume that the energy of each particle is kinetic energy, and ignore any other small amounts of energy that might be carried away by other emissions. In addition, ignore relativistic effects.

Explanation / Answer

By conservation of momentum 236.04556*v1 = 4.002603*v2 =>v1/v2= 0.017 Conservation of energy 5.6*1.6*10^-13 = 6.023*10*-23 *0.5( 236.04556*v1^2+ 4.002603*v2^2) =>29752.615*10^6 = 0.0682*v2^2+4.002603*v2^2 =>v2=85491.42 m/sec v1= 1453.35 m/sec a) 0.5* 236.04556*6.023*10^-23*145^2= 0.0938Mev b)5.5 Mev

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