`._(235)^(239)Pu._(94)` is undergoing `alpha-decay` according to the equation `._(94)^(235)Pu rarr (._(97)^(235)U) +._2^4 He` . The energy released in the process is mostly kinetic energy of the `alpha`-particle. However, a part of the energy is released as `gamma` rays. What is the speed of the emitted `alpha`-particle if the `gamma` rays radiated out have energy of `0.90 MeV`? Given: Mass of `._(94)^(239)Pu =239.05122 u`, mass of `(._(97)^(235)U)=235.04299 u` and mass of `._1^4He =4.002602 u (1u =931 MeV)`.

Answer

`144.10ms^(-1)`

Explanation

`Delta m=0.00563 u=5.24 MeV`
KE of `alpha`-particle` =5.24 -0.9=4.34 MeV`
`:. v=sqrt((2.434.10.1.6.10^(9))/(4.1.67.10))=1.44.10 m s^(-1)`.