Consider a nuclear reaction `A+B rarr C`. A nucleus `A` moving with kinetic energy of `5 MeV` collides with a nucleus `B` moving with kinetic energy of `3MeV` and forms a nucleus `C` in excited state. Find the kinetic energy of nucleus `C` just after its formation if it is formed in a state with excitation energy `10 MeV`. Take masses of nuclei of `A,B` and `C ` as `25.0,10.0,34.995 amu`, respectively.
`(1amu=930 MeV//c^(2))`.

Answer

`2.65 MeV`

Explanation

Applying conservation of energy:
`m_(A)c^(2)+K_(A)+m_(B)c^(2)+K_(B)c^(2)+K_(B)=mc^(c^(2))+K_(C)+excitation nergy `
`(m_(A)+m_(B)+m_(C)C^(2)+K_(A)+K_(B)=K_(C)+`excitation nergy
`4.65 +5+3=K_(C) +10`
or `K_(C) =2.65 MeV`.