Find the `Q` value of the reaction
`N^(14)+alpha rarr O^(17) +P`
The mass of are, respectively`, 14.00307 u, 4.00260 u`, and `16.99913u`. Find the total kinetic energy of the products if the striking `alpha` particles has the minimum kinetic energy required to initiate the reaction .

Answer

`-1.20MeV,0.34MeV`

Explanation

`N^(14) +alpha rarr O^(17) +proton`
`Q value =(14.00307 + 4.00260 -1.00783 -16.99913) 931.5=-1.20 MeV`
Let m and `M` be the mass of `alpha-` particles and nitrogen nucleus, respectively, and let minimum KE of `alpha-`particles be `(1)/(2) mu^(2)`. From energy equation ,
`(1)/(2) mu^(2) =|Q| + minimum KE of system`
`=|Q| + |Q| +(1)/(2) (m+M)[(m u)/(m+M)]^(2)`
`(1)/(2) m u^(2) ((M)/(m+M)) =|Q|`
`(1)/(2) m u^(2)=|Q|((m+M')/(M))`
`KE` of products `=(1)/(2)(m+M)[ (m u)/(m+M)]^(2)`
`=(1)/(2)m u^(2)((m)/(m+M))`
`|Q| (m)/(M)`
`=1.2 xx (4)/(14) =0.34 MeV` .