what kinetic energy must an `alpha`-particle possess to split a deuteron `H^(2)` whose binding energy is `E_(b)=2.2 MeV`?

Answer

`6.6 MeV`

Explanation

Loss in `K.E` will be used in splitting of Deuteron.
Let us calculate the minimum `K.E`. Required for the loss of `2.2 MeV`. For this we must assume perfectly inelastic collision.

maximum loss of `K.E.` will be in perfectly inelastic collision.
By energy conservation.
`rArr (1)/(2)xx4 m u^(2)-(1)/(2)xx6m((2u)/(3))^(2)=E_(b)`
`(1)/(2)xx4 m u^(2)[1-(2)/(3)]=E_(b)rArr(1)/(2)xx4m u^(2)=3E_(b)`
`:. KE gt 3E_(b)`
`KE gt 6.6 MeV`