Since the reaction `Li^(7)(p,n)Be^(7)(Q= -1.65MeV)` is initiated, the incident proton energy must be
`ge(1+(M_(p))/(M_(d)))xx1.65= 1.89MeV`
since the reaction `Be^(9)(p,n)B^(9)(Q= -1.85MeV)` is not initiated
`Tle(1+(M_(p))/(M_(ve)))xx1.85= 2.06MeV` Thus `1.89MeVleT_(p)le2.06MeV`