Making use of the tables of atomic masses, determine the energies of the following reaction:
(a) `Li^(7)(p,n)Be^(7)` ,
(b)`Be^(9)(n,gamma)Be^(10)`,
(c )`Li^(7)(alpha,n)B^(10)`,
(d) `O^(16)(d,alpha)N^(14)`.

Explanation

(a) The reaction is `Li^(7)(p,n)Be^(7)` and the energy of reaction is
`Q=(M_(Be)^(7)+M_(Li)^(7))c^(2)+(M_(p)-M_(n))c^(2)`
`=(Delta_(Li_(7))-Delta_(Be)^(7))c^(2)+Delta_(p)-Delta_(n)`
`=[0.01601+0.00738-0.01693-0.00867]am uxxc^(2)`
`= -1.64MeV`
(b) The reaction is `Be^(9)(n, gamma)Be^(10)`
Mass of `gamma` is taken zero.Then
`Q=(M_(Be)^(9)+M_(n)-M_(Be)^(10))c^(2)`
`=(Delta_(Be)^(9)+Delta_(n)-Delta_(Be)^(10))c^(2)`
`=(0.01219+0.00867-0.01354)overset//, am uoverset//xxc^(2)`
`=6.81MeV`
(c )The reaction is `Li^(7)(alpha,n)B^(10)`. The energy is
`Q=(Delta_(Li)^(7)+Delta_(alpha)-Delta_(n)-Delta_(B)^(10))c^(2)`
`=(0.01601+0.00260-0.00867-.01294)am uxxc^(2)`
`= -2.79MeV`
(d) The reaction is `O^(16)(d, alpha)N^(14)`. The energy of reaction is
`Q=(Delta_(O)^(16)+Delta_(d)-Delta_(alpha)-Delta_(N)^(14))c^(2)`
`=(-0.00509+0.01410-0.00260-0.00307)am uxxc^(2)`
`=2.79MeV`