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)`.

Making use of the tables of atomic masses, determine the energies of t

1.	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)`.
Answer» (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`

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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)`.

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