Making use of the tables of atomic masses, find:
(a) the mean binding energy per one nucleon in `O^(16)` nucleus
(b) The binding energy of neutron and an alpha-particle in a `B^(11)` nuclues.

Explanation

(a) Total binding energy of the `O^(16)` nucleus is
`E_(b)= 8xx.00867+8xx.00783+0.00509 am u`
`=0.13709am u= 127.6 MeV`
so `B.E` per nucleon is `7.98 Mev//n ucl eon`
(b) `B.E` of neturon in `B^(11)` nucleus
`=B.E of B^(11)-B.E. of B^(10)`
(since on removing a neutron from `B^(11)` we get `B^(10)`)
`=Delta_(n)-Delta_(B_(11))+Delta_(B_(10))= .00867-.00930+.01294`
`=00.01231 am u= 11.46MeV`
`B.E` of ( an `alpha`-particle in `B^(11)`)
`=B.E of B^(1)-B.E of Li^(7)-B.E of alpha`
(since on removing an `alpha from B^(11)` we get `li^(7)`)
`=-Delta_(B_(11))+Delta_(Li_(7))+Delta_(alpha)`
`= -0.00930+0.01601+0.00260`
`=0.00931 am u= 8.67MeV`
(c ) This energy is
`[B.E of O^(16)+4("B.E of " alpha "particles")]`
`=Delta_(o)^(16)+4Delta_(alpha)`
`=4xx0.00260+0.00509`
`=0.01549 am u= 14.42MeV`.