Show that the nuclide `.^8Be` has a positive binding energy but is unstable with respect to decay into two alpa particles, where masses of neutron, `.^1H`, and `.^8 Be` are `1.008665` u, `1.007825 u`, and `8.005305 u`, respectively.

Explanation

The binding energy of `(.^8Be)` is determined by the equation
`B(.^8Be) =[4m_(n) +4M(.^(1)H) -M(.^8Be)]c^(2)`
The binding energy
`B(.^8Be) =[4 xx 1.008665 +4 xx 1.007825-8.005305] xx 931.5`
`=56.5MeV`
Now, we calculated the binding energy `B` of the decay of `(.^8Be)` into two `alpha`-particles.
`(.^8Be) rarr 2 alpha`
`B=[2 M(.^4He)-M(.^8Be)]c^(2)`
`=[2 xx 4.002603 -8.005305]xx931.5`
`= -0.092 MeV`
Because `B` is negative for this reaction, hence `(.^8Be)` is unstable against decay to two alpha particles.