The `._(88)^(226)Ra` nucleus undergoes `alpha`-decay to `._(88)^(226)Rn`. Calculate the amount of energy liberated in this decay. Take the mass of `._(88)^(226)Ra` to be `226.025 402 u`, that of `._(88)^(226)Rn` to be `222.017 571 u`, and that of `._(2)^(4)He` to be `4.002 602 u`.

Explanation

After decay, the mass of the daughter `m_(d)` plus the mass of the alpha particle `m_(alpha)` is
`m_(d) +m_(alpha) =222.017 571 u +4.002 602 u`
`=226.020 173 u`
Thus, calling the mass of the parent nucleus `M_(p)`, we find that the mass lot during decay is
`Delta m=M_(p)-(m_(d) +m_(alpha))`
`=226.025 402 u-226.020 173 u=0.005229 u`
Using the relationship that 1 u corresponds to` 931.434 MeV`, we find that the energy liberated is
`(E=(0.005 229 u)(931. 494 MeV//u) =4.87 MeV`.