Under certain circumstances, a nucleus can decay by emitting a particle more massive than an `alpha`-particle. Consider the following decay processes:
`._(88)Ra^(223)to._(82)Pb^(209)+._(6)C^(14)`, `._(88)Ra^(223)to._(86)Rn^(219)+._(2)He^(4)`
(a) Calculate the Q-values for these decays and determine that both are energetically allowed.

Explanation

Take a `""_(6)^(14)C` emission nuclear reaction:
`""_(88)^(223)Ra rarr ""_(82)^(209)Pb+""_(6)^(14)C`
We know that :
Mass of `""_(88)^(223)Ra,m_(1)=223.01850 u`
Mass of `""_(82)^(209)Pb,m_(2)=208.98107 u`
Mass of `""_(6)^(14)C,m_(3)=14.00324 u`
Hence, the Q-value of the reaction is given as:
`Q=(m_(1)-m_(2)-m_(3))c^(2)`
`=223.01850-208.98107-14.00324)c^(2)`
`=(0.03419 c^(2))u`
But 1 u=931.5 MeV/`c^(2)`
`therefore` Q=0.03419xx931.5
=31.848 MeV
Hence, the Q-value of the nuclear reaction is 31. 848 MeV. Since the value is positive, the reaction is energetically allowed.
Now take a `""_(2)^(4)He` emission nuclear reaction:
`""_(88)^(223)Ra rarr ""_(86)^(229)Rn+""_(2)^(4)He`
We know that:
Mass of `""_(88)^(223)Ra,m_(1)=223.01850`
Mass of `""_(82)^(219)Rn,m_(2)=219.00948`
Mass of `""_(2)^(4),m_(3)=4.00260`
Q-value of this nuclear reaction is given as:
`Q=(m_(1)-m_(2)-m_(3))c^(2)`
`=(223.01850-219.00948-4.00260)C^(2)`
`=(0.00642 c^(2))u`
`=0.00642xx931.5 =5.98 MeV`
Hence, the Q value of the second nuclear reaction is 5.98 MeV. Since the value is positive, the reaction is energetically allowed.