Question

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

Answer 1

i) For the decay process
`._(88)Ra^(223)to ._(82)Pb^(209)+ ._(6)C^(14)+Q`
mass defect, `Delta M` = mass of `Ra^(223)` - (mass of `Pb^(209)` + mass of `C^(14)`)
`= 223.01850 - (208.98107+14.00324)`
= 0.03419 u
`Q = 0.03419 xx 931 MeV = 31.83 MeV`
ii) For the decay process
`._(88)Ra^(223)to ._(86)Rn^(219)+ ._(2)He^(4)+Q`
mass defect, `Delta M` = mass of `Ra^(223)` - (mass of `Rn^(219)+` mass of `He^(4)`)
`= 223.01850 - (219.00948 + 4.00260)`
`= 0.00642 u`
`therefore Q = 0.00642xx931 MeV = 5.98 MeV`
As Q values are positive in both the cases, therefore both the decays are energetically possible.