Explain with reason the nature of emitted particle by:

Options

A. `._(7)^(14)N (n, p)`
B. `._(18)^(35)Ar`
C. `._(32)^(80)Ge`
D. `._(20)^(40)Ca`

Answer

(a) `._(20)^(38)Ca` : It has n/p=18/20=0.9, Which lies below the belt of stability and thus positron emitter
`._(20)^(38)Ca rarr ._(19)^(38)K+_(+1)^(0)e`
(b) `._(18)^(35)Ar` : It has `n/p=17/18=0.994`, which lies below the belt of stability and thus, positron emitter
`._(18)^(35)Ar rarr ._(17)^(35)Cl+._(+1)^(0)e`
If `n//p lt 1` and nuclear charge is high the nuclide may show K-electron capture.
(c) `._(32)^(80)Ge` : It has `n/p=48/32=1.5`, which lies above the belt of stability and thus `beta`-emitter
`._(32)^(80)Ge rarr._(33)^(80)As +_(-1)^(0)e`
(d) `._(20)^(40)Ca` : It has both magic number `p=20, m=20` and thus, stable.