Potassium-40 can decay in three modes .It can decay by `beta^(-)` - emission,`beta^(+)` -emission or electron capature. (a) Write the equation showing the end products. (b) Find the Q-value in each of the three cases. Atomic masses of `Ar_18^40`,`K_19^40`and `Ca_20^40` are 39.9624 u,39.9640 u,and 39.9626 u respectively.

Answer

A, B, C, D

Explanation

(a) ``_(19)K^(40) rarr _(20)Ca^(40) + _(-1)e^(0) + _(0)barv^(0)`
``_(19)K^(40) + _(-1)e^(0) rarr _(18)Ar^(40) + _(0)barv^(0)`
``_(19)K^(40) + _(-1)e^(0) rarr _(18)Ar^(40)`
``_(19)K^(40) rarr _(20)Ca^(40) + _(-1)e^0 + _(0)barv^(0)`
(b) `Q = [ Mass of reactants
-Mass of products]`c^2`
`=[39.964u - 39.9526u]`
`=[39.964u - 39.964u`
`- 39.9626] uc^2`
`=(39.964 - 39.9626)931MeV`
`=1.3034 MeV`
`_(19)K^(40) rarr _(18)Ar^(40) + 5 _(-1)e^(0) + _(0)barv^(0)`
Energy = (39.9640 - 39.9624)`uc^2`
`= 1.4890 = 1.49 MeV`
`_(19)K^(40) + 4_(-1)e^0 rarr _(15)Ar^(40)`
`Q_(value) = (39.964 - 39.9624)uc^2` .