#### Question

Potassium-40 can decay in three modes. It can decay by β^{−}-emission, B*-emission of electron capture. (a) Write the equations showing the end products. (b) Find the *Q*-values in each of the three cases. Atomic masses of `""_18^40Ar` , `""_19^40K` and `""_20^40Ca` are 39.9624 u, 39.9640 u and 39.9626 u respectively.

(Use Mass of proton m_{p} = 1.007276 u, Mass of `""_1^1"H"` atom = 1.007825 u, Mass of neutron m_{n} = 1.008665 u, Mass of electron = 0.0005486 u ≈ 511 keV/c^{2},1 u = 931 MeV/c^{2}.)

#### Solution

(a) Decay of potassium-40 by β^{−}emission is given by

`""_19"K"^40` → `""_20"Ca"^40` + `β^-` + `bar"v"`

Decay of potassium-40 by β^{+} emission is given by

`""_19"K"^40 → ""_18"Ar"^40 + β^+ + "v"`

Decay of potassium-40 by electron capture is given by

`""_19"K"^40 + e^(-) → ""_18"Ar"^40 + "v"`

(b)

Q_{value} in the β^{−} decay is given by

Q_{value} = [m(_{19}K^{40}) − m(_{20}Ca^{40})]c^{2}

= [39.9640 u − 39.9626 u]c^{2}

= 0.0014 `xx` 931 MeV

= 1.3034 MeV

Q_{value} in the β^{+} decay is given by

Q_{value} = [m(_{19}K^{40}) − m(_{20}Ar^{40}) − 2m_{e}]c^{2}

= [39.9640 u − 39.9624 u − 0.0021944 u]c^{2}

= (39.9640 − 39.9624) 931 MeV − 1022 keV

= 1489.96 keV − 1022 keV

= 0.4679 MeV

Q_{value} in the electron capture is given by

Q_{value} = [ m(_{19}K^{40}) − m(_{20}Ar^{40})]c^{2}

= (39.9640 − 39.9624)uc^{2}

= 1.4890 = 1.49 MeV