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/c²,1 u = 931 MeV/c².)

Answer 1

(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(₁₉K⁴⁰) − m(₂₀Ca⁴⁰)]c²
           = [39.9640 u − 39.9626 u]c²
           = 0.0014  `xx` 931 MeV
           = 1.3034 MeV

Q_value in the β⁺ decay is given by
Q_value = [m(₁₉K⁴⁰) − m(₂₀Ar⁴⁰) − 2m_e]c²
           = [39.9640 u − 39.9624 u −  0.0021944 u]c²
           = (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(₁₉K⁴⁰) − m(₂₀Ar⁴⁰)]c²
          = (39.9640 − 39.9624)uc²
          = 1.4890 = 1.49 MeV