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 Ar4018, K4019 and Ca4020 are 39.9624 u, 39.9640 u and 39.9626 u respectively.

Answer 1

Explanation:

(a) By β−emission, the decay of potassium-40 is shown as

$\mathrm{K} 4019 \rightarrow 20 \mathrm{Ca} 40+\mathrm{v}^{-}+\beta$-

By β−emission, the decay of potassium-40 is shown as

$\mathrm{K} 4019 \rightarrow 18 \mathrm{Ar} 40+\mathrm{v}+\beta+$

By the capture of electron, the potassium-40 decay is shown as

$\mathrm{K} 4019+\mathrm{e}-\rightarrow 18 \mathrm{Ar} 40+\mathrm{v}$

(b) In the beta decay, Qvalue is shown as  

$Q_{\text {value }}=\left[-m\left(_{20} \mathrm{Ca}^{40}\right)+m\left(_{19} \mathrm{K}^{40}\right)\right] c^{2}$

$=931 \mathrm{MeV} \times 0.0014$

= 1.3034 MeV

In the beta decay, Qvalue is shown as

$Q_{\text {value }}=\left[m\left(_{19} K^{40}\right)-2 m_{e}-m\left(20 \mathrm{Ar}^{40}\right)\right] c^{2}$

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

$=-1022 \mathrm{keV}+1489.96 \mathrm{keV}$

$=0.4679 \mathrm{MeV}$

In the capture of electron, Qvalue is shown as

$Q_{\text {value }}=\left[-m\left(_{20} A r^{40}\right)+m\left(_{19} R^{40}\right)\right] c^{2}$

$=(-39.9624+39.9640) u c^{2}$

$=1.49 \mathrm{MeV}$