Question

Calcium-41 decays by electron capture.

(a) Find$\mathbf{Q}$ for the decay.

(b) Show that calcium-41 cannot decay by${\mathbf{\beta }}^{\mathbf{+}}$ emission.

Short answer

a) The $Q$ factor for this decay is, $\left|0.42\text{MeV}\right|$.

b) If calcium- 41 underwent ${\mathrm{\beta }}^{+}$decay, it would also become potassium-$41$, but according to the above calculation, the $\mathrm{Q}$ factor is $0.42\text{MeV}$, which is smaller than the electron's rest mass-energy.

This is impossible because electron cannot have a kinetic energy less than its rest-energy.

Step-by-step solution

Given data 

Calcium- $41$decays by electron capture.

Concept of Energy released during Electron Capture

During electron capture the amount of energy released is dependent on the mass difference between parent and the daughter nucleus.

Which is,$\mathbf{Q}\mathbf{=}\mathbf{(}{\mathbf{m}}_{\mathbf{\text{parent}}}\mathbf{-}{\mathbf{m}}_{\mathbf{\text{daughter}}}\mathbf{)}{\mathbf{c}}^{\mathbf{2}}$ .

Determine the Kinetic Energy

(a)

Calcium-$41$ undergoes electron capture, and it will lose a charge, becomes potassium- $41$.

${}_{20}^{41}\text{Ca}{+}_{-1}^0\text{e}{\to }_{19}^{41}\text{K}$

The amount of energy released in the electron capture is:

$\begin{array}{c}\mathrm{Q}=({\mathrm{m}}_{\text{parent}}-{\mathrm{m}}_{\text{daughter}}){\mathrm{c}}^2\\ =({\mathrm{m}}_{\text{Ca}}-{\mathrm{m}}_{\mathrm{K}})\left(931.5\frac{\text{MeV}}{\text{u}}\right)\\ =(40.962278\text{u}-40.961825\text{u})\left(931.5\frac{\text{MeV}}{\text{u}}\right)\\ =0.42\text{\hspace{0.17em}MeV}\end{array}$

Therefore, the $Q$ factor for this decay is$\left|0.42\text{\hspace{0.17em}MeV}\right|$ .

Explain how Calcium- 41 cannot decay by β+  emission

(b)

Now, calculate the released energy as:

$\begin{array}{l}\mathrm{Q}=(40.962591\text{\hspace{0.17em}u}-(40.961825\text{\hspace{0.17em}u}+2\cdot 0.0005486\text{\hspace{0.17em}u}))\cdot {\mathrm{c}}^2\\ \mathrm{Q}=-0.3\text{\hspace{0.17em}MeV}\end{array}$

Note the released energy is negative and it can be concluded that ${\mathrm{\beta }}^{+}$ ion has no decay.