91Ó°ÊÓ

The Given Element is ${}_{67}^{152}\text{Ho}$.

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

Which is:

$\mathbf{Q}\mathbf{=}\mathbf{(}{\mathbf{m}}_{\mathbf{\text{parent}}}\mathbf{−}{\mathbf{m}}_{\mathbf{\text{daughter}}}\mathbf{)}{\mathbf{c}}^{\mathbf{2}}$.

Where${\mathbf{m}}_{\mathbf{\text{parent}}}$is the atomic mass of the parent nucleus, and${\mathbf{m}}_{\mathbf{\text{daughter}}}$is the atomic mass daughter nucleus.

The alpha decay of ${}_{67}^{152}\text{Ho}$is:

${}_{67}^{152}\mathrm{Ho}{→}_2^4\mathrm{Î\pm }{+}_{65}^{148}\mathrm{Tb}$

The $Q$ factor can be calculated as:

$\begin{array}{c}\mathrm{Q}=({\mathrm{m}}_{\text{parent}}−{\mathrm{m}}_{\text{daughter}}){\mathrm{c}}^2\\ =(151.931580\mathrm{u}−4.002603\mathrm{u}−147.924140\mathrm{u})×(931.5\text{ M±ð³Õ}/\text{u})\\ =4.51\text{ M±ð³Õ}\end{array}$

Assume the speed of the recoil nucleus is${\mathrm{v}}_{\mathrm{Tb}}$then from momentum conservation, we have:

${\mathrm{m}}_{\mathrm{Tb}}{\mathrm{v}}_{\mathrm{Tb}}+{\mathrm{m}}_{\mathrm{Î\pm }}{\mathrm{v}}_{\mathrm{Î\pm }}=0$

Rearrange the above equation for ${\mathrm{v}}_{\mathrm{Î\pm }}$as:

${\mathrm{v}}_{\mathrm{Î\pm }}=−\frac{{\mathrm{m}}_{\mathrm{Tb}}{\mathrm{v}}_{\mathrm{Tb}}}{{\mathrm{m}}_{\mathrm{Î\pm }}}$

The total kinetic energy is given by the $Q$ factor, so we obtain:

$\begin{array}{c}\mathrm{Q}={\mathrm{KE}}_{\mathrm{Tb}}+{\mathrm{KE}}_{\mathrm{a}}\\ =\frac{1}{2}{\mathrm{m}}_{\mathrm{Tb}}{\mathrm{v}}_{\mathrm{Tb}}^2+\frac{1}{2}{\mathrm{m}}_{\mathrm{Î\pm }}{\mathrm{v}}_{\mathrm{Î\pm }}^2\\ =\frac{1}{2}{\mathrm{v}}_{\mathrm{Tb}}^2\left({\mathrm{m}}_{\mathrm{Tb}}+\frac{{\mathrm{m}}_{\mathrm{Tb}}^2}{{\mathrm{m}}_{\mathrm{a}}}\right)\end{array}$

Rearrange the above equation for ${\mathrm{v}}_{\mathrm{Tb}}$ as:

${\mathrm{v}}_{\mathrm{Tb}}=\sqrt{\frac{2\mathrm{Q}}{{\mathrm{m}}_{\mathrm{Tb}}+\frac{{\text{m}}_{\mathrm{Tb}}^2}{{\mathrm{m}}_{\mathrm{Î\pm }}}}}$

Substitute values in the above equation, the recoil speed of the daughter nucleus is given below as:

$\begin{array}{c}{\mathrm{v}}_{\mathrm{Tb}}=\sqrt{\frac{2\mathrm{Q}}{{\mathrm{m}}_{\mathrm{Tb}}+\frac{{\mathrm{m}}_{\mathrm{Tb}}^2}{{\mathrm{m}}_{\mathrm{a}}}}}\\ =\sqrt{\frac{2(4.51\text{ M±ð³Õ})}{(147.924140\mathrm{u})+\frac{{(147.922140\mathrm{u})}^2}{4.002603\text{u}}}}\\ =\sqrt{\frac{(9.02\text{ M±ð³Õ})}{(5614.75\mathrm{u})}\left(\frac{{\mathrm{c}}^2}{931.5\text{ M±ð³Õ/u}}\right)}\\ =1.313×{10}^{−3}\mathrm{c}\end{array}$

Calculate further as:

$\begin{array}{c}{\mathrm{v}}_{\mathrm{Tb}}=1.313×{10}^{−3}\mathrm{c}\\ =1.313×{10}^{−3}(3.0×{10}^8\text{″¾/²õ})\\ =3.94×{10}^5\text{″¾/²õ}\end{array}$

The recoil nucleus's speed is $3.94×{10}^5\text{″¾/²õ}$.