Question

What is the recoil speed of the daughter nucleus when ${}_{\mathbf{67}}^{\mathbf{152}}\mathbf{Ho}$$\mathbf{\alpha }$ decays.

Short answer

The recoil nucleus's speed is$3.94\times {10}^5\text{m/s}$.

Step-by-step solution

Given data

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

Concept of Energy released during Electron Capture

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.

Determine the Recoil Speed from the energy released during Electron Capture

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

${}_{67}^{152}\mathrm{Ho}{\to }_2^4\mathrm{\alpha }{+}_{65}^{148}\mathrm{Tb}$

The $Q$ factor can be calculated as:

$\begin{array}{c}\mathrm{Q}=({\mathrm{m}}_{\text{parent}}-{\mathrm{m}}_{\text{daughter}}){\mathrm{c}}^2\\ =(151.931580\mathrm{u}-4.002603\mathrm{u}-147.924140\mathrm{u})\times (931.5\text{\hspace{0.17em}MeV}/\text{u})\\ =4.51\text{\hspace{0.17em}MeV}\end{array}$

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

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

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

${\mathrm{v}}_{\mathrm{\alpha }}=-\frac{{\mathrm{m}}_{\mathrm{Tb}}{\mathrm{v}}_{\mathrm{Tb}}}{{\mathrm{m}}_{\mathrm{\alpha }}}$

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

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

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

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

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

$\begin{array}{c}{\mathrm{v}}_{\mathrm{Tb}}=\sqrt{\frac{2\mathrm{Q}}{{\mathrm{m}}_{\mathrm{Tb}}+\frac{{\mathrm{m}}_{\mathrm{Tb}}^2}{{\mathrm{m}}_{\mathrm{a}}}}}\\ =\sqrt{\frac{2(4.51\text{\hspace{0.17em}MeV})}{(147.924140\mathrm{u})+\frac{{(147.922140\mathrm{u})}^2}{4.002603\text{u}}}}\\ =\sqrt{\frac{(9.02\text{\hspace{0.17em}MeV})}{(5614.75\mathrm{u})}\left(\frac{{\mathrm{c}}^2}{931.5\text{\hspace{0.17em}MeV/u}}\right)}\\ =1.313\times {10}^{-3}\mathrm{c}\end{array}$

Calculate further as:

$\begin{array}{c}{\mathrm{v}}_{\mathrm{Tb}}=1.313\times {10}^{-3}\mathrm{c}\\ =1.313\times {10}^{-3}(3.0\times {10}^8\text{\hspace{0.17em}m/s})\\ =3.94\times {10}^5\text{\hspace{0.17em}m/s}\end{array}$

The recoil nucleus's speed is $3.94\times {10}^5\text{\hspace{0.17em}m/s}$.