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Q49E

Expert-verifiedFound in: Page 520

Book edition
2nd Edition

Author(s)
Randy Harris

Pages
633 pages

ISBN
9780805303087

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

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

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:

${}_{67}^{152}\mathrm{Ho}{\to}_{2}^{4}\mathrm{\alpha}{+}_{65}^{148}\mathrm{Tb}$

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 $3.94\times {10}^{5}\text{\hspace{0.17em}m/s}$ .

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