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Q49E

Expert-verified
Found in: Page 520

### Modern Physics

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

See the step by step solution

## Step 1: Given data

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

## Step 2: 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:

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

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.

## Step 3: 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:

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

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:

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

Calculate further as:

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

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

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