Book edition 2nd Edition

Author(s) Randy Harris

Pages 633 pages

ISBN 9780805303087

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Short Answer

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

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

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given data

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

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:

$Q = (m_{parent} - m_{daughter}) c^{2}$ .

Where $m_{parent}$ is the atomic mass of the parent nucleus, and $m_{daughter}$ is the atomic mass daughter nucleus.

Step 3: Determine the Recoil Speed from the energy released during Electron Capture

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

$_{67}^{152} Ho \to_{2}^{4} α +_{65}^{148} Tb$

The $Q$ factor can be calculated as:

$\begin{matrix} Q = (m_{parent} - m_{daughter}) c^{2} \\ = (151 .931580 u - 4 .002603 u - 147 .924140 u) \times (931 .5 MeV / u) \\ = 4 .51 MeV \end{matrix}$

Assume the speed of the recoil nucleus is $v_{Tb}$ then from momentum conservation, we have:

$m_{Tb} v_{Tb} + m_{α} v_{α} = 0$

Rearrange the above equation for $v_{α}$ as:

$v_{α} = - \frac{m_{Tb} v_{Tb}}{m_{α}}$

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

$\begin{matrix} Q = {KE}_{Tb} + {KE}_{a} \\ = \frac{1}{2} m_{Tb} v_{Tb}^{2} + \frac{1}{2} m_{α} v_{α}^{2} \\ = \frac{1}{2} v_{Tb}^{2} (m_{Tb} + \frac{m_{Tb}^{2}}{m_{a}}) \end{matrix}$

Rearrange the above equation for $v_{Tb}$ as:

$v_{Tb} = \sqrt{\frac{2 Q}{m_{Tb} + \frac{m_{Tb}^{2}}{m_{α}}}}$

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

$\begin{matrix} v_{Tb} = \sqrt{\frac{2 Q}{m_{Tb} + \frac{m_{Tb}^{2}}{m_{a}}}} \\ = \sqrt{\frac{2 (4 .51 MeV)}{(147 .924140 u) + \frac{{(147 .922140 u)}^{2}}{4 .002603 u}}} \\ = \sqrt{\frac{(9 .02 MeV)}{(5614 .75 u)} (\frac{c^{2}}{931 .5 MeV/u})} \\ = 1 .313 \times 10^{- 3} c \end{matrix}$

Calculate further as:

$\begin{matrix} v_{Tb} = 1.313 \times 10^{- 3} c \\ = 1.313 \times 10^{- 3} (3.0 \times 10^{8} m/s) \\ = 3.94 \times 10^{5} m/s \end{matrix}$

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

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Modern Physics

Answers without the blur.

Short Answer

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

Step by Step Solution

Step 1: Given data

Step 2: Concept of Energy released during Electron Capture

Step 3: Determine the Recoil Speed from the energy released during Electron Capture

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Modern Physics

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Short Answer

What is the recoil speed of the daughter nucleus when 67152Ho α decays.

Step by Step Solution

Step 1: Given data

Step 2: Concept of Energy released during Electron Capture

Step 3: Determine the Recoil Speed from the energy released during Electron Capture

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