Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

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Short Answer

**The spin- $\frac{3}{2} Σ^{* 0}$ baryon (see table in Problem 24) has a rest energy of ( $1385 M e V$ with an intrinsic uncertainty ignored here); the spin $\frac{1}{2} Σ^{0}$ - baryon has a rest energy of. $1192.5 MeV$ If each of these particles has a kinetic energy of $1000 MeV$ ,
(a) which is moving faster and
(b) by how much?**

(a) The spin- $\frac{1}{2} Σ^{0}$ baryon is moving faster.

(b) It is moving faster by. $7.5 \times 10^{6} \frac{m}{s}$

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given data

Rest energy of the spin- $\frac{3}{2} Σ^{* 0}$ baryon

$E_{0}^{*} = 1385 MeV$

Rest energy of the spin- $\frac{1}{2} Σ^{0}$ baryon

$E_{0} = 1192.5 MeV$

Kinetic energy of the two particles

$K = 1000 MeV$

Step 2:Determine the formula for the relativistic energy.

The relativistic kinetic energy of a particle moving with velocity $v$ and having rest energy $E_{0}$ is:

$K = (\frac{1}{\sqrt{1 - \frac{v^{2}}{c^{2}}}} - 1) E_{0}$ ..... (I)

Here, $c$ is the speed of light in vacuum with value

$c = 3 \times 10^{8} \frac{m}{s}$

Step 3: Determinethe particle with greater velocity

(a)

Let the velocity of the spin- $\frac{3}{2} Σ^{* 0}$ baryonbe $v^{*}$ . Then from equation (I)

$K = (\frac{1}{\sqrt{1 - \frac{v^{* 2}}{c^{2}}}} - 1) E_{0}^{*}$

Substitute the values and solve as:

$\begin{array}{l} 1000 MeV = (\frac{1}{\sqrt{1 - \frac{v^{* 2}}{c^{2}}}} - 1) \times 1385 MeV \\ \frac{1}{\sqrt{1 - \frac{v^{* 2}}{c^{2}}}} = 1.722 \\ \frac{v^{* 2}}{c^{2}} = 0.663 \\ v^{*} = 0.814 c \end{array}$

Let the velocity of the spin- $\frac{1}{2} Σ^{0}$ baryon be . $v$ Then from equation (I)

$K = (\frac{1}{\sqrt{1 - \frac{v^{2}}{c^{2}}}} - 1) E_{0}$

Substitute the values and solve as:

$\begin{array}{l} 1000 MeV = (\frac{1}{\sqrt{1 - \frac{v^{2}}{c^{2}}}} - 1) \times 1192.5 MeV \\ \frac{1}{\sqrt{1 - \frac{v^{2}}{c^{2}}}} = 1.839 \\ \frac{v^{2}}{c^{2}} = 0.704 \\ v = 0.839 c \end{array}$

Compare the two velocities:

$\begin{matrix} v^{*} = 0.814 c \\ v = 0.839 c \\ v > v^{*} \end{matrix}$

Thus, the speed of the spin- $\frac{1}{2} Σ^{0}$ is greater.

Step 4: Determine the difference in velocities of the two particles

$7.5 \times 10^{6} \frac{m}{s}$ (b)

The speed of the spin $\frac{3}{2} Σ^{* 0}$ - baryon is

$\begin{matrix} v^{*} = 0.814 c \\ = 0.814 \times 3 \times 10^{8} \frac{m}{s} \\ = 2.442 \times 10^{8} \frac{m}{s} \end{matrix}$

The speed of the spin- $\frac{1}{2} Σ^{0}$ baryon is

$\begin{matrix} v = 0.839 c \\ = 0.839 \times 3 \times 10^{8} \frac{m}{s} \\ = 2.517 \times 10^{8} \frac{m}{s} \end{matrix}$

The difference in their speeds is calculated as:

$\begin{matrix} v - v^{*} = (2.517 - 2.442) \times 10^{8} \frac{m}{s} \\ = 0.075 \times 10^{8} \frac{m}{s} \\ = 7.5 \times 10^{6} \frac{m}{s} \end{matrix}$

Thus, the difference is. $7.5 \times 10^{6} \frac{m}{s}$

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Fundamentals Of Physics

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Short Answer

Step by Step Solution

Step 1: Given data

Step 2:Determine the formula for the relativistic energy.

Step 3: Determinethe particle with greater velocity

Step 4: Determine the difference in velocities of the two particles

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Fundamentals Of Physics

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The spin- 32Σ*0baryon (see table in Problem 24) has a rest energy of (1385 MeVwith an intrinsic uncertainty ignored here); the spin12Σ0- baryon has a rest energy of.1192.5 MeVIf each of these particles has a kinetic energy of 1000 MeV, (a) which is moving faster and (b) by how much?

Step by Step Solution

Step 1: Given data

Step 2:Determine the formula for the relativistic energy.

Step 3: Determinethe particle with greater velocity

Step 4: Determine the difference in velocities of the two particles

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