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### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

# The spin- $\frac{\mathbf{3}}{\mathbf{2}}{{\mathbit{\Sigma }}}^{\mathbf{*}\mathbf{0}}$baryon (see table in Problem 24) has a rest energy of (${\mathbf{1385}}{\mathbf{\text{\hspace{0.17em}}}}{\mathbit{M}}{\mathbit{e}}{\mathbit{V}}$with an intrinsic uncertainty ignored here); the spin$\frac{1}{2}{\mathbf{\Sigma }}^{0}$- baryon has a rest energy of.${\mathbf{1192}}{\mathbf{.}}{\mathbf{5}}{\text{\hspace{0.17em}}}{\mathbf{MeV}}$If each of these particles has a kinetic energy of ${\mathbf{1000}}{\mathbf{\text{\hspace{0.17em}}}}{\mathbf{MeV}}$, (a) which is moving faster and (b) by how much?

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

(b) It is moving faster by.$7.5×{10}^{6}\text{\hspace{0.17em}}\frac{\text{m}}{\text{s}}$

See the step by step solution

## Step 1: Given data

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

${E}_{0}^{\ast }=1385\text{\hspace{0.17em}MeV}$

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

${E}_{0}=1192.5\text{\hspace{0.17em}MeV}$

Kinetic energy of the two particles

$K=1000\text{\hspace{0.17em}MeV}$

## 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:

${\mathbf{K}}{\mathbf{=}}\mathbf{\left(}\frac{1}{\sqrt{\mathbf{1}\mathbf{-}\frac{{\mathbf{v}}^{2}}{{\mathbf{c}}^{2}}}}\mathbf{-}\mathbf{1}\mathbf{\right)}{{\mathbf{E}}}_{0}$ ..... (I)

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

${\mathbf{c}}{\mathbf{=}}{\mathbf{3}}{\mathbf{×}}{{\mathbf{10}}}^{8}{\text{\hspace{0.17em}}}\frac{m}{s}$

## Step 3: Determinethe particle with greater velocity

(a)

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

$K=\left(\frac{1}{\sqrt{1-\frac{{v}^{\ast 2}}{{c}^{2}}}}-1\right){E}_{0}^{\ast }$

Substitute the values and solve as:

$\begin{array}{l}1000\text{\hspace{0.17em}MeV}=\left(\frac{1}{\sqrt{1-\frac{{v}^{\ast 2}}{{c}^{2}}}}-1\right)×1385\text{\hspace{0.17em}MeV}\\ \frac{1}{\sqrt{1-\frac{{v}^{\ast 2}}{{c}^{2}}}}=1.722\\ \frac{{v}^{\ast 2}}{{c}^{2}}=0.663\\ {v}^{\ast }=0.814c\end{array}$

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

$K=\left(\frac{1}{\sqrt{1-\frac{{v}^{2}}{{c}^{2}}}}-1\right){E}_{0}$

Substitute the values and solve as:

$\begin{array}{l}1000\text{\hspace{0.17em}MeV}=\left(\frac{1}{\sqrt{1-\frac{{v}^{2}}{{c}^{2}}}}-1\right)×1192.5\text{\hspace{0.17em}MeV}\\ \frac{1}{\sqrt{1-\frac{{v}^{2}}{{c}^{2}}}}=1.839\\ \frac{{v}^{2}}{{c}^{2}}=0.704\\ v=0.839c\end{array}$

Compare the two velocities:

$\begin{array}{c}{v}^{\ast }=0.814c\\ v=0.839c\\ v>{v}^{\ast }\end{array}$

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

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

$7.5×{10}^{6}\text{\hspace{0.17em}}\frac{\text{m}}{\text{s}}$(b)

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

$\begin{array}{c}{v}^{\ast }=0.814c\\ =0.814×3×{10}^{8}\text{\hspace{0.17em}}\frac{\text{m}}{\text{s}}\\ =2.442×{10}^{8}\text{\hspace{0.17em}}\frac{\text{m}}{\text{s}}\end{array}$

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

$\begin{array}{c}v=0.839c\\ =0.839×3×{10}^{8}\text{\hspace{0.17em}}\frac{\text{m}}{\text{s}}\\ =2.517×{10}^{8}\text{\hspace{0.17em}}\frac{\text{m}}{\text{s}}\end{array}$

The difference in their speeds is calculated as:

$\begin{array}{c}v-{v}^{\ast }=\left(2.517-2.442\right)×{10}^{8}\text{\hspace{0.17em}}\frac{\text{m}}{\text{s}}\\ =0.075×{10}^{8}\text{\hspace{0.17em}}\frac{\text{m}}{\text{s}}\\ =7.5×{10}^{6}\text{\hspace{0.17em}}\frac{\text{m}}{\text{s}}\end{array}$

Thus, the difference is.$7.5×{10}^{6}\text{\hspace{0.17em}}\frac{\text{m}}{\text{s}}$