Question

The spin- $\frac{\mathbf{3}}{\mathbf{2}}{\mathit{\Sigma }}^{\mathbf{*}\mathbf{0}}$baryon(see table in Problem 24) has a rest energy of ($\mathbf{1385}\mathit{M}\mathit{e}\mathit{V}$with an intrinsic uncertainty ignored here); the spin$\frac{1}{2}{\mathbf{\Sigma }}^0$- baryonhas a rest energyof.$\mathbf{1192}\mathbf{.}\mathbf{5}\mathbf{MeV}$If each of these particles has a kinetic energy of $\mathbf{1000}\mathbf{MeV}$,

(a) which is moving faster and

(b) by how much?

Short answer

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

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

Step-by-step solution

Given data

Rest energy of thespin-$\frac{3}{2}{\Sigma }^{\ast 0}$ 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:

$\mathbf{K}\mathbf{=}\mathbf{(}\frac{1}{\sqrt{\mathbf{1}\mathbf{-}\frac{{\mathbf{v}}^2}{{\mathbf{c}}^2}}}\mathbf{-}\mathbf{1}\mathbf{)}{\mathbf{E}}_0$ ..... (I)

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

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

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)

$K=(\frac{1}{\sqrt{1-\frac{v^{\ast 2}}{c^2}}}-1)E_0^{\ast }$

Substitute the values and solve as:

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

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

$K=(\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}-1)E_0$

Substitute the values and solve as:

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

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.

Determine the difference in velocities of the two particles

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

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

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

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

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

The difference in their speeds is calculated as:

$\begin{array}{c}v-v^{\ast }=(2.517-2.442)\times {10}^8\frac{\text{m}}{\text{s}}\\ =0.075\times {10}^8\frac{\text{m}}{\text{s}}\\ =7.5\times {10}^6\frac{\text{m}}{\text{s}}\end{array}$

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