91Ó°ÊÓ

Rest energy of thespin-$\frac{3}{2}{\mathrm{Î\pounds }}^{∗0}$ baryon

$E_0^{∗}=1385\text{ M±ð³Õ}$

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

$E_0=1192.5\text{ M±ð³Õ}$

Kinetic energy of the two particles

$K=1000\text{ M±ð³Õ}$

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

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

(a)

Let the velocity of the spin-$\frac{3}{2}{\mathrm{Î\pounds }}^{∗0}$ 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:

$\begin{array}{l}1000\text{ M±ð³Õ}=(\frac{1}{\sqrt{1−\frac{v^{∗2}}{c^2}}}−1)×1385\text{ M±ð³Õ}\\ \frac{1}{\sqrt{1−\frac{v^{∗2}}{c^2}}}=1.722\\ \frac{v^{∗2}}{c^2}=0.663\\ v^{∗}=0.814c\end{array}$

Let the velocity of the spin-$\frac{1}{2}{\mathrm{Î\pounds }}^0$ 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:

$\begin{array}{l}1000\text{ M±ð³Õ}=(\frac{1}{\sqrt{1−\frac{v^2}{c^2}}}−1)×1192.5\text{ M±ð³Õ}\\ \frac{1}{\sqrt{1−\frac{v^2}{c^2}}}=1.839\\ \frac{v^2}{c^2}=0.704\\ v=0.839c\end{array}$

Compare the two velocities:

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

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

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

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

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

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

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

The difference in their speeds is calculated as:

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

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