91Ó°ÊÓ

Kinetic energy is the difference between total energy and rest energy.

The total energyE of a particle is

$$\begin{gathered} E=K+m{c}^2 \\ =\frac{mc}{\sqrt{1-v^2/c^2}} \\ =\mathrm{γ}m{c}^2 \end{gathered}$$

Where,K is the relativistic kinetic energy. The expression ofK is:

$K=\frac{mc}{\sqrt{1-v^2/c^2}}-m{c}^2$

$\mathrm{γ}$is the Lorentz factor. The expression of$\mathrm{γ}$ is:

$\mathrm{γ}=\frac{1}{\sqrt{1-v^2/c^2}}$

The energy$\mathit{m}{\mathit{c}}^{\mathbf{2}}$ associated with rest massm rather than motion is known as rest energy of the particle.

According to the conservation of mass-energy, the sum of the kinetic energies of the protons before collision is equal to the total energies of the$n_0$ particle and the two protons after collision.

So,$2M{c}^2+mc=2\left(\mathrm{γ}M{c}^2\right)$

$$\begin{gathered} \mathrm{γ}=1+\frac{m}{2M} \\ =1+\frac{9.75×{10}^{-28}}{1.67×{10}^{-27}} \\ =1.292 \end{gathered}$$

So, initial speed of proton is:

$$\begin{gathered} \frac{1}{\sqrt{1-v^2/c^2}}=\frac{1}{1.292} \\ \frac{v}{c}=\sqrt{1-\frac{1}{{\left(1.292\right)}^2}} \\ v=0.6331c \end{gathered}$$

Hence the initial speed of protons is 0.6331c.

$$\begin{gathered} K=\left(\mathrm{γ}-1\right)M{c}^2 \\ =\left(1.292-1\right)\left(1.67×{10}^{-27}\right)\left(3.0×{10}^8\right)\left(\frac{1}{1.602×{10}^{-13}}\right) \\ =274MeV \end{gathered}$$The kinetic energy is:

Hence, kinetic energy of each proton is 274 MeV.

The rest energy of$n^0$ is:

$$\begin{gathered} E=m{c}^2 \\ =\left(9.75×{10}^{-28}\right)\left(3.0×{10}^8\right)\left(\frac{1}{1.602×{10}^{-13}}\right) \\ =548\mathrm{MeV} \end{gathered}$$

Hence, the rest energy ofrole="math" localid="1664107189723" $n^0$ is 548 MeV and the sum of the initial kinetic energies of the protons is equal to the rest energy of $n^0$.