91Ó°ÊÓ

The speed of proton is $v=0.8c$

The total energy of a particle moving at high speed consists of kinetic energy and rest energy. The rest energy is the energy same as produced due to complete conversion of mass of particle into energy. The momentum of moving particle is relativistic momentum.

The momentum of proton is given as:

$P=\frac{mv}{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}$

Here is the mass of proton and its value is $1.67×{10}^{-27}\text{kg}$, c is the speed of light and its value is $3×{10}^8\text{m}/\text{s}$

Substitute all the values in the equation.

$$\begin{gathered} P=\frac{\left(1.67×{10}^{-27}\text{kg}\right)\left(0.8c\right)\left(\frac{3×{10}^8\text{m}/\text{s}}{c}\right)}{\sqrt{1-{\left(\frac{0.8c}{c}\right)}^2}} \\ P=6.68×{10}^{-19}\text{kg}·\text{m}/\text{s} \end{gathered}$$

The energy of proton is given as:

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

Substitute all the values in the equation.

$$\begin{gathered} E=\frac{\left(1.67×{10}^{-27}\text{kg}\right){\left(3×{10}^8\text{m}/\text{s}\right)}^2}{\sqrt{1-{\left(\frac{0.8c}{c}\right)}^2}} \\ E=2.505×{10}^{-10}\text{J} \end{gathered}$$

The energy of proton is given as:

$K=E-m{c}^2$

Substitute all the values in the equation.

$$\begin{gathered} K=2.505×{10}^{-10}\text{J}-\left(1.67×{10}^{-27}\text{kg}\right){\left(3×{10}^8\text{m}/\text{s}\right)}^2 \\ K=2.505×{10}^{-10}\text{J}-1.503×{10}^{-10}\text{J} \\ K=1.002×{10}^{-10}\text{J} \end{gathered}$$

Therefore, the momentum, energy and kinetic energy of proton are $6.68×{10}^{-19}\text{kg}·\text{m}/\text{s}$, $2.505×{10}^{-10}\text{J}$ and $1.002×{10}^{-10}\text{J}$.