91Ó°ÊÓ

The given data can be listed below,

The relationship of mass and energy is,${E^2}_{particle}-{\left(pc\right)}^2={\left(m{c}^2\right)}^2$

The energy of the particle is, $E_{particle}=\frac{m{c}^2}{\sqrt{1-{\left(v/c\right)}^2}}$.

The momentum of the particle is,$p=\frac{mv}{\sqrt{1-{\left(v/c\right)}^2}}$

Mass is energy acquired by matter that makes it react to the attempted addition of more energy. mass energy relation allows us to quantify the energy that can be produced from a given mass and vice versa.

The Mass-Energy relation is given by,

${E^2}_{particle}-{\left(pc\right)}^2={\left(m{c}^2\right)}^2$

Here, p is the momentum of the particle, m is the mas of the particle, c is the speed of light whose value is $3×{10}^{8}m/s^2$.

Substitute the value in the left-hand side of the above equation.

$$\begin{gathered} {E^2}_{particle}-{\left(pc\right)}^2=\left(\frac{m{c}^2}{\sqrt{1-{\left(v/c\right)}^2}}\right)-{\left(\frac{mvc}{\sqrt{1-{\left(v/c\right)}^2}}\right)}^2 \\ =\frac{m^2{c}^4}{1-v^2/c^2}-\frac{m^2{v}^2{c}^2}{1-v^2/c^2} \\ =\frac{m^2{c}^2}{1-v^2/c^2}\left(c^2-v^2\right) \end{gathered}$$

On further solving the above equation, we get:

$$\begin{gathered} {E^2}_{particle}-{\left(pc\right)}^2=\frac{m^2{c}^4}{c^2-v^2}\left(c^2-v^2\right) \\ =m^2{c}^4 \\ ={\left(m{c}^2\right)}^2 \\ LHS=RHS \end{gathered}$$

Hence, LHS = RHS. So, it is verified that mass energy equation holds true for given value of Energy and momentum