91Ó°ÊÓ

The given data can be listed below as,

The speed of an electron and a proton is, $v_e=v_p=6000\text{m/s}$.

In this question, the concept of the kinetic energy of two different charged particles will be determined by using the mass of differently charged particles by considering the speed as the same.

The relation of kinetic energy of an electron is expressed as,

${\left(KE\right)}_e=\frac{1}{2}{m}_e{v_e}^2$

Here, m_e is the mass of an electron whose value is $9.1×{10}^{-31}\text{kg}$.

Substitute all the known values in the above equation.

$$\begin{gathered} {\left(KE\right)}_e=\frac{1}{2}\left(9.1×{10}^{-31}kg\right){\left(6000\text{m/s}\right)}^2 \\ ≈1.64×{10}^{-23}\text{kg}·{\text{m}}^{\text{2}}{\text{/s}}^{\text{2}} \\ ≈\left(1.64×{10}^{-23}\text{kg}·{\text{m}}^{\text{2}}{\text{/s}}^{\text{2}}\right)×\left(\frac{1\text{J}}{1\text{kg}·{\text{m}}^{\text{2}}{\text{/s}}^{\text{2}}}\right) \\ ≈1.64×{10}^{-23}\text{J} \end{gathered}$$

Thus, the kinetic energy of the electron is $1.64×{10}^{-23}\text{J}$.

The relation of kinetic energy of a proton is expressed as,

${\left(KE\right)}_p=\frac{1}{2}{m}_p{v_p}^2$

Here, m_p is the mass of a proton whose value is $1.67×{10}^{-27}\text{kg}$.

Substitute all the known values in the above equation.

$$\begin{gathered} {\left(KE\right)}_p=\frac{1}{2}\left(1.67×{10}^{-27}kg\right){\left(6000\text{m/s}\right)}^2 \\ ≈3.01×{10}^{-20}\text{kg}·{\text{m}}^{\text{2}}{\text{/s}}^{\text{2}} \\ ≈\left(3.01×{10}^{-20}\text{kg}·{\text{m}}^{\text{2}}{\text{/s}}^{\text{2}}\right)×\left(\frac{1\text{J}}{1\text{kg}·{\text{m}}^{\text{2}}{\text{/s}}^{\text{2}}}\right) \\ ≈3.01×{10}^{-20}\text{J} \end{gathered}$$

Thus, the kinetic energy of the proton is $3.01×{10}^{-20}\text{J}$.