Question

Question: At a particular instant an electron is traveling with speed (6000 m/s) . (a) What is the kinetic energy of the electron? (b) If a proton were traveling at the same speed (6000 m/s), what would be the kinetic energy of the proton?

Short answer

Answer

1. The kinetic energy of the electron is$1.64\times {10}^{-23}\text{J}$ .
2. The kinetic energy of the proton is

$3.01\times {10}^{-20}\text{J}$

Step-by-step solution

Identification of the given data

The given data can be listed below as,

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

Significance of kinetic energy of different particles

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.

(a) Determination the kinetic energy of an electron

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 $9.1\times {10}^{-31}\text{kg}$.

Substitute all the known values in the above equation.

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

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

(b) Determination the kinetic energy of a proton

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 $1.67\times {10}^{-27}\text{kg}$.

Substitute all the known values in the above equation.

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

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