Question

Question: If the kinetic energy of an electron is $\mathbf{4}\mathbf{.}\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{18}}\mathbf{J}$, what is the speed of the electron? You can use the approximate (non relativistic) equation here.

Short answer

Answer

The speed of the electron is $3.11\times {10}^6\text{m/s}$.

Step-by-step solution

Identification of the given data  

The given data can be listed below as,

• The kinetic energy of an electron is, $KE=4.4\times {10}^{-18}\text{J}$.

Significance of kinetic energy

In this problem, the concept of a non-relativistic equation of kinetic energy will be used in order to estimate an electron's speed. The relation between the electron’s kinetic energy and electron mass is direct linear.

Determination the speed of an electron

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

$KE=\frac{1}{2}{m}_e{v}^2$

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

Substitute all the known values in the above equation.

$$\begin{gathered} \left(4.4\times {10}^{-18}\text{J}\right)=\frac{1}{2}\left(9.1\times {10}^{-31}kg\right)v^2 \\ {v}^2=\left[\frac{2\left(4.4\times {10}^{-18}\text{J}\right)}{\left(9.1\times {10}^{-31}kg\right)}\right] \\ \approx \left(9.67032\times {10}^{12}\text{J/kg}\right)\left(\frac{1{\text{m}}^{\text{2}}{\text{/s}}^{\text{2}}}{1\text{J/kg}}\right) \\ v\approx 3.11\times {10}^6\text{m/s} \end{gathered}$$

Thus, the speed of the electron is $3.11\times {10}^6\text{m/s}$.