Question

Question: An electron passes through a region in which there is an electric field, and whiles it is in the region its kinetic energy decreases by $\mathbf{4}\mathbf{.}\mathbf{5}\times {\mathbf{10}}^{\mathbf{-}\mathbf{17}}\mathbf{J}$. Initially the kinetic energy of the electron was$\mathbf{4}\mathbf{.}\mathbf{5}\times {\mathbf{10}}^{\mathbf{-}\mathbf{17}}\mathbf{J}$ . What is the final speed of the electron? (You can use the approximate (nonrelativistic) equation here.)

Short answer

Answer

The final speed of the electron is $3.33\times {10}^6\text{m/s}$.

Step-by-step solution

Identification of the given data

The given data can be listed below as,

• The kinetic energy of an electron decreases by, $\Delta KE=4\times {10}^{-17}\text{J}$.
• The initial kinetic energy of the electron is, $KE=4\times {10}^{-17}\text{J}$.

 Step 2: Significance of kinetic energy

The term ‘kinetic energy’ exists when an object is in motion condition means moving with a specific speed. The kinetic energy of an object varies/ changes directly square to the object's speed.

Determination the initial speed of an electron

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

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

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

Substitute all the known values in the above equation.

$$\begin{gathered} \left(4.5\times {10}^{-17}\text{J}\right)=\frac{1}{2}\left(9.1\times {10}^{-31}kg\right){v_i}^2 \\ {v_i}^2=\left[\frac{2\left(4.5\times {10}^{-17}\text{J}\right)}{\left(9.1\times {10}^{-31}kg\right)}\right] \\ \approx \left(9.8901\times {10}^{13}\text{J/kg}\right)\left(\frac{1{\text{m}}^{\text{2}}{\text{/s}}^{\text{2}}}{1\text{J/kg}}\right) \\ {v}_i\approx 9.95\times {10}^6\text{m/s} \end{gathered}$$

Determination the final speed of an electron

The relation of kinetic energy decrement of the electron is expressed as,

$\Delta KE=\frac{1}{2}{m}_e\left({v_i}^2-{v_f}^2\right)$

Here, is the final speed of the electron.

Substitute all the known values in the above equation.

$$\begin{gathered} \left(4\times {10}^{-17}\text{J}\right)=\frac{1}{2}\left(9.1\times {10}^{-31}kg\right)\left[{\left(9.95\times {10}^6\text{m/s}\right)}^2-{v_f}^2\right] \\ \left[{\left(9.95\times {10}^6\text{m/s}\right)}^2-{v_f}^2\right]=\left[\frac{2\left(4\times {10}^{-17}\text{J}\right)}{\left(9.1\times {10}^{-31}kg\right)}\right] \\ \approx \left(8.7912\times {10}^{13}\text{J/kg}\right)\left(\frac{1{\text{m}}^{\text{2}}{\text{/s}}^{\text{2}}}{1\text{J/kg}}\right) \\ {v}_f\approx 3.33\times {10}^6\text{m/s} \end{gathered}$$

Thus, the final speed of the electron is $3.33\times {10}^6\text{m/s}$.