Question

An electron is accelerated from rest through a potential difference of \(370 .\) V. What is its final speed?

Short answer

Answer: The final speed of the electron after it's accelerated through a potential difference of 370 V is approximately \(1.14\times10^7\ \text{m/s}\).

Step-by-step solution

Understand the given information

An electron is accelerated from rest through a potential difference of 370 V. The electron is initially at rest, and we have to find the final speed.

Determine the charge and mass of an electron

In order to solve this problem, we need to know the charge (e) and mass (m) of an electron. The charge of an electron is given by \(e=1.6\times10^{-19}\ \text{C}\), and the mass of an electron is \(m=9.11\times10^{-31}\ \text{kg}\).

Calculate the work done on the electron

The work done on the electron is equal to the product of its charge (e) and the potential difference (V): Work = e × V Work = \((1.6\times10^{-19}\ \text{C}) (370 \ \text{V})\) Work = \(5.92\times10^{-17}\ \text{J}\)

Determine the final kinetic energy of the electron

As the electron is initially at rest, all the work done on it will be converted into kinetic energy. So, the final kinetic energy of the electron will be equal to the work done on it: Final Kinetic Energy = Work Final Kinetic Energy = \(5.92\times10^{-17}\ \text{J}\)

Use the kinetic energy formula to find the final speed

We know that the kinetic energy of an object is given by the formula \(\frac{1}{2}mv^2\), where m is the mass of the object and v is its speed. In our case, the final kinetic energy of the electron is \(5.92\times10^{-17}\ \text{J}\) and its mass is \(9.11\times10^{-31}\ \text{kg}\). We can rearrange the formula to find the final speed of the electron: \(v^2 = \frac{2\times \text{Final Kinetic Energy}}{m}\) \(v^2= \frac{2 \times (5.92\times10^{-17}\ \text{J})}{(9.11\times10^{-31}\ \text{kg})}\) \(v^2= \frac{1.184\times10^{-16}\ \text{J}}{(9.11\times10^{-31}\ \text{kg})}\) \(v^2= 1.3\times10^{14}\) \(v= \sqrt{1.3\times10^{14}}\) \(v= 1.14\times10^7\ \text{m/s}\)

Answer

The final speed of the electron after it's accelerated through a potential difference of 370 V is approximately \(1.14\times10^7\ \text{m/s}\).