Question

An electron is accelerated from rest through a potential of \(1.0 \cdot 10^{6} \mathrm{~V}\). What is its final speed?

Short answer

Answer: The final speed of the electron is \(1.87 \cdot 10^7 \ \mathrm{m/s}\).

Step-by-step solution

Define the given quantities

An electron has a charge \(q = -e = -1.6 \cdot 10^{-19} \mathrm{C}\) (Coulombs) and a mass \(m = 9.11 \cdot 10^{-31} \mathrm{kg}\). The potential difference through which the electron is accelerated is \(V = 1.0 \cdot 10^{6} \mathrm{V}\).

Calculate the change in electric potential energy

The change in electric potential energy, \(\Delta U\), is given by the product of the charge of the electron (\(q\)) and the potential difference (\(V\)): \(\Delta U = q \cdot V = (-1.6 \cdot 10^{-19} \mathrm{C})(1.0 \cdot 10^{6}\mathrm{~V}) = -1.6 \cdot 10^{-13} \mathrm{J}\) (Joules)

Apply the conservation of energy principle

Since the electron starts from rest, we can apply the conservation of energy principle, which states that the change in electric potential energy is equal to the change in kinetic energy: \(\Delta K = \Delta U\) The initial kinetic energy, \(K_{initial}\), is zero, and the final kinetic energy, \(K_{final}\), can be represented as: \(K_{final} = \frac{1}{2}mv^2\)

Calculate the final kinetic energy

We know that \(\Delta K = K_{final} - K_{initial}\), which simplifies to: \(K_{final} = \Delta U\) Plugging in the value of \(\Delta U\) calculated in Step 2, we get: \(K_{final} = -1.6 \cdot 10^{-13} \mathrm{J}\)

Calculate the final speed of the electron

Now that we have the final kinetic energy, we can solve for the final speed, \(v\), of the electron: \(\frac{1}{2}mv^2 = K_{final}\) Solving for \(v\), we get: \(v = \sqrt{\frac{2 \cdot K_{final}}{m}}\) Plugging in the values for \(K_{final}\) and \(m\), we get: \(v = \sqrt{\frac{2 \cdot (-1.6 \cdot 10^{-13} \mathrm{J})}{(9.11 \cdot 10^{-31} \mathrm{kg})}} \approx 1.87 \cdot 10^7 \ \mathrm{m/s}\) (meters per second) The final speed of the electron is \(1.87 \cdot 10^7 \ \mathrm{m/s}\).