Question

Suppose that an electron inside a cathode ray tube starts from rest and is accelerated by the tube's voltage of \(21.9 \mathrm{kV}\). What is the speed (in \(\mathrm{km} / \mathrm{s}\) ) with which the electron (mass \(\left.=9.11 \cdot 10^{-31} \mathrm{~kg}\right)\) hits the screen of the tube?

Short answer

Given the mass of an electron is \(9.11 \times 10^{-31} \mathrm{kg}\). Answer: To find the speed of the electron, follow the given steps and calculate the value of \(v\): $$v = \sqrt{\frac{2((-1.60 \times 10^{-19} \mathrm{C})(21.9 \times 10^3 \mathrm{V}))}{9.11 \times 10^{-31} \mathrm{kg}}}$$ Then, convert the final velocity to km/s using: $$v_{km/s} = v_{m/s} \frac{1}{1000}$$

Step-by-step solution

Identify known quantities

The known quantities in this problem are: - Electron mass: \(m = 9.11 \times 10^{-31} \mathrm{kg}\) - Voltage across the tube: \(V = 21.9 \times 10^3 \mathrm{V}\)

Use the work-energy theorem

The work-energy theorem states that the work done on an object is equal to the change in its kinetic energy. In this case, the work done on the electron by the electric force is equal to the change in the electron's kinetic energy: $$W = \Delta KE$$

Relate work, charge, and voltage

The work done on an object by an electric force is equal to the product of the charge of the object and the voltage through which the charge is moved: $$W = qV$$ For an electron, the charge is \(q = -e = -1.60 \times 10^{-19} \mathrm{C}\). Therefore, the work done on the electron is $$W = (-1.60 \times 10^{-19} \mathrm{C})(21.9 \times 10^3 \mathrm{V})$$

Calculate the change in kinetic energy

From steps 2 and 3, we can now calculate the change in the electron's kinetic energy: $$\Delta KE = (-1.60 \times 10^{-19} \mathrm{C})(21.9 \times 10^3 \mathrm{V})$$

Determine the final velocity

The kinetic energy of an object depends on its mass and velocity as follows: $$KE = \frac{1}{2} m v^2$$ Since the electron starts from rest, the initial kinetic energy is zero, and the change in kinetic energy is equal to the final kinetic energy: $$\Delta KE = \frac{1}{2} m v^2$$ Now, we can solve for the final velocity \(v\) using the equation above and the values calculated in previous steps: $$v^2 = \frac{2 \Delta KE}{m}$$ $$v = \sqrt{\frac{2((-1.60 \times 10^{-19} \mathrm{C})(21.9 \times 10^3 \mathrm{V}))}{9.11 \times 10^{-31} \mathrm{kg}}}$$ Calculate the value of \(v\).

Convert the final velocity to km/s

After calculating the speed of the electron in meters per second, take the value of \(v\) and convert it to kilometers per second: $$v_{km/s} = v_{m/s} \frac{1}{1000}$$ This will give the final speed of the electron in km/s.