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 \(=9.11 \cdot 10^{-31} \mathrm{~kg}\) ) hits the screen of the tube?

We can calculate the work done on the electron using the equation: $$W = qV$$ where \(W\) is the work, \(q\) is the charge of the electron, and \(V\) is the voltage of the cathode ray tube. The charge of an electron is approximately \(1.6 \cdot 10^{-19} \mathrm{C}\), and the voltage of the tube is \(21.9 \mathrm{kV}\), which is equal to \(21,900 \mathrm{V}\). Plugging in these values, we get: $$W = (1.6 \cdot 10^{-19} \mathrm{C})(21,900 \mathrm{V})$$

Since the electron starts from rest, all of the work done on the electron goes into increasing its kinetic energy. Using the work-energy theorem, we can write: $$W = \Delta KE$$ where \(\Delta KE\) is the change in kinetic energy, which is equal to the final kinetic energy, since the electron starts from rest. Also, we know that the kinetic energy of an object is given by the equation: $$KE = \frac{1}{2}mv^2$$ where \(m\) is the mass of the object, in this case, the electron, and \(v\) is the final speed. We can express the final kinetic energy in terms of the final speed as: $$\Delta KE = \frac{1}{2}m(v^2 - 0) = \frac{1}{2}mv^2$$ Since \(W = \Delta KE\), we have: $$\frac{1}{2}mv^2 = (1.6 \cdot 10^{-19} \mathrm{C})(21,900 \mathrm{V})$$

Now, we can solve for the final speed, \(v\): $$v^2 = \frac{2(1.6 \cdot 10^{-19} \mathrm{C})(21,900 \mathrm{V})}{(9.11 \cdot 10^{-31} \mathrm{~kg})}$$ Solving for \(v\), we get: $$v = \sqrt{\frac{2(1.6 \cdot 10^{-19} \mathrm{C})(21,900 \mathrm{V})}{(9.11 \cdot 10^{-31} \mathrm{~kg})}}$$ Calculating the value for \(v\), we obtain: $$v \approx 1.825 \cdot 10^7 \mathrm{m/s}$$ Finally, we need to convert this speed to \(\mathrm{km/s}\), so we have: $$v \approx\frac{1.825 \cdot 10^7 \mathrm{m/s}}{(10^3 \mathrm{m/km})} = 18.25 \mathrm{km/s}$$ Thus, the electron hits the screen of the cathode ray tube with a speed of approximately \(18.25\ \mathrm{km/s}\).