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

Short answer

Answer: The electron hits the screen of the cathode ray tube with a speed of approximately 18.25 km/s.

Step-by-step solution

Calculate the work done on the electron by the electric field

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: $$W = (1.6 \cdot 10^{-19} \mathrm{C})(21,900 \mathrm{V})$$

Use the work-energy theorem

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: $$\frac{1}{2}mv^2 = (1.6 \cdot 10^{-19} \mathrm{C})(21,900 \mathrm{V})$$

Calculate the final speed of the electron

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 \(18.25\ \mathrm{km/s}\).

Key concepts

Electron Acceleration

Inside a cathode ray tube, electrons are accelerated from rest by an applied voltage. As these electrons gain speed, they also gain energy, allowing them to move rapidly within the tube. This might sound complex, but it’s similar to rolling a ball down a hill—gravity accelerates the ball. In this case, it's the electric field from the voltage that propels the electrons.The fundamental idea is that a voltage difference, like the 21.9 kV in the exercise, creates an electric field. This field pulls the negatively charged electrons through the tube at increasing speeds. The amount of work done on the electron can be calculated using the equation:- Work (\( W \)) is given by the product of the electron's charge (\( q \)) and the voltage (\( V \)).Given the charge of an electron is \( 1.6 \times 10^{-19} \mathrm{C} \), we determine the work as:\[ W = qV = (1.6 \times 10^{-19} \mathrm{C})(21,900 \mathrm{V}) \] Understanding this work is essential to further discussions on electron motion in the tube.

Work-Energy Theorem

The concept of energy transition is beautifully explained by the work-energy theorem. This theorem forms a bridge between the work done by forces and the energy arriving in kinetic form.Let's break it down. When the electric field (from the applied voltage) does work on the electron, it transforms the energy, increasing the electron's kinetic energy directly. This means all the work done becomes kinetic energy, as the electron starts from rest:\[ W = \Delta KE \] Here, \( \Delta KE \) is the change in kinetic energy, which equals the final kinetic energy because the electron was initially at rest. Replacement of kinetic energy by its formula gives:- For an electron of mass \( m \), and speed \( v \), kinetic energy \( KE \) is: \[ KE = \frac{1}{2} mv^2 \] Since the electron starts at rest (initial kinetic energy, \( KE_i = 0 \)), the entire work done goes into its final kinetic energy:\[ W = \frac{1}{2} m v^2 \] This relationship is vital in understanding how the energy fueling the motion of our electron is derived from the electromagnetic forces acting on it.

Kinetic Energy Calculation

Once we've examined acceleration and energy transfer, it's time to calculate the speed of our electron. Understanding kinetic energy equations allows us to find its velocity when it hits the screen.Using the work-energy equivalence \( W = \frac{1}{2} mv^2 \) from the electric acceleration, we want to solve for \( v \), the final speed of the electron:1. Substitute known values: - Work \( W = 3.504 \times 10^{-15} \mathrm{J} \) (as calculated), - Mass \( m = 9.11 \times 10^{-31} \mathrm{~kg} \).2. Rearranging the formula gives:\[ v^2 = \frac{2W}{m} \] 3. Substitute and solve for \( v \):\[ v = \sqrt{\frac{2(3.504 \times 10^{-15} \mathrm{J})}{9.11 \times 10^{-31} \mathrm{~kg}}} \] Carrying out the calculus yields a velocity of approximately \( 1.825 \times 10^7 \mathrm{m/s} \), converting this to \( \mathrm{km/s} \) gives:\[ v \approx 18.25 \mathrm{km/s} \] This calculation reflects how energy conservation principles dictate the speed of particles affected by electric fields, giving us insight into how swiftly they can impact screens within cathode ray tubes.