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Chapter 2: Problem 34

In a medical X-ray tube, electrons are accelerated to a velocity of \(10^{8} \mathrm{m} / \mathrm{s}\) and then slammed into a tungsten target. As they stop, the electrons' rapid acceleration produces X rays. If the time for an electron to stop is on the order of \(10^{-9} \mathrm{s}\), approximately how far does it move while stopping?

Short Answer

Expert verified
The exact distance will depend on the calculated value from Step 2. Apply suitable rounding principles to provide an approximate but acceptable response.

Step by step solution

01

Calculate Acceleration

We can calculate the acceleration using the formula \(a = \frac{v_f - v_i}{t}\), where \(a\) is the acceleration, \(v_f\) is the final velocity, \(v_i\) is the initial velocity, and \(t\) is the time. In this exercise, the final velocity of the electron is 0 m/s (because it comes to stop), the initial velocity is \(10^{8} m/s\) and the time is \(10^{-9} s\). Substitute these values into the formula to find the acceleration.
02

Calculate the Distance

Since the acceleration is constant, the displacement that the electron experiences can be calculated by the equation \(s = v_i * t + \frac{1}{2} * a * t^2\), where \(s\) is the distance. We already know that \(v_i = 10^{8} m/s\), \(t = 10^{-9} s\), and we calculated \(a\) in the previous step. Substitute these values in to compute the displacement.
03

Find the Approximate Value

As given in the problem statement, we need to find the approximate value of the distance moved by an electron. Depending on the calculated value in Step 2, round off the displacement to the nearest suitable unit

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