Chapter 2: Q47P (page 86)

SLAC, the Standford Linear Accelerator Center, located at Stanford University in Palo Alto California, accelerates electrons through a vacuum tube 2 mi long (it can be seen from an overpass of the Junipero Serra freeway that goes right over the accelerator). Electrons that are initially at rest are subjected to a continuous force of ${2×10}^{-12}$ newton along the entire length of 2 mi (1 mi is 1.6 km) and reach speeds very near the speed of light. (a) Determine how much time is required to increase the electron's speed from 0.93c to 0.99c. (that is the quantity role="math" localid="1657119037340" $| \vec{ν} | / c$ increases from0.93c to 0.99c .) (b) Approximately how far does the electron go in this time? What is approximate about your result?

Short Answer

Expert verified

Time is required to increase the electron's speed from 0.93c to 0.99c is $8 .19 \times 10^{- 12} s$
The electron goes in this time is $7 .62 \times 10^{- 12} m$

Step by step solution

Identification of the given data

Force acting on the electron is $F = {2×10}^{-12} N$
The initial velocity is 0.93c
The final velocity is 0.99c
The mass of the electron is $m = 9.1 \times 10^{- 31} kg$

(a) Determination of the time required to increase the electron's speed from 0.93c to 0.99c

The time is calculated by using Impulse formula.

Impulse is the equivalent to the force as the change in momentum,

$\begin{matrix} F \times t = m v - m u \cdot \cdot \cdot \cdot \cdot \cdot (1) \\ where, \\ F = Force acting on the electron \Rightarrow 2 \times 10^{- 12} N \\ t =Time required \\ m = mass of the proton \Rightarrow 9.1 \times 10^{- 31} kg \\ v = final velocity \Rightarrow 0.99 c (where, c = speed of the light \Rightarrow 3 \times 10^{8} m/s) \\ v = 0.99 \times 3 \times 10^{8} \\ = 2.97 \times 10^{8} m/s \\ u = initial velocity \Rightarrow 0.93 c (where, c = speed of the light \Rightarrow 3 \times 10^{8} m/s) \\ u = 0.994 \times 3 \times 10^{8} \\ = 2.79 \times 10^{8} m/s \end{matrix}$

From Equation (1),

$\begin{matrix} t = \frac{m v - m u}{F} \\ = \frac{(9 .1 \times 10^{- 31} kg \times 2 .97 \times 10^{8} m/s) - (9 .1 \times 10^{- 31} kg \times 2 .79 \times 10^{8} m/s)}{2 \times 10^{- 12} N} \\ = \frac{(9 .1 \times 10^{- 31} \times 2 .97 \times 10^{8}) - (9 .1 \times 10^{- 31} \times 2 .79 \times 10^{8})}{2 \times 10^{- 12}} (\frac{1 kg \cdot 1 m}{1 s} - \frac{1 kg \cdot 1 m}{1 s}) / 1 N \\ = \frac{(9 .1 \times 10^{- 31} \times 2 .97 \times 10^{8}) - (9 .1 \times 10^{- 31} \times 2 .79 \times 10^{8})}{2 \times 10^{- 12}} (\frac{1 kg \cdot 1 m}{1 s \cdot 1 N}) \\ = \frac{(9 .1 \times 10^{- 31} \times 2 .97 \times 10^{8}) - (9 .1 \times 10^{- 31} \times 2 .79 \times 10^{8})}{2 \times 10^{- 12}} (\frac{1 kg \cdot 1 m \cdot 1 s^{2}}{1 s \cdot 1 kg \cdot 1 m}) \\ = 8 .19 \times 10^{- 12} s \end{matrix}$