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 ${\mathbf{\text{2×10}}}^{\mathbf{\text{-12}}}$ 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" $\left|\overrightarrow{\nu }\right|\mathbf{/}\mathit{c}$ increases from0.93c to 0.99c .) (b) Approximately how far does the electron go in this time? What is approximate about your result?

• Force acting on the electron is$F={\text{2×10}}^{\text{-12}}\text{\hspace{0.33em}N}$
• The initial velocity is 0.93c
• The final velocity is 0.99c
• The mass of the electron is$m=9.1\times {10}^{-31}\text{\hspace{0.33em}kg}$

The time is calculated by using Impulse formula.

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

$\begin{array}{c}F\times t=mv-mu\cdot \cdot \cdot \cdot \cdot \cdot \left(1\right)\\ \text{where,}\\ F=\text{Force\hspace{0.33em}acting\hspace{0.33em}on\hspace{0.33em}the\hspace{0.33em}electron}\Rightarrow 2\times {10}^{-12}\text{\hspace{0.33em}N}\\ \text{t\hspace{0.33em}=Time\hspace{0.33em}required}\\ m=\text{mass\hspace{0.33em}of\hspace{0.33em}the\hspace{0.33em}proton}\Rightarrow 9.1\times {10}^{-31}\mathrm{kg}\\ v=\text{final\hspace{0.33em}velocity}\Rightarrow 0.99\text{c\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\left(\text{where},c=\text{speed\hspace{0.33em}of\hspace{0.33em}the\hspace{0.33em}light}\Rightarrow 3\times {10}^8\text{\hspace{0.33em}m/s}\right)\\ v=0.99\times 3\times {10}^8\\ =2.97\times {10}^8\text{\hspace{0.33em}m/s}\\ u=\text{initial\hspace{0.33em}velocity}\Rightarrow 0.93\text{c\hspace{0.33em}}\left(\text{where},c=\text{speed\hspace{0.33em}of\hspace{0.33em}the\hspace{0.33em}light}\Rightarrow 3\times {10}^8\text{\hspace{0.33em}m/s}\right)\\ u=0.994\times 3\times {10}^8\\ =2.79\times {10}^8\text{\hspace{0.33em}m/s}\end{array}$

From Equation (1),

$\begin{array}{c}t=\frac{mv-mu}{F}\\ =\frac{\left(9.1\times {10}^{-31}\mathrm{kg}\times 2.97\times {10}^8\text{\hspace{0.33em}m/s}\right)\text{\hspace{0.33em}-}\left(9.1\times {10}^{-31}\mathrm{kg}\times 2.79\times {10}^8\text{\hspace{0.33em}m/s}\right)}{2\times {10}^{-12}\text{\hspace{0.33em}N}}\\ =\frac{\left(9.1\times {10}^{-31}\times 2.97\times {10}^8\right)\text{-}\left(9.1\times {10}^{-31}\times 2.79\times {10}^8\right)}{2\times {10}^{-12}}\left(\frac{1\mathrm{kg}\cdot 1\text{\hspace{0.33em}m\hspace{0.33em}}}{1\text{\hspace{0.33em}s}}-\frac{1\mathrm{kg}\cdot 1\text{\hspace{0.33em}m\hspace{0.33em}}}{1\text{\hspace{0.33em}s}}\right)/1\text{\hspace{0.33em}N}\\ =\frac{\left(9.1\times {10}^{-31}\times 2.97\times {10}^8\right)\text{-}\left(9.1\times {10}^{-31}\times 2.79\times {10}^8\right)}{2\times {10}^{-12}}\left(\frac{1\mathrm{kg}\cdot 1\text{\hspace{0.33em}m\hspace{0.33em}}}{1\text{\hspace{0.33em}s}\cdot 1\text{\hspace{0.33em}N}}\right)\\ =\frac{\left(9.1\times {10}^{-31}\times 2.97\times {10}^8\right)\text{-}\left(9.1\times {10}^{-31}\times 2.79\times {10}^8\right)}{2\times {10}^{-12}}\left(\frac{1\mathrm{kg}\cdot 1\text{\hspace{0.33em}m\hspace{0.33em}}\cdot 1{\text{\hspace{0.33em}s}}^2}{1\text{\hspace{0.33em}s}\cdot 1\mathrm{kg}\cdot 1\text{\hspace{0.33em}m}}\right)\\ =8.19\times {10}^{-12}\text{\hspace{0.33em}s}\end{array}$

Hence, Time is required to increase the electron's speed from 0.93c to 0.99c is $\mathbf{8}\mathbf{.19}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{12}}\mathbf{\text{\hspace{0.33em}s}}$

To find the distance,

$\begin{array}{c}{x}_f=x_i+v_{xi}t\\ =0+0.93\times 8.19\times {10}^{-12}\\ =7.62\times {10}^{-12}\text{\hspace{0.33em}m}\end{array}$

Hence, the electron goes in this time is $\mathbf{7}\mathbf{.62}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{12}}\mathbf{\text{\hspace{0.33em}m}}$