Question

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?

Short answer

1. Time is required to increase the electron's speed from 0.93c to 0.99c is$\mathbf{8}\mathbf{.19}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{12}}\mathbf{\text{\hspace{0.33em}s}}$
2. The electron goes in this time is $\mathbf{7}\mathbf{.62}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{12}}\mathbf{\text{\hspace{0.33em}m}}$

Step-by-step solution

Identification of the given data

• 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$m=9.1\times {10}^{-31}\text{\hspace{0.33em}kg}$

(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,

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

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 $\mathbf{8}\mathbf{.19}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{12}}\mathbf{\text{\hspace{0.33em}s}}$

(b) Determination of the distance traveled by the electron in this time

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 $\mathbf{7}\mathbf{.62}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{12}}\mathbf{\text{\hspace{0.33em}m}}$