The velocity \(v\) of electrons from a high energy accelerator is very near the velocity \(c\) of light. Given the voltage \(V\) of the accelerator, we often want to calculate the ratio \(v / c .\) The relativistic formula for this calculation is (approximately, for \(V \gg 1\) ) $$\frac{v}{c}=\sqrt{1-\left(\frac{0.511}{V}\right)^{2}}, \quad V=\text { number of million volts. }$$ Use two terms of the binomial series (13.5) to find 1 \(-v / c\) in terms of \(V\). Use your result to find \(1-v / c\) for the following values of \(V\). Caution: \(V=\) the number of million volts. (a) \(\quad V=100\) million volts (b) \(\quad V=500\) million volts (c) \(\quad V=25,000\) million volts (d) \(\quad V=100\) gigavolts \(\left(100 \times 10^{9} \text { volts }=10^{5}\) million volts) \right.

Short Answer

Expert verified
For (a) 0.00001305605, (b) 0.000000522241, (c) 0.000000000208896, (d) 0.00000000001305605.

Step by step solution

01

Use the Binomial Series Approximation

To approximate the expression \( \frac{v}{c} = \sqrt{1-\frac{(0.511)^2}{V^2}} \), we use the binomial series approximation for \( (1 - x)^a \) when \( x \) is very small. The binomial series for \( (1 - x)^{1/2} \) is approximately \( 1 - \frac{x}{2} \). In our case, \( x = \frac{(0.511)^2}{V^2} \). So, \( \frac{v}{c} \approx 1 - \frac{(0.511)^2}{2V^2} \).
02

Calculate 1 - \( \frac{v}{c} \)

Now we rearrange the approximation from Step 1 to find \( 1 - \frac{v}{c} \): \[ 1-\frac{v}{c} \approx \frac{(0.511)^2}{2V^2} = \frac{0.261121}{2V^2} = \frac{0.1305605}{V^2} \]
03

Compute for \( V = 100 \) million volts

Substitute \( V = 100 \): \[ 1-\frac{v}{c} = \frac{0.1305605}{(100)^2} = \frac{0.1305605}{10000} = 0.00001305605 \]
04

Compute for \( V = 500 \) million volts

Substitute \( V = 500 \): \[ 1-\frac{v}{c} = \frac{0.1305605}{(500)^2} = \frac{0.1305605}{250000} = 0.000000522241 \]
05

Compute for \( V = 25,000 \) million volts

Substitute \( V = 25000 \): \[ 1-\frac{v}{c} = \frac{0.1305605}{(25000)^2} = \frac{0.1305605}{625000000} = 0.000000000208896 \]
06

Compute for \( V = 100 \times 10^9 \) volts

Since \( 100 \times 10^9 \text{ volts} = 10^5 \text{ million volts} \), substitute \( V = 100000 \): \[ 1-\frac{v}{c} = \frac{0.1305605}{(100000)^2} = \frac{0.1305605}{10000000000} = 0.00000000001305605 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Binomial Series Approximation
The binomial series is a powerful tool in mathematics used to approximate functions that are not easily solvable otherwise. In this context, we use it to simplify complex expressions involving high powers. The series expansion for \( (1 - x)^a \) is particularly useful when \( x \) is very small. For our problem, we considered \( (1 - x)^{1/2} \), which can be approximated by the first term of its binomial expansion: \( 1 - \frac{x}{2} \). This works well for high energy accelerators because \( 0.511/V \) is very small, allowing us to use the approximation effectively. In essence, we turn a complex square root problem into a simple linear equation.
High Energy Accelerators
High energy accelerators are devices used to propel charged particles, such as electrons, to speeds close to that of light. These accelerators are crucial in many physics experiments and have practical applications in medical and industrial fields. The enormous voltages involved, often measured in millions or even billions of volts, necessitate relativistic calculations to predict particle velocities. Because the electron velocity approaches the speed of light (c), relativistic effects become significant, and we must use specific formulas to accurately describe the motion. Understanding these accelerators involves not just the physics of high-speed particles but also sophisticated technology to achieve and control such high energies.
Relativity in Physics
Relativity fundamentally changes how we understand motion at speeds close to the speed of light. Traditional Newtonian mechanics fail to accurately describe such high-speed phenomena, leading to the development of Einstein's theory of relativity. According to relativity, as objects move faster, their masses effectively increase, and time dilates from the perspective of a stationary observer. This is why we need to use the relativistic formula to determine the velocity of electrons in high energy accelerators. The formula \( \frac{v}{c} = \sqrt{1 - \frac{(0.511)^2}{V^2}} \) helps account for these relativistic effects by showing how the velocity fraction depends on the accelerator voltage, ensuring our calculations remain precise even at speeds close to light.
Electron Velocity
To find the velocity of electrons in a high energy accelerator, we need to consider both the applied voltage and relativistic formulas. Electrons, having a small mass, can be accelerated to extremely high velocities under high voltages. When we use the formula \( \frac{v}{c} \), we calculate the electron's velocity as a fraction of the speed of light. In our example, using the binomial series expansion, this ratio is simplified to \( 1 - \frac{(0.511)^2}{2V^2} \). By computing values for different voltages like 100 million, 500 million, 25,000 million, and 100 billion volts, we can see how close to the speed of light electrons can get. These calculations illustrate the profound impact of voltage on electron velocity and underscore the nearly light-speed velocities achievable in advanced accelerators.

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Most popular questions from this chapter

Use the ratio test to find whether the following series converge or diverge: $$\sum_{n=0}^{\infty} \frac{e^{n}}{\sqrt{n !}}$$

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