Problem 82 - Fundamentals of Physics

Question

A linear accelerator produces a pulsed beam of electrons. The pulse current is \(0.50 \mathrm{~A},\) and the pulse duration is \(0.10 \mu \mathrm{s}\). (a) How many electrons are accelerated per pulse? (b) What is the average current for a machine operating at 500 pulses/s? If the electrons are accelerated to an energy of \(50 \mathrm{MeV},\) what are the (c) average power and (d) peak power of the accelerator?

Step-by-Step Solution

Answer

(a) Approx. \(3.1 \times 10^{11}\) electrons, (b) \(2.5 \times 10^{-5}\) A, (c) 2 W, (d) 40 MW.

Step 1: Calculate Charge per Pulse

To find the number of electrons accelerated per pulse, we need to determine the total charge in a single pulse. The relation between charge, current, and time is given by: \(Q = I \cdot t\), where \(Q\) is charge in coulombs, \(I\) is current in amperes, and \(t\) is time in seconds. Given the pulse current \(I = 0.50 \text{ A}\) and pulse duration \(t = 0.10 \times 10^{-6} \text{ s}\), we can calculate the charge per pulse: \(Q = 0.50 \cdot 0.10 \times 10^{-6} = 5.0 \times 10^{-8} \text{ C}\).

Step 2: Calculate Number of Electrons per Pulse

Since the charge of a single electron is \(1.6 \times 10^{-19} \text{ C}\), the number of electrons \(n\) accelerated per pulse is \(n = \frac{Q}{e}\), where \(e\) is the elementary charge. Substituting the known values, \(n = \frac{5.0 \times 10^{-8}}{1.6 \times 10^{-19}} \approx 3.1 \times 10^{11} \text{ electrons}\).

Step 3: Calculate Average Current

The average current over a second for 500 pulses per second can be calculated using: \(\text{Average Current} = \text{Number of Pulses per Second} \times \text{Charge per Pulse}\). Given the result from Step 1 and 500 pulses/s, \(\text{Average Current} = 500 \times 5.0 \times 10^{-8} = 2.5 \times 10^{-5} \text{ A}\).

Step 4: Calculate Average Power

The average power is calculated by multiplying the average current by the energy per electron. Energy per electron is given as \(50 \text{ MeV} = 50 \times 10^{6} \times 1.6 \times 10^{-13} \text{ joules}\). Thus, \(\text{Average Power} = \text{Average Current} \times \text{Voltage per Pulse}\). Substituting the values, \(\text{Average Power} = 2.5 \times 10^{-5} \times 50 \times 10^{6} \times 1.6 \times 10^{-13}\). The average power is approximately \(2.0 \text{ W}\).

Step 5: Calculate Peak Power

Peak power occurs during the pulse when the current is at its maximum value of \(0.50 \text{ A}\) for a pulse duration \(0.10 \mu\text{s}\). The peak power is given by \(\text{Peak Power} = \text{Peak Current} \times \text{Energy per Pulse Mover}\). Substituting the values: \(\text{Peak Power} = 0.50 \text{ A} \times 50 \times 10^{6} \text{ eV} \times 1.6 \times 10^{-13} \frac{\text{J}}{\text{eV}} \approx 40 \text{ MW}\).

Key Concepts

Current Calculation

Current is an essential concept when dealing with electron beams in linear accelerators. It's a measure of how much charge is flowing per unit of time. In the context of short electron pulses, as we find in linear accelerators, we need to calculate both the charge per pulse and the average current.

To determine the charge carried by each pulse, we use the formula:

• \( Q = I \cdot t \)

Here, \( Q \) is the total charge, \( I \) is the current, and \( t \) is the time duration of the pulse. If we take an example of a pulse with \( I = 0.50 \ \text{A} \) lasting \( t = 0.10 \ \mu\text{s} \), the charge \( Q \) evaluates to \( 5.0 \times 10^{-8} \text{ C} \).

Once we know the charge per pulse, we can find the number of electrons per pulse. This depends on the charge of a single electron, \( e = 1.6 \times 10^{-19} \text{ C} \), so the number of electrons \( n = \frac{Q}{e} \). Each pulse here, therefore, contains around \( 3.1 \times 10^{11} \) electrons.

Finally, calculating the average current involves multiplying the number of pulses per second by the charge per pulse. For a machine firing 500 pulses every second, the average current is \( 2.5 \times 10^{-5} \text{ A} \). This average gives insight into the steady state of the current during regular operation.

Electron Acceleration

In a linear accelerator, electrons gain energy as they travel through an electric field. This process is crucial since the final energy level determines the application and effectiveness of the beam produced.

The energy achieved by each electron is often represented in electron volts (eV), a unit that makes it easier to deal with the vast numbers involved. In our given situation, each electron is accelerated to reach an energy of \(50 \ \text{MeV}\), or 50 million electron volts. To convert this into joules (as energy calculations often require), we multiply by the charge of an electron: \(1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}\).

Therefore, each electron gains:

• \( 50 \times 10^6 \times 1.6 \times 10^{-13} \text{ J} \)

That equates to approximately \( 8.0 \times 10^{-12} \text{ J} \) per electron. Understanding how energy is imparted to particles aids in designing accelerators and predicting beam behavior.

Power Calculation

Calculating power, especially in linear accelerators, involves understanding both average and peak power. Power, in simple terms, is how quickly energy is used or converted.

For average power, we multiply the average current by the energy rate per electron. Given our data:

• Average Current: \( 2.5 \times 10^{-5} \text{ A} \)
• Energy per electron: \( 50 \ \text{MeV} = 8.0 \times 10^{-12} \text{ J} \)

The average power calculation becomes:

• \( ext{Average Power} = 2.5 \times 10^{-5} \times 50 \times 10^6 \times 1.6 \times 10^{-13} \approx 2.0 \text{ W} \)

Peak power is determined when the full pulse current (\(0.50 \ \text{A}\)) delivers high energy simultaneously. Thus, calculating peak power involves:

• \( ext{Peak Power} = 0.50 \ \text{A} \times 50 \times 10^6 \times 1.6 \times 10^{-13} \approx 40 \text{ MW} \)

This calculation indicates the potential maximum energy flow during the pulse, a critical factor for assessing system performance and energy needs.