/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 83 A particle accelerator produces ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A particle accelerator produces a beam with a radius of \(1.25 \mathrm{mm}\) with a current of \(2.00 \mathrm{mA}\). Each proton has a kinetic energy of \(10.00 \mathrm{MeV}\). (a) What is the velocity of the protons? (b) What is the number (n) of protons per unit volume? (b) How many electrons pass a cross sectional area each second?

Short Answer

Expert verified
(a) The velocity of the protons is \(v_p = 1.438 \times 10^7 m/s\). (b) The number of protons per unit volume is \(n = 8.079 \times 10^{14} m^{-3}\). (c) The number of electrons passing the cross-sectional area each second is equal to the number of protons, which is \(5.082 \times 10^{10} electrons/s\).

Step by step solution

01

a. Finding the velocity of the protons

: First, we will convert the kinetic energy of the protons from MeV to Joules by multiplying it with the conversion factor (1 MeV = 1.602 × 10^-13 J). Then, we can use the kinetic energy formula to find the velocity of the protons: \(KE = \frac{1}{2}m_pv_p^2\) Where \(KE\) is the kinetic energy, \(m_p\) is the mass of a proton (\(1.673\times10^{-27} kg\)), and \(v_p\) is the velocity of the protons. We will solve for \(v_p\).
02

b. Calculating the number (n) of protons per unit volume

: We are given the current (\(I\)), which is the flow of charge per unit time, and the charge of a proton (\(q_p = 1.602\times10^{-19} C\)). We can calculate the number of protons passing through the cross-sectional area each second using the formula: \(I=nq_pAv_p\) Where \(n\) is the number of protons per unit volume, \(A\) is the cross-sectional area of the beam (\(\pi r^2\), where \(r=1.25\times10^{-3}\;m\)), and \(v_p\) is the velocity of the protons. We will solve for \(n\).
03

c. Finding the number of electrons passing the cross-sectional area each second

: Since the beam is neutral, the number of electrons passing the cross-sectional area each second must be equal to the number of protons. Therefore, we can say: Number of electrons passing the cross-sectional area each second = Number of protons passing the cross-sectional area each second We will use the number of protons calculated in part (b) as the number of electrons.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

kinetic energy
The concept of kinetic energy is crucial when discussing the motion of particles like protons in a particle accelerator. Kinetic energy is the energy that a particle possesses due to its motion. In the formula form, it is expressed as:
\[ KE = \frac{1}{2}mv^2 \]
where:
  • \( m \) is the mass of the proton
  • \( v \) is the velocity of the proton
To find the kinetic energy of a proton in common particle accelerator problems, you often need to convert its given energy to units compatible with the International System of Units (SI). Protons in accelerators typically have their energies provided in mega-electron volts (MeV). For calculations, this must be converted to joules, where 1 MeV is equivalent to \( 1.602 \times 10^{-13} \text{ J} \). Understanding and converting these units is essential for precise scientific calculation and analysis.
protons
Protons, which are positively charged particles, play an important role in atomic structure and particle physics. They are part of the nucleus of an atom, along with neutrons, and have specific properties that can be calculated and predicted in particle physics experiments.
In a particle accelerator, protons are accelerated to high speeds, dramatically increasing their kinetic energy. Properties like their mass (\(1.673 \times 10^{-27} \text{ kg}\)) and charge (\(1.602 \times 10^{-19} \text{ C}\)) are critical in calculations related to motion and interaction with electric and magnetic fields.
Understanding protons' behavior in particle accelerators allows scientists to investigate fundamental forces and particles, simulate cosmic processes, and even seek practical applications such as developing cancer treatment with proton therapy.
current
The term "current" in the context of a particle accelerator refers to the flow of charged particles, such as protons, through a cross-sectional area over time. It is commonly measured in amperes (A), where 1 ampere is equivalent to 1 coulomb of charge passing through a point per second.
In this problem, the current is given as 2.00 mA (milliamperes), or 0.002 A. We use this information to calculate how many charged particles pass through the beam per second using the formula:
\[ I = nq_pAv_p \]
where:
  • \( I \) is the current
  • \( n \) is the number of protons per unit volume
  • \( q_p \) is the charge of a proton
  • \( A \) is the cross-sectional area
  • \( v_p \) is the velocity of the protons
This equation shows how the current is directly related to the velocity of protons and the cross-sectional area they pass through, offering insights into the dynamics within a particle accelerator.
velocity
Velocity is a key parameter when discussing the motion of particles such as protons in a particle accelerator. It essentially describes the speed and direction of the protons as they move through the accelerator.
To calculate the velocity of protons, we make use of the kinetic energy equation:
\[ KE = \frac{1}{2}m_pv_p^2 \]
Solving for velocity, the equation becomes:
\[ v_p = \sqrt{\frac{2KE}{m_p}} \]
In the context of a particle accelerator, the velocity helps determine how protons will interact within the accelerator and with any targets they might hit. This value is critical for experimental accuracy and achieving the desired particle collisions.The velocity of protons not only contributes to calculations about their kinetic energy but also helps determine the magnetic and electric field settings needed to guide and focus the particle beam.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

An aluminum wire 1.628 mm in diameter (14-gauge) carries a current of 3.00 amps. (a) What is the absolute value of the charge density in the wire? (b) What is the drift velocity of the electrons? (c) What would be the drift velocity if the same gauge copper were used instead of aluminum? The density of copper is \(8.96 \mathrm{g} / \mathrm{cm}^{3}\) and the density of aluminum is \(2.70 \mathrm{g} / \mathrm{cm}^{3} .\) The molar mass of aluminum is \(26.98 \mathrm{g} / \mathrm{mol}\) and the molar mass of copper is \(63.5 \mathrm{g} / \mathrm{mol} .\) Assume each atom of metal contributes one free electron.

Consider a resistor made from a hollow cylinder of carbon as shown below. The inner radius of the cylinder is \(R_{i}=0.20 \mathrm{mm}\) and the outer radius is \(R_{0}=0.30 \mathrm{mm}\) The length of the resistor is \(L=0.90 \mathrm{mm}\). The resistivity of the carbon is \(\rho=3.5 \times 10^{-5} \Omega \cdot \mathrm{m} \cdot\) (a) Prove that the resistance perpendicular from the axis is \(R=\frac{\rho}{2 \pi L} \ln \left(\frac{R_{0}}{R_{i}}\right)\) (b) What is the resistance?

A physics student uses a \(115.00-\mathrm{V}\) immersion heater to heat 400.00 grams (almost two cups) of water for herbal tea. During the two minutes it takes the water to heat, the physics student becomes bored and decides to figure out the resistance of the heater. The student starts with the assumption that the water is initially at the temperature of the room \(T_{i}=25.00^{\circ} \mathrm{C}\) and reaches \(T_{f}=100.00^{\circ} \mathrm{C}\) The specific heat of the water is \(c=4180 \frac{\mathrm{J}}{\mathrm{kg} \cdot \mathrm{K}} \cdot\) What is the resistance of the heater?

A physics student has a single-occupancy dorm room. The student has a small refrigerator that runs with a current of \(3.00 \mathrm{A}\) and a voltage of \(110 \mathrm{V},\) a lamp that contains a \(100-\mathrm{W}\) bulb, an overhead light with a \(60-\mathrm{W}\) bulb, and various other small devices adding up to \(3.00 \mathrm{W}\). (a) Assuming the power plant that supplies \(110 \mathrm{V}\) electricity to the dorm is \(10 \mathrm{km}\) away and the two aluminum transmission cables use 0-gauge wire with a diameter of 8.252 mm, estimate the percentage of the total power supplied by the power company that is lost in the transmission. (b) What would be the result is the power company delivered the electric power at \(110 \mathrm{kV} ?\)

The quantity of charge through a conductor is modeled as \(Q=4.00 \frac{\mathrm{C}}{\mathrm{s}^{4}} t^{4}-1.00 \frac{\mathrm{C}}{\mathrm{s}} t+6.00 \mathrm{mC}\)

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.