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Chapter 19: Problem 17

In a mass spectrometer, a doubly charged ion having a particular velocity is selected by using a magnetic field of \(100 \mathrm{mT}\) perpendicular to an electric field of \(1.0 \mathrm{kV} / \mathrm{m}\). This same magnetic field is then used to deflect the ion in a circular path with a radius of \(15 \mathrm{~mm}\). Find (a) the mass of the ion and (b) the kinetic energy of the ion. (c) Does the kinetic energy of the ion increase in the circular path? Explain.

Short Answer

Expert verified
(a) Mass: \(4.8 \times 10^{-25} \text{kg}\); (b) KE: \(2.4 \times 10^{-17} \text{J}\); (c) KE does not increase.

Step by step solution

01

Determine Ion's Velocity

In the velocity selector, the electric force is balanced by the magnetic force. The forces can be equated as:\[ qE = qvB \]where \( q \) is the charge of the ion, \( E = 1000 \, \text{V/m} \), and \( B = 0.1 \, \text{T} \). Solve for \( v \):\[ v = \frac{E}{B} = \frac{1000}{0.1} = 10,000 \, \text{m/s} \].This is the velocity of the ion in the selector.
02

Calculate Mass of the Ion

Once the ion moves in a circular path, the centripetal force is provided by the magnetic force, given by:\[ qvB = \frac{mv^2}{r} \]Simplify to solve for the mass \( m \) of the ion:\[ m = \frac{qBr}{v} \]The ion is doubly charged, so \( q = 2e = 2 \times 1.6 \times 10^{-19} \, \text{C} \). The magnetic field \( B = 0.1 \, \text{T} \), the radius \( r = 0.015 \, \text{m} \), and \( v = 10,000 \, \text{m/s} \):\[ m = \frac{2 \times 1.6 \times 10^{-19} \times 0.1 \times 0.015}{10,000} \approx 4.8 \times 10^{-25} \, \text{kg} \].
03

Calculate Kinetic Energy of the Ion

Kinetic energy (KE) is calculated using the formula:\[ KE = \frac{1}{2}mv^2 \]Substitute the values \( m = 4.8 \times 10^{-25} \, \text{kg} \) and \( v = 10,000 \, \text{m/s} \):\[ KE = \frac{1}{2} \times 4.8 \times 10^{-25} \times (10,000)^2 \approx 2.4 \times 10^{-17} \, \text{J} \].
04

Evaluate Change in Kinetic Energy

The kinetic energy remains constant as the ion travels in a circular path because there is no work done by the magnetic field. Magnetic fields do not change the speed (kinetic energy) of a charged particle, they only change the direction of its motion.

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Key Concepts

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

kinetic energy
Kinetic energy is the energy a particle has due to its motion. For an ion in a mass spectrometer, this depends on its mass and velocity. The formula used to calculate kinetic energy (KE) is:
  • KE = \(\frac{1}{2}mv^2\)
  • \(m\) is the mass of the ion.
  • \(v\) is the velocity.
Kinetic energy is measured in joules (J). In this scenario, after gaining velocity from a velocity selector, the ion's kinetic energy can be calculated using its speed and mass. It's important to note that kinetic energy is not influenced by the ion's path, allowing us to affirm that it remains unchanged as long as the velocity remains constant.
magnetic field
A magnetic field is a powerful force used in devices like mass spectrometers to manipulate charged particles. This field is typically described by its direction and strength, measured in Tesla (T). A magnetic field can exert a force on a moving charge, given by:
  • \(F = qvB\), where \(F\) is the force, \(q\) is the charge, and \(B\) is the magnetic field strength.
For instance, in a mass spectrometer, the magnetic field plays a crucial role by bending the path of ions. This deflection is utilized to analyze and measure the properties of ions, as the field doesn't change their speed but redirects their trajectory, enabling the study of their mass and charge.
circular motion
Particles moving in a magnetic field can experience circular motion, a key feature in the operation of mass spectrometers. When a charged ion enters a perpendicular magnetic field, it follows a circular trajectory due to the magnetic force acting as the centripetal force:
  • The centripetal force equation is \(F = \frac{mv^2}{r}\).
  • \(m\) is the mass, \(v\) is velocity, and \(r\) is the radius of the circular path.
This relationship helps in determining the mass of the ion when other factors are known, enabling the categorization and analysis of ions in science and industry. Understanding this motion highlights the role of magnetic fields in directing particle travel within technological apparatuses.
ion charge
The charge of an ion is pivotal in determining how it behaves under electric and magnetic fields. Charge is quantified in coulombs (C), and ions can be singly, doubly, or multiply charged, affecting their interactions significantly. In mass spectrometry:
  • The charge impacts how the ion is deflected by the fields.
  • Doubly charged ions (2e) experience twice the influence compared to singly charged ones (e).
The value of \(q\) determines the strength and direction of the force on an ion. Higher charges mean greater response to magnetic forces, resulting in different paths for ions of different charge states. Analyzing these variations provides insights into the constitution of the ion itself.
velocity selector
A velocity selector is an essential component in a mass spectrometer that helps to identify ions with a specific velocity. It uses both electric and magnetic fields to achieve this separation:
  • Electric field (E) exerts force on the ion. \(F = qE\).
  • Magnetic field (B) counteracts this force. \(F = qvB\).
  • When forces balance, \(v = \frac{E}{B}\).
The velocity selector ensures that only ions with a predetermined speed pass through to the next stage of the spectrometer. By fine-tuning these fields, it effectively filters out particles, making it a critical tool for isolating ions for further analysis in various experimental settings.

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Most popular questions from this chapter

A particle with charge \(q\) and mass \(m\) moves in a horizontal plane at right angles to a uniform vertical magnetic field \(B\). (a) What are the period \(T\) and frequency \(f\) of the particle's circular motion in terms of \(q, B\), and \(m\) ? (This frequency is called the cyclotron frequency.) You result should verify that the time for one orbit for any charged particle in a uniform magnetic field is independent of its speed and radius. (b) Compute the path radius and the cyclotron frequency if the particle is an electron with a speed of \(1.0 \times 10^{5} \mathrm{~m} / \mathrm{s}\) traveling in a region where the field strength is \(0.10 \mathrm{mT}\)

A positive charge moves horizontally to the right across this page and enters a magnetic field directed vertically downward in the plane of the page. (a) What is the direction of the magnetic force on the charge: (1) into the page, (2) out of the page, (3) downward in the plane of the page, or (4) upward in the plane of the page? Explain. (b) If the charge is \(0.25 \mathrm{C}\), its speed is \(2.0 \times 10^{2} \mathrm{~m} / \mathrm{s},\) and it is acted on by a force of \(20 \mathrm{~N},\) what is the magnetic field strength?

A circular loop of wire with a radius of \(5.0 \mathrm{~cm}\) carries a current of 1.0 A. Another circular loop of wire is concentric with (that is, has a common center with) the first and has a radius of \(10 \mathrm{~cm}\). The magnetic field at the center of the loops is double what the field would be from the first one alone, but oppositely directed. What is the current in the second loop?

A horizontal magnetic field of \(1.0 \times 10^{-4} \mathrm{~T}\) is at an angle of \(30^{\circ}\) to the direction of the current in a straight, horizontal wire \(75 \mathrm{~cm}\) long. If the wire carries a current of \(15 \mathrm{~A}\), (a) what is the magnitude of the force on the wire? (b) What angle(s) would be required for the force to be half the value found in part (a), assuming nothing else is changed?

A wire carries a current of 10 A in the \(+x\) -direction in a uniform magnetic field of \(0.40 \mathrm{~T}\). Find the magnitude of the force per unit length and the direction of the force on the wire if the magnetic field is (a) in the \(+x\) -direction, (b) in the \(+y\) -direction, \((\mathrm{c})\) in the \(+z\) -direction, \((\mathrm{d})\) in the \(-y\) -direction, \((\mathrm{e})\) in the \(-z\) -direction, and \((\mathrm{f})\) at an angle of \(45^{\circ}\) above the \(+x\) -axis and in the \(x-y\) plane.
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