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Chapter 21: Problem 21

Two of the isotopes of carbon, carbon- 12 and carbon- \(13,\) have masses of \(19.93 \times 10^{-27} \mathrm{kg}\) and \(21.59 \times 10^{-27} \mathrm{kg},\) respectively. These two isotopes are singly ionized \((+e),\) each given a speed of \(6.667 \times 10^{5} \mathrm{m} / \mathrm{s} .\) The ions then enter the bending region of a mass spectrometer where the magnetic field is \(0.8500 \mathrm{T} .\) Determine the spatial separation between the two isotopes after they have traveled through a half-circle.

Short Answer

Expert verified
The spatial separation between carbon-12 and carbon-13 isotopes is 1.61 cm.

Step by step solution

01

Understanding the Problem

We are tasked with finding the spatial separation between two isotopes of carbon, carbon-12 and carbon-13, after they travel through a magnetic field in a mass spectrometer. When these charged ions move in the magnetic field, they will follow a circular path based on their masses, charge, speed, and the magnetic field strength.
02

Applying the Formula for Radius

The radius of the circular path of a charged particle in a magnetic field is given by the formula \( r = \frac{mv}{qB} \), where \( m \) is the mass of the ion, \( v \) is the speed, \( q \) is the charge of the ion, and \( B \) is the magnetic field strength. Here, both ions are singly ionized, so \( q = e = 1.602 \times 10^{-19} \mathrm{C} \).
03

Calculate Radius for Carbon-12

Substitute the values into the formula for carbon-12:\[ r_{12} = \frac{(19.93 \times 10^{-27} \text{ kg}) \times (6.667 \times 10^{5} \text{ m/s})}{1.602 \times 10^{-19} \text{ C} \times 0.8500 \text{ T}} \]Solve to find \( r_{12} \approx 0.09701 \text{ m} \).
04

Calculate Radius for Carbon-13

Now use the formula for carbon-13:\[ r_{13} = \frac{(21.59 \times 10^{-27} \text{ kg}) \times (6.667 \times 10^{5} \text{ m/s})}{1.602 \times 10^{-19} \text{ C} \times 0.8500 \text{ T}} \]Solve to find \( r_{13} \approx 0.10506 \text{ m} \).
05

Find the Spatial Separation

Since the ions travel through exactly half of a circular path, the spatial separation is the difference in diameter of these two paths:\[ ext{Separation} = 2(r_{13} - r_{12}) = 2(0.10506 \text{ m} - 0.09701 \text{ m}) \]This gives us a separation of approximately \( 0.0161 \text{ m} \) or \( 1.61 \text{ cm} \).

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Key Concepts

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

Isotopes
Isotopes are different forms of the same element that contain the same number of protons but different numbers of neutrons. This causes the isotopes to have different masses, even though they are chemically similar. Carbon, for example, has several isotopes, with carbon-12 and carbon-13 being the most common. These numbers refer to the total number of protons and neutrons in the nucleus.
In the context of a mass spectrometer, the difference in mass allows us to separate isotopes. When ionized, these different isotopes can be manipulated by magnetic fields to travel in paths that depend on their mass. This is because the mass directly affects the radius of their path in a magnetic field, thus allowing scientists to discern between isotopes based on mass differences.
Magnetic Field
A magnetic field is a region of space around a magnetized object where the force of magnetism acts. In physics, it's represented by the symbol "B" and is measured in teslas (T). Magnetic fields exert forces on moving charges, such as those found in ions.
In a mass spectrometer, a magnetic field is used to exert an influence on charged particles, causing them to move in circular paths. The strength and direction of the magnetic field determine the curvature and direction of this path. A stronger magnetic field will bend the path more dramatically. This characteristic is vital in separating isotopes, as it helps determine the radius of their semi-circular paths within the mass spectrometer's bending region.
Circular Motion
Circular motion occurs when an object moves in a path that forms a circle. The movement is caused by a centripetal force that acts perpendicular to the direction of motion. In the case of charged particles in a magnetic field, this centripetal force is provided by the magnetic field itself, which makes the particles move in a circular path.
For mass spectrometry, this circular motion is crucial because it allows for the calculation of the radius of the particle's path. By using the formula \( r = \frac{mv}{qB} \), it's possible to determine how the mass, velocity, charge, and magnetic field influence the circular trajectory. The larger the mass, the larger the radius of the circular path; thus, heavier isotopes will travel in larger circles. This property helps in assessing the spatial separation between different isotopes.
Charged Particles
Charged particles, such as ions, have an electric charge due to the loss or gain of one or more electrons. In physics problems like those solved in a mass spectrometer, they typically refer to ions. These particles can interact dynamically with electric and magnetic fields.
When placed in a magnetic field, charged particles experience a force, known as the Lorentz force, which causes them to accelerate in a curved path if their trajectory isn't parallel to the field. For ions in a mass spectrometer, this results in circular motion. The extent of curvature of their path depends on several factors, including the magnitude of their charge, the strength of the magnetic field, and their mass and speed. Understanding the behavior of charged particles in fields is fundamental to the working of devices like mass spectrometers.

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Most popular questions from this chapter

Two circular coils are concentric and lie in the same plane. The inner coil contains 140 turns of wire, has a radius of \(0.015 \mathrm{m},\) and carries a current of \(7.2 \mathrm{A}\). The outer coil contains 180 turns and has a radius of \(0.023 \mathrm{m}\). What must be the magnitude and direction (relative to the current in the inner coil) of the current in the outer coil, so that the net magnetic field at the common center of the two coils is zero?

A copper rod of length \(0.85 \mathrm{m}\) is lying on a frictionless table (see the drawing). Each end of the rod is attached to a fixed wire by an unstretched spring that has a spring constant of \(k=75 \mathrm{N} / \mathrm{m} .\) A magnetic field with a strength of \(0.16 \mathrm{T}\) is oriented perpendicular to the surface of the table. (a) What must be the direction of the current in the copper rod that causes the springs to stretch? (b) If the current is 12 A, by how much does each spring stretch?

The drawing shows a parallel plate capacitor that is moving with a speed of \(32 \mathrm{m} / \mathrm{s}\) through a \(3.6-\mathrm{T}\) magnetic field. The velocity \(\overrightarrow{\mathbf{v}}\) is perpendicular to the magnetic field. The electric field within the capacitor has a value of \(170 \mathrm{N} / \mathrm{C},\) and each plate has an area of \(7.5 \times 10^{-4} \mathrm{m}^{2} .\) What is the magnetic force (magnitude and direction) exerted on the positive plate of the capacitor?

Multiple-Concept Example 7 discusses how problems like this one can be solved. A \(+6.00 \mu \mathrm{C}\) charge is moving with a speed of \(7.50 \times 10^{4} \mathrm{m} / \mathrm{s}\) parallel to a very long, straight wire. The wire is \(5.00 \mathrm{cm}\) from the charge and carries a current of \(67.0 \mathrm{A}\) in a direction opposite to that of the moving charge. Find the magnitude and direction of the force on the charge.

A positively charged particle of mass \(7.2 \times 10^{-8} \mathrm{kg}\) is traveling due east with a speed of \(85 \mathrm{m} / \mathrm{s}\) and enters a \(0.31-\mathrm{T}\) uniform magnetic field. The particle moves through one- quarter of a circle in a time of \(2.2 \times 10^{-3} \mathrm{s},\) at which time it leaves the field heading due south. All during the motion the particle moves perpendicular to the magnetic field. (a) What is the magnitude of the magnetic force acting on the particle? (b) Determine the magnitude of its charge.
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