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When beryllium-7 ions \(\left(m=11.65 \times 10^{-27} \mathrm{kg}\right)\) pass through a mass spectrometer, a uniform magnetic field of \(0.283 \mathrm{T}\) curves their path directly to the center of the detector (see Figure 21.14 ). For the same accelerating potential difference, what magnetic field should be used to send beryllium-10 ions \(\left(m=16.63 \times 10^{-27} \mathrm{kg}\right)\) to the same location in the detector? Both types of ions are singly ionized \((q=+e)\).

Short Answer

Expert verified
Use a magnetic field of approximately 0.338 T for beryllium-10 ions.

Step by step solution

01

Understand the Problem - Mass Spectrometer Basics

A mass spectrometer is used to separate ions of different masses through their circular motion in a magnetic field. The radius of curvature, to which ions are directed to the detector, depends on the mass \(m\), charge \(q\), velocity \(v\), and magnetic field \(B\) via the relation \( r = \frac{mv}{qB}\). Our goal is to find the required magnetic field for different ions given their mass while pointing them to the same detector location.
02

Determine Path Radius with Known Parameters

For beryllium-7 ions, the magnetic field \( B_1 = 0.283 \mathrm{T} \) is such that the path radius \( r \) is directed to the detector. Given the mass \( m_1 = 11.65 \times 10^{-27} \mathrm{kg} \) and charge \( q = e \), use \( r = \frac{m_1v_1}{qB_1} \) to represent the radius.
03

Apply Energy Conservation to Determine Velocity

Ions accelerated through a potential \( V \) gain kinetic energy. Equating energy terms, \( \frac{1}{2}mv^2 = qV \), gives velocity as \( v = \sqrt{\frac{2qV}{m}} \). This formula will be used for both beryllium-7 and beryllium-10 ions.
04

Write Expression for Both Ions and Equate Radii

Let the magnetic field for beryllium-10 be \( B_2 \). For the same detector location, radii must be equal: \( \frac{m_1v_1}{qB_1} = \frac{m_2v_2}{qB_2} \).Substitute velocities from Step 3, leading to: \[ \frac{m_1 \sqrt{\frac{2qV}{m_1}}}{qB_1} = \frac{m_2 \sqrt{\frac{2qV}{m_2}}}{qB_2} \].
05

Simplify and Solve for Magnetic Field \(B_2\)

Cancel common factors and simplify: \[ \frac{\sqrt{m_1}}{B_1} = \frac{\sqrt{m_2}}{B_2} \].Rearrange to find \( B_2 = \frac{B_1 \sqrt{m_2}}{\sqrt{m_1}} \).Substitute given masses and magnetic field: \[ B_2 = \frac{0.283 \mathrm{T} \times \sqrt{16.63 \times 10^{-27} \mathrm{kg}}}{\sqrt{11.65 \times 10^{-27} \mathrm{kg}}} \].
06

Calculate Numerical Value

Calculate: \( B_2 \approx \frac{0.283 \mathrm{T} \times 4.078 \times 10^{-14}}{3.413 \times 10^{-14}} \approx 0.338 \mathrm{T} \). Thus, the magnetic field should be approximately \(0.338 \mathrm{T}\) to direct beryllium-10 ions to the detector.

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Key Concepts

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

Magnetic Field
In a mass spectrometer, the magnetic field plays a crucial role in the separation of ions. The magnetic field is denoted by the symbol \( B \). It applies a perpendicular force to moving charged particles, causing them to travel in a circular path. The strength and uniformity of the magnetic field determine the extent of the curvature. The main equation governing this phenomenon is the Lorentz force, which is \( F = qvB \), where \( q \) is the charge of the ion and \( v \) is its velocity. This force causes the particles to move in a circular path with a radius, \( r \), defined by the equation \( r = \frac{mv}{qB} \).
For ions, the magnetic field is essential because it dictates path curvature and ensures ions of different masses can be separated and detected accurately. Adjustments to the magnetic field allow scientists to manipulate the focal point of ions into a specific location in a detector. Understanding and manipulating this field become necessary for accurately pinpointing the ions' paths.
Ions
Ions are charged particles that play a foundational role in the function of a mass spectrometer. Essentially, ions are atoms or molecules that have gained or lost electrons, resulting in either a positive or negative charge. In the context of this exercise, we focus on singly ionized beryllium ions. This means each ion has a single positive charge, often denoted as \( q = +e \), where \( e \) is the elementary charge, approximately \( 1.602 \times 10^{-19} \) C.
Ions are created by ionizing the sample, usually within a special chamber of the mass spectrometer. Post-ionization, these charged particles can be accelerated by electric fields to gain kinetic energy before entering the magnetic field. The mass of each ion, such as \( m_1 = 11.65 \times 10^{-27} \mathrm{kg} \) for beryllium-7 and \( m_2 = 16.63 \times 10^{-27} \mathrm{kg} \) for beryllium-10, is essential for determining their behavior under the influence of a magnetic field. The magnetic field affects their trajectories similarly, which allows their separation based on the differing radii of curvature.
Radius of Curvature
The radius of curvature is a pivotal concept in the working of a mass spectrometer. This radius \( r \) determines the path that the ions take when influenced by the magnetic field. The fundamental equation here is \( r = \frac{mv}{qB} \). It reveals how the path radius is related to an ion's mass \( m \), velocity \( v \), magnetic field \( B \), and charge \( q \).
In a mass spectrometer, ions of different masses will have different radii of curvature even when subjected to the same velocity and magnetic field. This is because the mass spectrometer operates under a principle where masses can be deduced by measuring the radius of their path curvature. Thus, by analyzing their distinct radii, it becomes possible to distinguish between ions of different masses, separating them effectively for analysis. It’s crucial to adjust the magnetic field correctly so that ions reach the detector together, particularly when the aim is to compare ions like beryllium-7 and beryllium-10.
Velocity
Velocity \( v \) of the ions is another critical factor that helps determine the curvature of their path in a mass spectrometer. Before entering the magnetic field, ions are usually accelerated through an electric potential difference \( V \). By conserving energy, we use the formula \( \frac{1}{2}mv^2 = qV \) to determine velocity, leading to \( v = \sqrt{\frac{2qV}{m}} \).
The velocity is essential because it directly influences the radius of curvature. Higher velocities result in larger radii, while lower velocities create tighter curves, assuming the magnetic field and charge remain constant. Therefore, the velocity in combination with the magnetic field and mass is central to the precise control of ion paths within the spectrometer. When it comes to comparing different ions, ensuring they are accelerated by the same potential difference is vital for making accurate comparisons and computations regarding their trajectories and final positions in the detector.

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Most popular questions from this chapter

Review Conceptual Example 2 as an aid in understanding this problem. A velocity selector has an electric field of magnitude \(2470 \mathrm{N} / \mathrm{C},\) directed vertically upward, and a horizontal magnetic field that is directed south. Charged particles, traveling east at a speed of \(6.50 \times 10^{3} \mathrm{m} / \mathrm{s},\) enter the velocity selector and are able to pass completely through without being deflected. When a different particle with an electric charge of \(+4.00 \times 10^{-12} \mathrm{C}\) enters the velocity selector traveling east, the net force (due to the electric and magnetic fields) acting on it is \(1.90 \times 10^{-9} \mathrm{N},\) pointing directly upward. What is the speed of this particle?

The two conducting rails in the drawing are tilted upward so they each make an angle of \(30.0^{\circ}\) with respect to the ground. The vertical magnetic field has a magnitude of 0.050 T. The 0.20-kg aluminum rod (length \(=1.6 \mathrm{m})\) slides without friction down the rails at a constant velocity. How much current flows through the rod?

The maximum torque experienced by a coil in a 0.75-T magnetic field is \(8.4 \times 10^{-4} \mathrm{N} \cdot \mathrm{m} .\) The coil is circular and consists of only one turn. The current in the coil is 3.7 A. What is the length of the wire from which the coil is made?

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.

A long solenoid has 1400 turns per meter of length, and it carries a current of 3.5 A. A small circular coil of wire is placed inside the solenoid with the normal to the coil oriented at an angle of \(90.0^{\circ}\) with respect to the axis of the solenoid. The coil consists of 50 turns, has an area of \(1.2 \times 10^{-3} \mathrm{m}^{2},\) and carries a current of 0.50 A. Find the torque exerted on the coil.

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