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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
The magnetic field for beryllium-10 should be approximately 0.339 T.

Step by step solution

01

Understand the problem

We are given two isotopes, Beryllium-7 and Beryllium-10, each moving through a magnetic field in a spectrometer. We need to determine the magnetic field required to direct Beryllium-10 ions to the same location as Beryllium-7 ions, given their masses and initial conditions.
02

Recognize key equations

The motion of a charged particle in a magnetic field is circular. The radius of curvature is given by the formula: \[ r = \frac{mv}{qB} \] where \( m \) is mass, \( v \) is velocity, \( q \) is charge, and \( B \) is magnetic field. Additionally, kinetic energy gained by the ions from the potential difference \( V \) is given by: \[ \frac{1}{2}mv^2 = qV \] allowing us to express velocity \( v \) as \( v = \sqrt{\frac{2qV}{m}} \).
03

Solve for the velocity

Using the energy equation \( v = \sqrt{\frac{2qV}{m}} \), both ions accelerate through the same potential \( V \) with the same charge \( q \), but different masses. This means their velocities are different.
04

Express the radius for beryllium-7

For beryllium-7, we substitute \( m_1 = 11.65 \times 10^{-27} \text{ kg} \), \( q = e \), and \( B_1 = 0.283 \text{ T} \) into radius formula: \[ r_1 = \frac{m_1 \cdot v_1}{q \cdot B_1} \]. Use velocity expression to substitute \( v_1 = \sqrt{\frac{2qV}{m_1}} \).
05

Solve for the radius in terms of known variables

\[ r_1 = \frac{m_1 \cdot \sqrt{\frac{2qV}{m_1}}}{q \cdot B_1} = \frac{\sqrt{2m_1 qV}}{qB_1} \]. Since \( r_1 \) is known for beryllium-7 ions, this expression is our reference point.
06

Equate radii of two isotopes

Since we want beryllium-10 to have the same radius, \( r_1 = r_2 \), where \[ r_2 = \frac{\sqrt{2m_2 qV}}{qB_2} \]. Equate \( r_1 \) and \( r_2 \) to solve for \( B_2 \): \[ \frac{\sqrt{2m_1 qV}}{qB_1} = \frac{\sqrt{2m_2 qV}}{qB_2} \].
07

Solve for the new magnetic field

By canceling appropriate terms (\( qV \)), we simplify: \[ B_2 = B_1 \cdot \sqrt{\frac{m_2}{m_1}} \]. Substitute known masses \( m_1 = 11.65 \times 10^{-27} \text{ kg} \) and \( m_2 = 16.63 \times 10^{-27} \text{ kg} \) to find \( B_2 \).
08

Calculation of magnetic field for beryllium-10

Perform the calculation: \[ B_2 = 0.283 \text{ T} \cdot \sqrt{\frac{16.63 \times 10^{-27}}{11.65 \times 10^{-27}}} \]. Compute the value to get \( B_2 \approx 0.339 \text{ T} \).

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Key Concepts

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

Beryllium Isotopes
Beryllium isotopes are different types of beryllium atoms that have the same number of protons but different numbers of neutrons. The number of protons in all beryllium isotopes is 4, which is the atomic number of beryllium. However, the isotopes differ in their mass number due to varying neutrons. For example, beryllium-7 has 3 neutrons while beryllium-10 has 6 neutrons.

These isotopes are significant in various scientific studies. Beryllium-7 is often used in studying solar neutrinos, while beryllium-10 is commonly used in geological dating and climate studies due to its longer half-life.

In a mass spectrometer, these isotopes can be distinguished by their masses, making the spectrometer an essential tool for isotope analysis. Each isotope, because of its unique mass, will follow a different path when subjected to a magnetic field.
Magnetic Field Calculation
Calculating the magnetic field required to guide charged particles accurately is central to the functioning of a mass spectrometer. When charged particles like ions move through a magnetic field, they experience a force that causes them to curve along a circular path. This path's radius is a critical factor and depends on multiple factors:
  • The mass of the particle
  • The velocity of the particle
  • The charge on the particle
  • The strength of the magnetic field

The relationship between these variables is given by the formula: \[ r = \frac{mv}{qB} \]where \(r\) is the radius, \(m\) is the mass, \(v\) is the velocity, \(q\) is the charge, and \(B\) is the magnetic field.

To determine the correct magnetic field for a different isotope, like from beryllium-7 to beryllium-10, we need to maintain the same path radius. Using the mass and charge of the new isotope, along with the known initial conditions, we can adjust the magnetic field accordingly using the formula \(B_2 = B_1 \cdot \sqrt{\frac{m_2}{m_1}}\). This ensures that despite the difference in mass, the new isotope will reach the same target on the detector.
Charged Particle Motion
The motion of charged particles in a magnetic field is a fascinating aspect of physics often explored in mass spectrometers. Charged particles such as ions will experience a force perpendicular to both their velocity and the magnetic field, described by the Lorentz force equation: \[ \vec{F} = q \vec{v} \times \vec{B} \]Consequently, the particles move in circular paths when subjected to a magnetic field.

The particles' speed as they enter the magnetic field is derived from the energy they gain from the potential difference they have been accelerated through. This relationship is given by: \[ \frac{1}{2}mv^2 = qV \]which allows velocity \(v\) to be calculated as \(v = \sqrt{\frac{2qV}{m}}\).

Understanding this motion helps us manipulate the conditions in a mass spectrometer, enabling precise control over the paths of different isotopes and the determination of their relative masses. The interplay between the magnetic field strength, particle charge and mass, and path radius provides the fundamental basis for mass spectrometry.

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Most popular questions from this chapter

A very long, straight wire carries a current of 0.12 A. This wire is tangent to a single-turn, circular wire loop that also carries a current. The directions of the currents are such that the net magnetic field at the center of the loop is zero. Both wires are insulated and have diameters that can be neglected. How much current is there in the loop?

A charged particle enters a uniform magnetic field and follows the circular path shown in the drawing. (a) Is the particle positively or negatively charged? Why? (b) The particle's speed is \(140 \mathrm{m} / \mathrm{s},\) the magnitude of the magnetic field is \(0.48 \mathrm{T}\), and the radius of the path is \(960 \mathrm{m}\). Determine the mass of the particle, given that its charge has a magnitude of \(8.2 \times 10^{-4} \mathrm{C}\).

A piece of copper wire has a resistance per unit length of \(5.90 \times 10^{-3} \Omega / \mathrm{m} .\) The wire is wound into a thin, flat coil of many turns that has a radius of \(0.140 \mathrm{m}\). The ends of the wire are connected to a \(12.0-\mathrm{V}\) battery. Find the magnetic field strength at the center of the coil.

The maximum torque experienced by a coil in a \(0.75-\mathrm{T}\) 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 wire carries a current of 0.66 A. This wire makes an angle of \(58^{\circ}\) with respect to a magnetic field of magnitude \(4.7 \times 10^{-5}\) T. The wire experiences a magnetic force of magnitude \(7.1 \times 10^{-5} \mathrm{N} .\) What is the length of the wire?

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