Ch. 27 - Magnetism

Giancoli Douglas - Physics for Scientists and Engineers 5th edition

Giancoli Douglas5th editionPhysics for Scientists and EngineersISBN: 9780137488179Not the one you use?Change textbook

Giancoli Douglas 5th edition

Ch. 27 - Magnetism

Problem 61

Chapter 26, Problem 61

A mass spectrometer is monitoring air pollutants. It is difficult, however, to separate molecules of nearly equal mass such as CO (28.0106 u) and N₂ (28.0134 u). How large a radius of curvature must a spectrometer have (Fig. 27–34) if these two molecules are to be separated on the detector by 0.50 mm?

Verified step by step guidance

Understand the problem: A mass spectrometer separates particles based on their mass-to-charge ratio (m/q) by bending their paths in a magnetic field. The radius of curvature (r) of the path depends on the mass, charge, velocity, and magnetic field strength. The goal is to calculate the radius of curvature required to separate CO and N₂ molecules by 0.50 mm on the detector.

Write the formula for the radius of curvature in a mass spectrometer:

r = \frac{mv}{qB}

, where

m

is the mass of the particle,

v

is its velocity,

q

is its charge, and

B

is the magnetic field strength. Since the charge and velocity are the same for both molecules, the difference in their radii is due to the difference in their masses.

Express the separation on the detector in terms of the difference in radii: The separation on the detector is given by

d = 2(r_{N} - r_{CO})

, where

r_{N}

and

r_{CO}

are the radii of curvature for N₂ and CO, respectively. The factor of 2 accounts for the fact that the particles travel in a circular arc.

Substitute the mass difference into the radius formula: The difference in radii can be expressed as

r_{N} - r_{CO} = \frac{v}{qB} (m_{N} - m_{CO})

. Use the given mass values for CO and N₂ to calculate the mass difference:

m_{N} - m_{CO} = 28.0134 - 28.0106

(in atomic mass units).

Solve for the radius of curvature: Rearrange the separation formula to find the radius of curvature:

r = \frac{d}{2 (m_{N} - m_{CO})}

. Substitute the given separation distance (0.50 mm) and the calculated mass difference into the equation. Ensure all units are consistent (e.g., convert mm to meters and atomic mass units to kilograms) before solving.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Mass Spectrometry

Mass spectrometry is an analytical technique used to measure the mass-to-charge ratio of ions. In this process, molecules are ionized and accelerated through an electric field, causing them to travel in a curved path within a magnetic field. The radius of curvature of their paths depends on their mass and charge, allowing for the separation of different species based on their mass.

Radius of Curvature

The radius of curvature in a mass spectrometer refers to the radius of the circular path that ions take when subjected to a magnetic field. This radius is influenced by the mass of the ions and their velocity, as described by the Lorentz force. A larger radius indicates a greater separation between ions of different masses, which is crucial for distinguishing between closely related molecules like CO and N₂.

Resolution in Mass Spectrometry

Resolution in mass spectrometry is the ability to distinguish between two ions of similar mass. It is defined as the smallest mass difference that can be detected and is influenced by factors such as the geometry of the spectrometer and the radius of curvature. In the context of the question, achieving a resolution that allows for a separation of 0.50 mm between CO and N₂ requires careful consideration of the spectrometer's design and operational parameters.

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Related Practice

Textbook Question

Near the equator, the Earth’s magnetic field points almost horizontally to the north and has magnitude B = 0.50 x 10⁻⁴ T. What should be the magnitude and direction for the velocity of an electron if its weight is to be exactly balanced by the magnetic force?

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Textbook Question

Suppose the electric field between the electric plates in the mass spectrometer of Fig. 27–34 is 2.84 x 10⁴ V/m and the magnetic fields are B = B'= 0.58 T. The source contains carbon isotopes of mass numbers 12, 13, and 14 from a long-dead piece of a tree. (To estimate atomic masses, multiply by 1.67 x 10⁻²⁷ kg.) Does it matter if the ion charge is positive (lost electrons) or negative (gained electrons)? Explain.

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Textbook Question

One form of mass spectrometer accelerates ions by a voltage V before they enter a magnetic field B. The ions are assumed to start from rest. Show that the mass of an ion is m = qB²R²/2V, where R is the radius of the ions’ path in the magnetic field and q is their charge.

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Textbook Question

Suppose the electric field between the electric plates in the mass spectrometer of Fig. 27–34 is 2.84 x 10⁴ V/m and the magnetic fields are B = B'= 0.58 T. The source contains carbon isotopes of mass numbers 12, 13, and 14 from a long-dead piece of a tree. (To estimate atomic masses, multiply by 1.67 x 10⁻²⁷ kg .) How far apart are the marks formed by the singly charged ions of each type on a detector or photographic film? What if the ions were doubly charged?

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Textbook Question

A uniform conducting rod of length ℓ and mass m sits atop a fulcrum, which is placed a distance ℓ/4 from the rod’s left-hand end and is immersed in a uniform magnetic field of magnitude B directed into the page (Fig. 27–54). An object whose mass M is 7.0 times greater than the rod’s mass is hung from the rod’s left-hand end. What current (direction and magnitude) should flow through the rod in order for it to be “balanced” (i.e., be at rest horizontally) on the fulcrum? (Flexible connecting wires which exert negligible force on the rod are not shown.)

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Textbook Question

The magnetic field B at the center of a circular coil of wire carrying a current I (as in Fig. 27–9) is B = (μ₀NI) / 2r where N is the number of loops in the coil and r is its radius. Imagine a simple model in which the Earth’s magnetic field of about 1 G ( = 1 x 10⁻⁴ T) near the poles is produced by a single current loop around the equator. Roughly estimate the current this loop would carry.

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