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Physicist S. A. Goudsmit devised a method for measuring the mass of heavy ions by timing their period of revolution in a known magnetic field. A singly charged ion of iodine makes \(7.00\) rev in a \(45.0 \mathrm{mT}\) field in \(1.29 \mathrm{~ms}\). Calculate its mass in atomic mass units.

Short Answer

Expert verified
The mass of the iodine ion is approximately 127.1 amu.

Step by step solution

01

Understand the given values

The problem provides the following values: the magnetic field strength \(B = 45.0 \text{ mT} = 0.045 \text{ T}\), the number of revolutions \(n = 7\), and the time \(t = 1.29 \text{ ms} = 1.29 \times 10^{-3} \text{ s}\).
02

Calculate the frequency of revolution

The frequency of revolution \(f\) is the number of revolutions per second. Calculate it using \( f = \frac{n}{t} = \frac{7}{1.29 \times 10^{-3}} \). This gives \( f \approx 5426.36 \text{ Hz}\).
03

Calculate the angular frequency

The angular frequency \( \omega \) is given by \( \omega = 2\pi f \). Substitute the frequency to get \( \omega = 2\pi \times 5426.36 \approx 34109.03 \text{ rad/s}\).
04

Use cyclotron formula

The mass of the ion can be found using the formula for cyclotron frequency: \( \omega = \frac{qB}{m} \). Rearrange for \( m \): \( m = \frac{qB}{\omega} \), where \( q = 1.60 \times 10^{-19} \text{ C} \) for a singly charged ion.
05

Calculate the mass in kg

Substitute \( q = 1.60 \times 10^{-19} \text{ C} \), \( B = 0.045 \text{ T} \), and \( \omega = 34109.03 \text{ rad/s} \) into \( m = \frac{qB}{\omega} \) to find \( m_{\text{ion}} \approx \frac{1.60 \times 10^{-19} \times 0.045}{34109.03} \) kg. This results in \( m_{\text{ion}} \approx 2.11 \times 10^{-25} \text{ kg}\).
06

Convert the mass to atomic mass units

Convert the mass from kilograms to atomic mass units using \( 1 \text{ amu} = 1.66 \times 10^{-27} \text{ kg} \). Thus, \( m_{\text{amu}} = \frac{2.11 \times 10^{-25}}{1.66 \times 10^{-27}} \) amu. This gives \( m_{\text{amu}} \approx 127.1 \text{ amu} \).

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Key Concepts

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

Cyclotron Formula
The cyclotron formula is crucial in mass spectrometry for determining the mass of charged particles. It connects the mass of the particle, the charge, the strength of the magnetic field, and the angular frequency of its motion. The formula is given by:
  • \( \omega = \frac{qB}{m} \)
Here, \( \omega \) is the angular frequency, \( q \) is the charge of the ion, \( B \) is the magnetic field strength, and \( m \) is the mass of the ion.
This formula is derived from the principle that a charged particle moving perpendicularly through a magnetic field experiences a centripetal force, keeping it in circular motion. By equating this force to the magnetic Lorentz force, we get the relationship between the angular frequency and mass.
In our example, rearranging to solve for mass, \( m = \frac{qB}{\omega} \), allows us to plug in known values of charge, magnetic field strength, and angular frequency to find the ion's mass. This process provides a pathway to measure unknown masses with high precision in a controlled field.
Angular Frequency
Angular frequency, denoted as \( \omega \), is a measure of how fast an object rotates or revolves. It is expressed in radians per second (rad/s).
This is particularly useful in cyclotron applications because it relates directly to the number of revolutions per time unit.
  • Formula: \( \omega = 2\pi f \)
In this formula, \( f \) represents the frequency of revolution in hertz (Hz), or cycles per second. By multiplying the linear frequency by \( 2\pi \), we convert cycles into radians, providing a more complete picture of motion.
In our iodine ion example, we start by determining its revolution frequency with \( f = \frac{n}{t} \), where \( n \) is the number of revolutions and \( t \) is the time. After computing \( f \), we use it to find \( \omega \). This angular frequency then becomes crucial in using the cyclotron formula to determine mass.
Magnetic Field Strength
Magnetic field strength, symbolized as \( B \), is a measure of the magnetic force exerted in a given area. It is typically measured in teslas (T) or milliteslas (mT).
The strength of the magnetic field influences how charged particles move within it, particularly affecting their circular path in instruments like cyclotrons.
In mass spectrometry, a known, uniform magnetic field helps facilitate precise calculations of particle mass. In our example:
  • \( B = 45.0 \text{ mT} = 0.045 \) T
This value is an essential input for the cyclotron formula. A more substantial magnetic field will increase the force on the charged particles, leading to a smaller radius of rotation, allowing for better separation and measurement resolution of ions. Understanding this concept is vital for realizing how instruments can be tuned for specific analytical tasks.

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Most popular questions from this chapter

In a nuclear experiment a proton with kinetic energy \(1.0 \mathrm{MeV}\) moves in a circular path in a uniform magnetic field. What energy must (a) an alpha particle \((q=+2 e, m=4.0 \mathrm{u})\) and (b) a deuteron \((q=+e, m=2.0 \mathrm{u})\) have if they are to circulate in the same circular path?

At time \(t=0\), an electron with kinetic energy \(12 \mathrm{keV}\) moves through \(x=0\) in the positive direction of an \(x\) axis that is parallel to the horizontal component of Earth's magnetic field \(\vec{B}\). The field's vertical component is downward and has magnitude \(55.0 \mu \mathrm{T}\). (a) What is the magnitude of the electron's acceleration due to \(\vec{B}\) ? (b) What is the electron's distance from the \(x\) axis when the electron reaches coordinate \(x=20 \mathrm{~cm} ?\)

A proton moves through a uniform magnetic field given by \(\vec{B}=(10 \hat{\mathrm{i}}-20 \hat{\mathrm{j}}+30 \hat{\mathrm{k}}) \mathrm{mT}\). At time \(t_{1}\), the proton has a velocity given by \(\vec{v}=v_{x} \hat{i}+v_{y} \hat{j}+(2.0 \mathrm{~km} / \mathrm{s}) \hat{\mathrm{k}}\) and the magnetic force on the proton is \(\vec{F}_{B}=\left(4.0 \times 10^{-17} \mathrm{~N}\right) \hat{\mathrm{i}}+\left(2.0 \times 10^{-17} \mathrm{~N}\right) \hat{\mathrm{j}} .\) At that instant, what are (a) \(v_{x}\) and (b) \(v_{y}\) ?

A wire \(1.80 \mathrm{~m}\) long carries a current of \(13.0 \mathrm{~A}\) and makes an angle of \(35.0^{\circ}\) with a uniform magnetic field of magnitude \(\underline{B}=\) \(1.50 \mathrm{~T}\). Calculate the magnetic force on the wire.

An electron moves in a circle of radius \(r=5.29 \times 10^{-11} \mathrm{~m}\) with speed \(2.19 \times 10^{6} \mathrm{~m} / \mathrm{s}\). Treat the circular path as a current loop with a constant current equal to the ratio of the electron's charge magnitude to the period of the motion. If the circle lies in a uniform magnetic field of magnitude \(B=7.10 \mathrm{mT}\), what is the maximum possible magnitude of the torque produced on the loop by the field?

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