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Chapter 28: Problem 59

A mass spectrometer (Fig. 28-12) is used to separate uranium ions of mass \(3.92 \times 10^{-25} \mathrm{~kg}\) and charge \(3.20 \times 10^{-19} \mathrm{C}\) from related species. The ions are accelerated through a potential difference of \(180 \mathrm{kV}\) and then pass into a uniform magnetic field, where they are bent in a path of radius \(1.00 \mathrm{~m}\). After traveling through \(180^{\circ}\) and passing through a slit of width \(1.00 \mathrm{~mm}\) and height \(1.00 \mathrm{~cm}\), they are collected in a cup. (a) What is the magnitude of the (perpendicular) magnetic field in the separator? If the machine is used to separate out \(100 \mathrm{mg}\) of material per hour, calculate (b) the current of the desired ions in the machine and (c) the thermal energy produced in the cup in \(1.00 \mathrm{~h}\).

Short Answer

Expert verified
The magnetic field is 0.104 T, the current is 22.7 A, and thermal energy is about 1.47 脳 10鈦� J.

Step by step solution

01

Calculate the velocity of the ions

The ions are accelerated through a potential difference, which can be used to find their velocity. The kinetic energy gained by the ions is equal to the electric potential energy, so \( \frac{1}{2}mv^2 = qV \). Solving for \( v \), \( v = \sqrt{\frac{2qV}{m}} \).Substitute \( q = 3.20 \times 10^{-19} \mathrm{~C} \), \( V = 180 \times 10^3 \mathrm{~V} \), and \( m = 3.92 \times 10^{-25} \mathrm{~kg} \):\[ v = \sqrt{\frac{2 \times 3.20 \times 10^{-19} \times 180 \times 10^3}{3.92 \times 10^{-25}}} \approx 8.52 \times 10^6 \mathrm{~m/s} \].
02

Determine the magnetic field strength

The ions are subjected to a magnetic force which provides the centripetal force necessary for circular motion: \( qvB = \frac{mv^2}{r} \).Solving for \( B \), we have \( B = \frac{mv}{qr} \).Using \( m = 3.92 \times 10^{-25} \mathrm{~kg} \), \( v = 8.52 \times 10^6 \mathrm{~m/s} \), \( q = 3.20 \times 10^{-19} \mathrm{~C} \), and \( r = 1.00 \mathrm{~m} \):\[ B = \frac{3.92 \times 10^{-25} \times 8.52 \times 10^6}{3.20 \times 10^{-19} \times 1.00} \approx 0.104 \mathrm{~T} \].
03

Calculate the ion current

First, find the number of ions needed to separate 100 mg of material per hour. The number of ions \( N \) is calculated by \( N = \frac{m_{total}}{m_{ion}} \), where \( m_{total} = 100 \times 10^{-3} \mathrm{~kg} \) and the mass of each ion is \( m_{ion} = 3.92 \times 10^{-25} \mathrm{~kg} \).\[ N = \frac{100 \times 10^{-3}}{3.92 \times 10^{-25}} \approx 2.55 \times 10^{23} \text{ ions/hour} \].Convert this to ions per second for current calculation: \[ \frac{2.55 \times 10^{23}}{3600} \approx 7.08 \times 10^{19} \text{ ions/s} \].The current \( I \) is given by \( I = nq \), where \( n = 7.08 \times 10^{19} \text{ ions/s} \):\[ I = 7.08 \times 10^{19} \times 3.20 \times 10^{-19} \approx 22.7 \mathrm{~A} \].
04

Calculate thermal energy produced

The thermal energy produced in the cup in one hour comes from the kinetic energy of the ions. The energy per ion is \( qV \), and for \( N = 2.55 \times 10^{23} \) ions:Total energy \( E = N \cdot qV \).\[ E = 2.55 \times 10^{23} \times 3.20 \times 10^{-19} \times 180 \times 10^3 \approx 1.47 \times 10^7 \mathrm{~J} \].

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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 is crucial for determining the path of ions. When ions enter the magnetic field perpendicularly, they experience a force that causes them to move in a circular path. This force is the magnetic force and is calculated using the formula \( F = qvB \), where \( q \) is the charge of the ion, \( v \) is the velocity, and \( B \) is the magnetic field strength.

The radius of the ion's path depends on the balance between this magnetic force and the centripetal force needed to maintain the circular motion. As given by the formula \( qvB = \frac{mv^2}{r} \), where \( m \) is mass and \( r \) is the radius.

By rearranging the formula to solve for the magnetic field, we see \[ B = \frac{mv}{qr} \]. When ions of known velocity and mass are present, as in a mass spectrometer, we can manipulate this formula to find the essential magnetic field needed to ensure they follow a specified path radius. This calculation is crucial for separating ions based on mass and charge effectively.
Ion Current
Ion current is a measure of the flow of ions through the mass spectrometer. It is determined by the number of ions that pass per second and their charge. This can be expressed with the formula \( I = nq \), where \( n \) is the number of ions per second, and \( q \) is the charge of each ion.

To calculate the ion current for separating a given mass of material, such as the 100 mg per hour in the exercise, we first determine how many ions this mass represents by using \( N = \frac{m_{total}}{m_{ion}} \), where \( m_{total} \) is the overall mass, and \( m_{ion} \) is the mass of a single ion.

Converting this number to a per-second basis allows us to compute the ion current, which is a critical consideration in evaluating the efficiency and operational parameters of the mass spectrometer. Understanding ion current is key to optimizing the separation process and achieving accurate and consistent results.
Thermal Energy
The thermal energy in a mass spectrometer is generated as the ions are collected. This occurs when the kinetic energy of the ions is transferred to the collection cup, effectively converting to thermal energy upon impact.

The energy per ion is given by the potential energy \( qV \), with \( q \) being the charge of the ion and \( V \) being the accelerating potential difference. To find the total thermal energy produced, we calculate \( E = N \cdot qV \), where \( N \) is the total number of ions impacting the collector.

In practice, this energy is significant, as it dictates the thermal management needs in the system. For applications like uranium ion separation, where high quantities of ions are processed, understanding and controlling thermal energy is essential to maintain the integrity and precision of the instrument.
Kinetic Energy
Kinetic energy in a mass spectrometer context refers to the energy that ions gain as they are accelerated by the electric potential difference. This energy is crucial for determining the velocity and subsequent path of the ions as they traverse the magnetic field.

The fundamental formula for kinetic energy is \( \,\frac{1}{2}mv^2 \), where \( m \) is the mass of the ion and \( v \) is its velocity. In the spectrometer, this kinetic energy equals the electric potential energy, given by \( qV \). Solving for velocity allows us to understand how ions accelerate through the system.

This energy transfer from electric to kinetic is a cornerstone principle enabling the precise measurement and separation of distinct ions. Knowing the kinetic energy helps in configuring the system to achieve desired motion parameters and separation criteria.
Potential Difference
The potential difference in a mass spectrometer is a driving factor that accelerates ions from a resting state to the speeds necessary for separation.

Defined as the voltage applied across the ion source and acceleration stage, it provides the energy required to move ions into the magnetic field. The work done on an ion by this potential difference is given by \( qV \), where \( q \) is the charge and \( V \) is the potential difference.

This quantity directly affects the kinetic energy of the ions, as the potential energy gained is converted into kinetic energy. The magnitude of the potential difference has a profound impact on the velocity and trajectory of the ions, thus playing a critical role in the accuracy and efficiency of the mass spectrometry process. Understanding and adjusting the potential difference is essential for tuning the instrument to the specific needs of a particular analysis or separation task.

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Most popular questions from this chapter

In Fig. 28-42, an electron with an initial kinetic energy of \(5.0 \mathrm{keV}\) enters region 1 at time \(t=0\). That region contains a uniform magnetic field directed into the page, with magnitude \(0.010 \mathrm{~T}\). The electron goes through a half-circle and then exits region 1, headed toward region 2 across a gap of \(25.0 \mathrm{~cm}\). There is an electric potential difference \(\Delta V=2000 \mathrm{~V}\) across the gap, with a polarity such that the electron's speed increases uniformly as it traverses the gap. Region 2 contains a uniform magnetic field directed out of the page, with magnitude \(0.020 \mathrm{~T}\). The electron goes through a half-circle and then leaves region 2. At what time \(t\) does it leave?

A current loop, carrying a current of \(7.5 \mathrm{~A}\), is in the shape of a right triangle with sides 30,40 , and \(50 \mathrm{~cm}\). The loop is in a uniform magnetic field of magnitude \(120 \mathrm{mT}\) whose direction is parallel to the current in the \(50 \mathrm{~cm}\) side of the loop. Find the magnitude of (a) the magnetic dipole moment of the loop and (b) the torque on the loop.

A proton circulates in a cyclotron, beginning approximately at rest at the center. Whenever it passes through the gap between dees, the electric potential difference between the dees is \(350 \mathrm{~V}\). (a) By how much does its kinetic energy increase with each passage through the gap? (b) What is its kinetic energy as it completes 100 passes through the gap? Let \(r_{100}\) be the radius of the proton's circular path as it completes those 100 passes and enters a dee, and let \(r_{101}\) be its next radius, as it enters a dee the next time. (c) By what percentage does the radius increase when it changes from \(r_{100}\) to \(r_{101}\) ? That is, what is percentage increase \(=\frac{r_{101}-r_{100}}{r_{100}} 100 \% ?\)

An electron that has an instantaneous velocity of $$ \vec{v}=\left(-5.0 \times 10^{6} \mathrm{~m} / \mathrm{s}\right) \hat{\mathrm{i}}+\left(3.0 \times 10^{6} \mathrm{~m} / \mathrm{s}\right) \hat{\mathrm{j}} $$ is moving through the uniform magnetic field \(\vec{B}=(0.030 \mathrm{~T}) \hat{\mathrm{i}}-\) \((0.15 \mathrm{~T}) \hat{j}\). (a) Find the force on the electron due to the magnetic field. (b) Repeat your calculation for a proton having the same velocity.

An electron moves in a circle of radius \(r=5.29 \times 10^{-11} \mathrm{~m}\) with speed \(4.12 \times 10^{6} \mathrm{~m} / \mathrm{s}\). 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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