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ssm 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 needed is approximately 0.366 T.

Step by step solution

01

Understand the Relationship

The mass spectrometer deflection is determined by the radius of curvature of the ions' path, which depends on the magnetic field, the charge of the ion, the velocity of the ion, and the mass of the ion.
02

Use the Formula for Radius of Curvature

The radius of curvature for an ion in a magnetic field is given by the formula \( r = \frac{mv}{qB} \), where \( m \) is the mass, \( v \) is the velocity, \( q \) is the charge, and \( B \) is the magnetic field.
03

Set Equations for Equal Radii

Since both types of ions must be detected at the same location, their path radii must be equal. Thus: \( \frac{m_1v_1}{qB_1} = \frac{m_2v_2}{qB_2} \).
04

Relate Velocities Using Energy

The ions are accelerated through the same potential difference, meaning \( \frac{1}{2}m_1v_1^2 = qV \) and \( \frac{1}{2}m_2v_2^2 = qV \), hence \( v_1 = \sqrt{\frac{2qV}{m_1}} \) and \( v_2 = \sqrt{\frac{2qV}{m_2}} \).
05

Simplify the Relationship

Substitute these velocities back into the radius formula: \( \frac{m_1\sqrt{\frac{2qV}{m_1}}}{qB_1} = \frac{m_2\sqrt{\frac{2qV}{m_2}}}{qB_2} \). Simplifying gives \( B_2 = B_1 \frac{m_2}{m_1} \sqrt{\frac{m_1}{m_2}} \).
06

Calculate Required Magnetic Field

Using the given masses and original magnetic field, calculate \( B_2 \): \( B_2 = 0.283 \times 16.63/11.65 \times (\sqrt{11.65/16.63}) \). Compute the exact numerical value.

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Key Concepts

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

Magnetic Field
A magnetic field is a region around a magnetic material or a moving electric charge within which the force of magnetism acts. In a mass spectrometer, the magnetic field is crucial as it bends the path of ions, allowing for their separation based on mass or charge.
The strength of this field, typically measured in teslas (T), determines the force acting on ions. The formula that relates the magnetic field to the motion of the ions is given by the force equation:
  • Magnetic Force = Ion Charge x Velocity x Magnetic Field
The curvature of the ion's path is directly influenced by this force. If the field is too strong, the ions curve too much; if too weak, they may not curve sufficiently to reach the detector. Thus, adjusting the magnetic field helps to target specific ions at the detector.
Ion Charge
The charge of an ion (
  • denoted by \( q \)
  • measured in coulombs (C)
) is the quantity of excess or deficient electrons that an atom carries. In the context of the mass spectrometer, ions are typically singly ionized, meaning they have a single positive charge, \( q = +e \), where \( e \) is the elementary charge, approximately \( 1.602 imes 10^{-19} \) C.
The charge affects how much an ion will be deflected when it passes through a magnetic field. More charge leads to a stronger interaction with the magnetic field, increasing the curvature of the ion's path. This aspect is fundamental when determining how different ions will behave and be detected in a mass spectrometer.
Radius of Curvature
In a mass spectrometer, the radius of curvature describes the path that an ion follows as it is deflected by a magnetic field. The formula for this radius is:\[ r = \frac{mv}{qB} \]Here, \( r \) is the radius, \( m \) is the ion's mass, \( v \) is its velocity, \( q \) is the charge, and \( B \) is the magnetic field.
The radius of curvature is crucial because it determines where an ion will strike the detector. For ions to be detected at the same spot, their path's radii must be identical. This means adjusting other factors, such as the magnetic field strength, to ensure consistent detection across different ions.
Accelerating Potential Difference
The accelerating potential difference is the voltage that accelerates ions in a mass spectrometer, influencing their velocity \( v \). It is related to the kinetic energy imparted to the ions: \[ \frac{1}{2}mv^2 = qV \]Where \( V \) is the potential difference, \( m \) is the ion mass, and \( q \) is the charge. This equation shows that the velocity of an ion is directly proportional to the square root of the voltage applied.
In simple terms, a higher potential difference means ions are accelerated to a higher velocity. This accelerated speed impacts the radius of curvature, as faster ions are less deflected by the magnetic field. By managing potential difference, we control the speed at which ions travel through the magnetic field, thereby influencing their trajectory within the mass spectrometer.

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Most popular questions from this chapter

A horizontal wire of length \(0.53 \mathrm{m},\) carrying a current of \(7.5 \mathrm{A},\) is placed in a uniform external magnetic field. When the wire is horizontal, it experiences no magnetic force. When the wire is tilted upward at an angle of \(19^{\circ},\) it experiences a magnetic force of \(4.4 \times 10^{-3} \mathrm{N}\) . Determine the magnitude of the external magnetic field.

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

A proton is projected perpendicularly into a magnetic field that has a magnitude of 0.50 T. The ficld is then adjusted so that an electron will follow a circular path of the same radius when it is projected perpendicularly into the field with the same velocity that the proton had. What is the magnitude of the field used for the electron?

ssm A charge of \(4.0 \times 10^{-6} \mathrm{C}\) is placed on a small conducting sphere that is located at the end of a thin insulating rod whose length is 0.20 \(\mathrm{m}\) . The rod rotates with an angular speed of \(\omega=150 \mathrm{rad} / \mathrm{s}\) about an axis that passes perpendicularly through its other end. Find the magnetic moment of the rotating charge. (Hint: The charge travels around a circle in a time equal to the period of the motion.)

A wire has a length of \(7.00 \times 10^{-2} \mathrm{m}\) and is used to make a a circular coil of one turn. There is a current of 4.30 \(\mathrm{A}\) in the wire. In the presence of a \(2.50-\mathrm{T}\) magnetic field, what is the maximum torque that this coil can experience?

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