Problem 26 - University Physics with Modern Physics

Question

A singly charged ion of \(^7\)Li (an isotope of lithium) has a mass of 1.16 \(\times\) 10\(^{-26}\) kg. It is accelerated through a potential difference of 220 V and then enters a magnetic field with magnitude 0.874 T perpendicular to the path of the ion. What is the radius of the ion's path in the magnetic field?

Step-by-Step Solution

Answer

The radius of the ion's path is approximately 0.0241 meters.

Step 1: Calculate the ion's velocity

To calculate the velocity of the ion, we use the fact that the kinetic energy acquired by the ion equals the electrical potential energy given by the voltage. The formula for kinetic energy (KE) is \( KE = \frac{1}{2}mv^2 \), and the electrical potential energy is \( qV \), where \( q \) is the charge of the ion (equal to the elementary charge \( e = 1.602 \times 10^{-19} \) C) and \( V \) is the potential difference. Therefore, equating the two gives:\[ \frac{1}{2}mv^2 = qV \]Solving for \( v \):\[ v = \sqrt{\frac{2qV}{m}} \]Substitute \( q = 1.602 \times 10^{-19} \) C, \( V = 220 \) V, and \( m = 1.16 \times 10^{-26} \) kg into the equation:\[ v = \sqrt{\frac{2 \times 1.602 \times 10^{-19} C \times 220 V}{1.16 \times 10^{-26} kg}} \approx 2.94 \times 10^4 \, m/s \]

Step 2: Determine the radius of the path

The centripetal force required to keep the ion in a circular path is provided by the magnetic force, which can be calculated using the equation \( F = qvB \), where \( B \) is the magnetic field. The centripetal force is also given by \( F = \frac{mv^2}{r} \), where \( r \) is the radius of the path. Equating these two expressions gives:\[ qvB = \frac{mv^2}{r} \]Solving for \( r \):\[ r = \frac{mv}{qB} \]Substitute the known values \( m = 1.16 \times 10^{-26} \), \( v = 2.94 \times 10^4 \), \( q = 1.602 \times 10^{-19} \), and \( B = 0.874 \) T:\[ r = \frac{1.16 \times 10^{-26} kg \times 2.94 \times 10^4 m/s}{1.602 \times 10^{-19} C \times 0.874 T} \approx 0.0241 \, m \]

Step 3: Conclusion

The radius of the ion's path in the magnetic field is approximately 0.0241 meters (or 2.41 cm).

Key Concepts

Singly Charged Ion

A singly charged ion means that the ion has acquired or lost a single fundamental charge, which is equal to the charge of an electron. In the case of a singly charged ion, like our Li isotope ion, it carries an elementary charge denoted as \( e \), which is approximately \( 1.602 \times 10^{-19} \; \text{C} \).
This charge is crucial, as it determines how the ion interacts with electric fields and magnetic fields.
In this example, the Li isotope has a charge resulting from either losing an electron (positively charged) or gaining one (negatively charged). Since the exercise involves ions, we assume the ion is positive, making it "+1" charged. Understanding the charge of the ion helps explain its behavior once it encounters a magnetic field, as we'll delve into in subsequent sections.

Kinetic Energy and Potential Energy

These two forms of energy are fundamental in understanding how the ion moves before entering the magnetic field.
When the ion is accelerated through a potential difference of 220 V, it gains kinetic energy, which is the energy due to its motion.
This gain in kinetic energy originates from the decrease in electrical potential energy thanks to the voltage.
The kinetic energy \( KE \) can be expressed as \( \frac{1}{2}mv^2 \), where \( m \) is the mass of the ion and \( v \) is its velocity.
Meanwhile, the potential energy \( U \) associated with the 220 V is given by the equation \( qV \), with \( q \) being the charge of the ion and \( V \) the potential difference.

By equating kinetic and potential energies, we can find the ion's velocity as:

• \( v = \sqrt{\frac{2qV}{m}} \)

This relationship is important because it bridges between static and dynamic states of the ion, allowing us to calculate how fast the ion moves as it enters the magnetic field.

Magnetic Force and Centripetal Force

When the ion enters the magnetic field, it is subjected to a magnetic force that acts perpendicularly to its velocity; this leads to a centripetal force, making it travel in a circular path.
The magnetic force \( F_{B} \) experienced by a moving charge in a magnetic field is described with \( F_{B} = qvB \), where \( B \) is the magnetic field strength.
For circular motion, the centripetal force \( F_{c} \) is given by \( F_{c} = \frac{mv^2}{r} \), where \( r \) represents the radius of the path.

Equating these forces allows us to solve for radius \( r \), which is key to understanding how the path curve is determined in the field:

• \( r = \frac{mv}{qB} \)

By substituting the known values from the exercise, the precise path radius is computed, showing how these forces dynamically maintain the ion's circular motion within the magnetic field.

Li Isotope Ion Path

The path followed by our \(^7\)Li isotope ion is greatly influenced by the balance between the kinetic forces and the magnetic field.
This balance results in a circular trajectory, whose radius depends on the mass of the ion, its velocity, its charge, and the magnetic field it encounters.
For the Li isotope with a given charge and mass, the step-by-step calculations show that its radius in a magnetic field of 0.874 T is approximately 0.0241 meters.
This path calculation is essential to various applications, such as mass spectrometry in identifying isotypes. The unique paths based on different masses allow accurate identification of isotopes and ions, thus showcasing the fundamental principles of physics and their practical applications.