Problem 26 - University Physics with Modern Physics

Question

A singly charged ion of \(^{7} \mathrm{Li}\) (an isotope of lithium) has a mass of \(1.16 \times 10^{-26} \mathrm{kg}\) . It is accelerated through a potential dif- ference of 220 \(\mathrm{V}\) and then enters a magnetic field with magnitude 0.723 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 2.47 cm.

Step 1: Understand the Concepts and Equations

The problem involves the motion of a charged ion in a magnetic field. The radius of the ion's path in the magnetic field can be determined using the formula for the radius of the circular motion, given by \( r = \frac{mv}{qB} \), where \( m \) is the mass, \( v \) is the velocity, \( q \) is the charge, and \( B \) is the magnetic field strength.

Step 2: Calculate the Ion's Velocity

When the ion is accelerated through the potential difference \( V \), it gains kinetic energy equal to the electrical energy: \( qV = \frac{1}{2}mv^2 \). Rearranging gives \( v = \sqrt{\frac{2qV}{m}} \). For a singly charged ion, \( q = 1.6 \times 10^{-19} \) C. Substitute the given values: \( V = 220 \) V, \( m = 1.16 \times 10^{-26} \) kg, and \( q = 1.6 \times 10^{-19} \) C into the formula.

Step 3: Substitute Values to Find Velocity

\[ v = \sqrt{\frac{2 \times 1.6 \times 10^{-19} \times 220}{1.16 \times 10^{-26}}} \] Calculate to get \( v = \sqrt{\frac{7.04 \times 10^{-17}}{1.16 \times 10^{-26}}} = \sqrt{6.068 \times 10^{9}} \approx 2.46 \times 10^5 \) m/s.

Step 4: Calculate the Radius of the Ion's Path

Substitute the values of \( m, v, q, \) and \( B \) into the formula for the radius: \( r = \frac{mv}{qB} \). \( m = 1.16 \times 10^{-26} \) kg, \( v = 2.46 \times 10^5 \) m/s, \( q = 1.6 \times 10^{-19} \) C, and \( B = 0.723 \) T.

Step 5: Substitute Values and Simplify

\[ r = \frac{1.16 \times 10^{-26} \times 2.46 \times 10^5}{1.6 \times 10^{-19} \times 0.723} \] Calculate to get \( r = \frac{2.8536 \times 10^{-21}}{1.1568 \times 10^{-19}} \approx 2.47 \times 10^{-2} \) m or 2.47 cm.

Key Concepts

Circular Motion

When particles like ions travel in a magnetic field, they usually follow a circular path. This happens because the magnetic force acts as a centripetal force, pulling the ion towards the center of its circular path.
The formula for the radius of this circular motion is:

• \( r = \frac{mv}{qB} \)

Here, \( r \) is the radius of the circle, \( m \) is the mass of the ion, \( v \) is its velocity, \( q \) is its charge, and \( B \) is the magnetic field strength.
This equation comes from equating magnetic force with centripetal force \( F = \frac{mv^2}{r} \). Solving for \( r \) shows how mass, velocity, and field strength affect the size of the ion's path.

Kinetic Energy

Kinetic energy is a form of energy that an object has due to its motion. When an ion is accelerated by a potential difference \( V \), it gains kinetic energy equal to the energy it absorbs from the electric field.
The change in kinetic energy can be calculated using the equation:

• \( KE = \frac{1}{2}mv^2 \)

But in this scenario, the energy gain from a potential difference \( V \) is given by:

• \( qV = \frac{1}{2}mv^2 \).

By rearranging this equation to find \( v \), the ion's velocity, we get:

• \( v = \sqrt{\frac{2qV}{m}} \)

Understanding how kinetic energy transforms and determines velocity helps explain the ion's behavior in magnetic fields.

Potential Difference

Potential difference, often measured in volts, is the amount of electrical energy provided per charge. It's the driving force that accelerates ions, giving them energy to change speed and direction.
In this exercise, the ion is accelerated through a potential difference of \( 220 \) volts.

• The potential difference transforms into kinetic energy: \( qV \).
• For a singly charged ion: \( q = 1.6 \times 10^{-19} \) C, thus \( qV = 1.6 \times 10^{-19} \times 220 \).

This equation links potential energy to the kinetic energy the ion gains, showing how potential difference influences the speed of charged particles. Understanding this concept is crucial for predicting particle motion.

Magnetic Field Strength

Magnetic field strength is represented by \( B \) and is measured in Tesla (T). It is a key factor that affects how a moving charged particle behaves in a magnetic field.
The magnetic field exerts a force on the charged particle, redirecting its path into a circular motion.

• The formula for magnetic force is \( F = qvB \), leading to circular motion \( F = \frac{mv^2}{r} \).
• \( B \) directly influences how tight or wide the path is; stronger fields lead to tighter circles.

In this problem, \( B \) is given as \( 0.723 \) Tesla. Understanding magnetic field strength is essential to predicting the path and radius of charged particles like ions in magnetic fields.