Problem 9 - University Physics with Modern Physics

Question

Deuterons in a cyclotron travel in a circle with radius Deuterons in a cyclotron travel in a circle with radius 32.0 \(\mathrm{cm}\) just before emerging from the dees. The frequency of the applied alternating voltage is 9.00 \(\mathrm{MHz}\) . Find (a) the magnetic field and (b) the kinetic energy and speed of the deuterons upon emergence.

Step-by-Step Solution

Answer

(a) Magnetic field is 1.56 T. (b) Kinetic energy is 5.50 x 10^-13 J; speed is 18.15 x 10^6 m/s.

Step 1: Formula for Magnetic Field

The magnetic force acting as centripetal force on a charged particle in a cyclotron is given by the formula: \( F = q v B = \frac{mv^2}{r} \), where \( q \) is the charge of the deuteron, \( v \) is the velocity, \( B \) is the magnetic field, \( m \) is the mass, and \( r \) is the radius. Rearranging gives us \( B = \frac{mv}{qr} \).

Step 2: Calculate Velocity Using Cyclotron Frequency

The cyclotron frequency formula is \( f = \frac{qB}{2\pi m} \). Given \( f = 9.00 \times 10^6 \mathrm{Hz} \) and knowing \( q = 1.6 \times 10^{-19} \mathrm{C} \) and \( m = 3.34 \times 10^{-27} \mathrm{kg} \), solve for \( B \): \( B = \frac{2\pi mf}{q} \). Substituting the values, we find \( B \).

Step 3: Substitute Values to Solve for Magnetic Field

After calculating the constants and substituting all values into the formula, we find \( B = 1.56 \, \mathrm{T} \) (Tesla).

Step 4: Calculate Kinetic Energy

The kinetic energy of the deuterons is given by \( K = \frac{1}{2}mv^2 \). Use \( v = r \cdot 2\pi f \) to find velocity and then calculate the kinetic energy using \( m \) and the found \( v \).

Step 5: Calculate Speed

Using the expression \( v = r \cdot 2\pi f \) where \( r = 0.32 \mathrm{m} \) and \( f = 9.00 \times 10^6 \mathrm{Hz} \), substitute into the formula to find \( v = 18.15 \times 10^6 \mathrm{m/s} \).

Step 6: Substitute into Kinetic Energy Formula

With \( m = 3.34 \times 10^{-27} \mathrm{kg} \) and \( v = 18.15 \times 10^6 \mathrm{m/s} \), substitute into \( K = \frac{1}{2}mv^2 \) to calculate \( K = 5.50 \times 10^{-13} \mathrm{J} \).

Key Concepts

Magnetic Field

In a cyclotron, a magnetic field plays a crucial role in keeping the charged particles like deuterons moving in circular paths. This magnetic field exerts a force on the charged deuterons, which provides the necessary centripetal force for circular motion. The formula used to calculate the magnetic field in a cyclotron is \[ B = \frac{mv}{qr} \]where:

• \( B \) is the magnetic field strength in Tesla (T),
• \( m \) is the mass of the particle (e.g., deuteron),
• \( v \) is the velocity of the particle,
• \( q \) is the charge of the particle, and
• \( r \) is the radius of the circular path.

By rearranging this equation, we can determine how strong the magnetic field needs to be to keep the particles on the desired path.Using these relationships, once all values are substituted, the resultant magnetic field in the described cyclotron comes out to be \( 1.56 \mathrm{T} \).
This demonstrates the power required to maintain the particles' motion within the cyclotron's dees.

Kinetic Energy

Kinetic energy is the energy an object possesses due to its motion. In the context of a cyclotron, it is essential to know because it tells us how much energy the deuterons have after being accelerated. The formula for kinetic energy (\( K \)) is:\[ K = \frac{1}{2}mv^2 \]where:

• \( m \) is the mass of the deuterons,
• \( v \) is the speed of the deuterons.

In this scenario, after calculating the speed using given parameters, the kinetic energy of the deuterons was found to be \( 5.50 \times 10^{-13} \mathrm{J}\).
This high level of energy is representative of how effective cyclotrons are at accelerating particles to high speeds necessary for nuclear research or medical applications like proton therapy.

Deuterons

Deuterons are the nuclei of deuterium, an isotope of hydrogen. They consist of one proton and one neutron, thus carrying a positive charge. In a cyclotron, deuterons are commonly used in particle acceleration due to their simple structure and significant mass. Understanding deuterons' properties is essential:

• They have a charge of \( 1.6 \times 10^{-19} \mathrm{C} \).
• Their mass is approximately \( 3.34 \times 10^{-27} \mathrm{kg} \), which is greater than that of simple hydrogen nuclei.
• They are accelerated to high velocities in cyclotrons using magnetic fields.

Their heavier mass than protons makes them advantageous in applications needing higher energy impact, such as in inducing nuclear reactions. Using deuterons, cyclotrons can probe the interactions at nuclear levels, giving insights into the structural properties of atoms.

Cyclotron Frequency

The cyclotron frequency is a fundamental concept in understanding how cyclotrons accelerate particles like deuterons. It denotes the rate at which deuterons complete cycles within the magnetic field. The formula to relate cyclotron frequency (\( f \)) with other parameters is:\[ f = \frac{qB}{2\pi m} \]where:

• \( q \) is the charge of the particle,
• \( B \) is the magnetic field strength,
• \( m \) is the mass of the particle.

For deuterons within the cyclotron, the given frequency is \( 9.00 \times 10^6 \mathrm{Hz} \), illustrating how it cycles 9 million times per second. Such high frequency ensures that particles receive rapid and repeated accelerations each time they pass through the electric field present at the gap between the cyclotron's dees. This fine-tuning of frequency is crucial to synchronizing the deuterons' motion with the cyclotron's electric field to maximize their energy gain.