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In a nuclear experiment a proton with kinetic energy \(1.2 \mathrm{MeV}\) moves in a circular path in a uniform magnetic field. What energy must (a) an alpha particle \((q=+2 e, m=4.0 \mathrm{u})\) and (b) a deuteron \((q=+e, m=2.0 \mathrm{u})\) have if they are to circulate in the same circular path?

Short Answer

Expert verified
An alpha particle needs a kinetic energy of 2.4 MeV, and a deuteron needs 0.6 MeV.

Step by step solution

01

Formula for Radius in Magnetic Field

The radius of the circular path in a uniform magnetic field can be calculated using 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 strength. Since the path radius remains the same for different particles, we equate their expressions for radius.
02

Express Velocity in Terms of Kinetic Energy

For each particle, the kinetic energy \( K = \frac{1}{2}mv^2 \) can be rewritten to express the velocity: \( v = \sqrt{\frac{2K}{m}} \). This equation will allow us to find the kinetic energy needed to maintain the same radius for different particles.
03

Equate Radii for Proton and Alpha Particle

The radius for the proton and the alpha particle must be the same. Using \( r = \frac{m_{p}v_{p}}{q_{p}B} = \frac{m_{\alpha}v_{\alpha}}{q_{\alpha}B} \), we substitute \( v \) from the energy expression. Simplifying gives: \( \frac{m_{p}olinebreak\sqrt{\frac{2K_{p}}{m_{p}}}}{q_{p}} = \frac{m_{\alpha}\sqrt{\frac{2K_{\alpha}}{m_{\alpha}}}}{q_{\alpha}} \).
04

Solve for Alpha Particle's Kinetic Energy

Rearranging and simplifying from Step 3: \( q_{\alpha}m_{p}^2 = q_{p}m_{\alpha}^2 \frac{K_{\alpha}}{K_{p}} \). Solve for \( K_{\alpha} \): \( K_{\alpha} = K_{p} \cdot \frac{q_{\alpha}m_{p}^2}{q_{p}m_{\alpha}^2} = 1.2 \cosity \cdot \frac{2(1.0)}{(4.0)} = 0.6 \cosity \cdot 4 = 2.4 \text{ MeV} \).
05

Equate Radii for Proton and Deuteron

Similarly, for the deuteron: \( \frac{m_{p}v_{p}}{q_{p}} = \frac{m_{d}v_{d}}{q_{d}} \). Substitute \( v \) from the energy expression: \( \frac{m_{p}\sqrt{\frac{2K_{p}}{m_{p}}}}{q_{p}} = \frac{m_{d}\sqrt{\frac{2K_{d}}{m_{d}}}}{q_{d}} \).
06

Solve for Deuteron's Kinetic Energy

Simplifying from the expression above: \( q_{d}m_{p}^2 = q_{p}m_{d}^2 \frac{K_{d}}{K_{p}} \). Solve for \( K_{d} \): \( K_{d} = K_{p} \cdot \frac{q_{d}m_{p}^2}{q_{p}m_{d}^2} = 1.2 \cosity \cdot \frac{(1.0)}{(2.0)} = 0.6 \text{ MeV} \).

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Key Concepts

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

Kinetic Energy
Kinetic energy is a crucial concept in understanding the movement of particles in nuclear physics. It is the energy that a particle possesses due to its motion. The equation for kinetic energy is given as:
  • \( K = \frac{1}{2}mv^2 \)
where \( K \) is the kinetic energy, \( m \) is the mass of the particle, and \( v \) is its velocity. In the context of nuclear experiments, particles such as protons, alpha particles, and deuterons are often accelerated to high speeds, imparting them with significant kinetic energy.
Understanding kinetic energy is essential when calculating how particles move and interact within a magnetic field, especially when attempting to maintain a specified path. In our exercise, the goal is to find out what kinetic energy different particles require to follow the same circular path as an initial reference particle.
Circular Motion
Circular motion is the motion of a particle moving along the circumference of a circle. When particles like protons or deuterons are subject to a uniform magnetic field, they travel in circular paths. This happens because the magnetic force acts perpendicular to the velocity of the particles, changing their direction but not their speed.
The radius of this circular path can be calculated using the formula:
  • \( r = \frac{mv}{qB} \)
Here, \( r \) is the radius, \( m \) is the mass, \( v \) is the velocity, \( q \) is the charge of the particle, and \( B \) is the magnetic field strength. This formula allows us to equate the radii of different particles moving in the same magnetic field, making it possible to compare their kinetic energies to maintain the same circular path.
Magnetic Field
A magnetic field is a force field that surrounds magnets and moving electric charges. It affects charged particles, like protons or alpha particles, causing them to move in circular paths when they enter the field.
  • This field applies a force perpendicular to the direction of the particle's motion, causing circular motion.
  • The strength of this field is denoted as \( B \), and it's a crucial factor in determining the radius of the particle's path.
In our exercise, the magnetic field is uniform, creating consistent conditions that allow different particles to be compared and calculated regarding their kinetic energy requirements. This consistency is key to solving problems where particles need to maintain a specific path within a field.
Alpha Particle
An alpha particle consists of two protons and two neutrons, giving it a charge of \(+2e\) and a mass of \(4.0 \mathrm{u}\). Alpha particles are one of the common particles studied in nuclear physics because of their high interaction cross-section and distinct charge and mass properties.
In the context of a magnetic field, an alpha particle will curve according to its charge and mass. Compared to a proton, because it is heavier, it will require a different kinetic energy to circulate the same path within the same magnetic field.
  • They are significantly more massive than protons, which factors into the equations governing their motion in fields.
  • Understanding these factors is key when adapting solutions for different particles in similar scenarios, as seen in our exercise solution.
Deuteron
A deuteron is the nucleus of deuterium, an isotope of hydrogen, and consists of one proton and one neutron. It has a charge of \(+e\) and a mass of \(2.0 \mathrm{u}\). In nuclear physics experiments, deuterons are often used due to their simple structure and the fact that they are slightly heavier than protons.
In our exercise, we calculated how the deuteron needs a different kinetic energy compared to a proton to maintain the same circular path. Because of its unique mass and charge, the equations for circular motion in a magnetic field need adjustment to get accurate results for the deuteron.
  • Understanding these characteristics helps when calculating motion and energy requirements for different particles.
  • This opens up a deeper understanding of how alterations in mass and charge influence a particle's trajectory in magnetic fields.

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Most popular questions from this chapter

A long, rigid conductor, lying along an \(x\) axis, carries a current of \(7.0 \mathrm{~A}\) in the negative \(x\) direction. A magnetic field \(\vec{B}\) is present, given by \(\vec{B}=3.0 \mathrm{i}+8.0 x^{2} \hat{\mathrm{j}}\), with \(x\) in meters and \(\vec{B}\) in milliteslas. Find, in unit-vector notation, the force on the \(2.0 \mathrm{~m}\) segment of the conductor that lies between \(x=1.0 \mathrm{~m}\) and \(x=3.0 \mathrm{~m}\).

(a) What uniform magnetic field, applied perpendicular to a beam of electrons moving at \(1.30 \times 10^{6} \mathrm{~m} / \mathrm{s}\), is required to make the electrons travel in a circular arc of radius \(0.500 \mathrm{~m} ?(\mathrm{~b})\) What is the period of the motion?

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.

A horizontal power line carries a current of 7000 A from south to north. Earth's magnetic field \((60.0 \mu \mathrm{T})\) is directed toward the north and inclined downward at \(70.0^{\circ}\) to the horizontal. Find the (a) magnitude and (b) direction of the magnetic force on \(100 \mathrm{~m}\) of the line due to Earth's field.

A wire \(2.30 \mathrm{~m}\) long carries a current of \(13.0 \mathrm{~A}\) and makes an angle of \(35.0^{\circ}\) with a uniform magnetic field of magnitude \(B=1.50 \mathrm{~T}\). Calculate the magnetic force on the wire.

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