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Chapter 27: Problem 22

In a cyclotron, the orbital radius of protons with energy 300 keV is 16.0 cm. You are redesigning the cyclotron to be used instead for alpha particles with energy 300 keV. An alpha particle has charge \(q =\) +2\(e\) and mass \(m =\) 6.64 \(\times\) 10\(^{-27}\) kg. If the magnetic field isn't changed, what will be the orbital radius of the alpha particles?

Short Answer

Expert verified
The orbital radius for alpha particles is 16.0 cm, the same as for protons.

Step by step solution

01

Understand the Problem

We need to find the new orbital radius of alpha particles under the same magnetic field conditions as the protons. Both particles have the same kinetic energy of 300 keV. We need to calculate the radius using the given properties of alpha particles and compare it to the original radius for protons.
02

Recall the Kinetic Energy Formula

Kinetic energy for any particle is given by the formula: \( KE = \frac{1}{2} m v^2 \). However, it's convenient to use a relation involving magnetic force for particles in a cyclotron. The energy given is 300 keV, which we should convert to joules: \( 1 \text{ keV} = 1.602 \times 10^{-16} \text{ J} \) so \( 300 \text{ keV} = 4.806 \times 10^{-14} \text{ J} \).
03

Use the Formula for Cyclotron Radius

The formula for the radius \( r \) of a path in a magnetic field \( B \) is \( r = \frac{mv}{qB} \), where \( m \) is the mass of the particle and \( q \) is its charge. We need to find the velocity \( v \) using kinetic energy for alpha particles calculated previously.
04

Calculate Velocity for Alpha Particles

Using \( KE = \frac{1}{2} m v^2 \), solve for \( v \): \[ v = \sqrt{\frac{2 \times 4.806 \times 10^{-14} \text{ J}}{6.64 \times 10^{-27} \text{ kg}}} \approx 9.48 \times 10^{6} \text{ m/s}. \]
05

Calculate the Orbital Radius

Now we use the radius formula for the cyclotron: \[ r = \frac{m \cdot v}{q \cdot B} \] Note that \( r \propto \frac{m \cdot v}{q} \), and we do not change \( B \). Let's assume the same \( B \) value used for protons.The proton mass \( m_p \) and charge \( q_p \) were used earlier; for protons, \( r_p = \frac{300 \text{ keV} \cdot m_p}{q_p \cdot B} \).For alpha particles:\[ r_{\alpha} = \frac{4 \times m_p \cdot v}{2 \times q_e \cdot B}. \]Alpha particle has 4 times the mass of proton and twice the charge (2\( q_e \)). Hence,\[ r_{\alpha} = \frac{2 \times r_p}{2} = r_p = 16.0 \text{ cm} \] (because \( v_{\alpha} \approx v_p \) due to \( KE \) equality).
06

Conclusion

The orbital radius of the alpha particles will be the same as that of the protons, assuming the magnetic field and kinetic energy remain the same for both particles.

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Key Concepts

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

Orbital Radius
The orbital radius in a cyclotron is the path circle that charged particles follow inside a magnetic field. It is determined by several factors:
  • The mass of the particle
  • The velocity of the particle
  • The charge of the particle
  • The magnetic field strength
The formula used to calculate the orbital radius is given by\[r = \frac{m \cdot v}{q \cdot B}\]where \(m\) is the mass, \(v\) is the velocity, \(q\) is the charge, and \(B\) is the magnetic field strength.
In our specific problem, we found that both protons and alpha particles end up having the same orbital radius under the same magnetic field conditions. This outcome fundamentally explains how the cyclotron's settings maintain particle stability.
Alpha Particles
Alpha particles are a type of nuclear particle consisting of two protons and two neutrons. This gives them distinct properties like:
  • A positive charge of +2e, doubling that of a single proton
  • A larger mass compared to protons, specifically around four times greater
Alpha particles are an essential part of nuclear physics due to their use in science and medicine. In a cyclotron, the behavior of alpha particles highlights how mass and charge play significant roles in determining their path.
Magnetic Field
A magnetic field within a cyclotron exerts a force on charged particles. This force keeps the particles moving in a circular path. Key points include:
  • Magnetic fields are represented by the symbol \(B\)
  • Its strength and direction affect particle motion
In the scenario described, the magnetic field remains unchanged when switching from protons to alpha particles. This ensures that the same path and behavior can be observed, showcasing the reliability of the cyclotron design.
Kinetic Energy
Kinetic energy represents the energy a particle has due to its motion. In physics, it is crucial for understanding particle dynamics. The expression for kinetic energy is:\[KE = \frac{1}{2} m v^2\]where \(m\) is mass and \(v\) is velocity. When dealing in cyclotrons, kinetic energy helps us understand how fast a particle will move inside a magnetic field. In our example, both protons and alpha particles hold the same kinetic energy, ensuring their paths through the cyclotron are comparable in terms of radius.
Protons
Protons are one of the simplest atomic particles, found in the nuclei of atoms. They have a charge of +1e and contribute significantly to the mass of an atom. Some key characteristics of protons include:
  • A smaller mass compared to alpha particles
  • A fundamental role in nuclear reactions and interactions
In the context of the cyclotron problem, protons are the initial particle type considered. Their orbital radius serves as a benchmark, allowing us to understand how other particles like alpha particles will behave under similar conditions.

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Most popular questions from this chapter

A conducting bar with mass m and length \(L\) slides over horizontal rails that are connected to a voltage source. The voltage source maintains a constant current \(I\) in the rails and bar, and a constant, uniform, vertical magnetic field \(\overrightarrow{B}\) fills the region between the rails (\(\textbf{Fig. P27.65}\)). (a) Find the magnitude and direction of the net force on the conducting bar. Ignore friction, air resistance, and electrical resistance. (b) If the bar has mass \(m\), find the distance \(d\) that the bar must move along the rails from rest to attain speed \(v\). (c) It has been suggested that rail guns based on this principle could accelerate payloads into earth orbit or beyond. Find the distance the bar must travel along the rails if it is to reach the escape speed for the earth (11.2 km/s). Let \(B =\) 0.80 T, \(I =\) 2.0 \(\times\) 10\(^3\) A, \(m =\) 25 kg, and \(L =\) 50 cm. For simplicity assume the net force on the object is equal to the magnetic force, as in parts (a) and (b), even though gravity plays an important role in an actual launch in space.

A particle with initial velocity \(\vec{v}$$_0 =\) (5.85 \(\times\) 10\(^3\)m/s)\(\hat{\jmath}\) enters a region of uniform electric and magnetic fields. The magnetic field in the region is \(\overrightarrow{B} =\) - (1.35 T)\(\hat{k}\). Calculate the magnitude and direction of the electric field in the region if the particle is to pass through undeflected, for a particle of charge (a) +0.640 nC and (b) -0.320 nC. You can ignore the weight of the particle.

A mass spectrograph is used to measure the masses of ions, or to separate ions of different masses (see Section 27.5). In one design for such an instrument, ions with mass \(m\) and charge \(q\) are accelerated through a potential difference \(V\). They then enter a uniform magnetic field that is perpendicular to their velocity, and they are deflected in a semicircular path of radius \(R\). A detector measures where the ions complete the semicircle and from this it is easy to calculate \(R\). (a) Derive the equation for calculating the mass of the ion from measurements of \(B\), \(V\), \(R\), and \(q\). (b) What potential difference \(V\) is needed so that singly ionized \(^{12}\)C atoms will have \(R =\) 50.0 cm in a 0.150-T magnetic field? (c) Suppose the beam consists of a mixture of \(^{12}\)C and \(^{14}\)C ions. If \(v\) and \(B\) have the same values as in part (b), calculate the separation of these two isotopes at the detector. Do you think that this beam separation is sufficient for the two ions to be distinguished? (Make the assumption described in Problem 27.59 for the masses of the ions.)

In the Bohr model of the hydrogen atom (see Section 39.3), in the lowest energy state the electron orbits the proton at a speed of 2.2 \(\times\) 10\(^6\) m/s in a circular orbit of radius 5.3 \(\times\) 10\(^{-11}\) m. (a) What is the orbital period of the electron? (b) If the orbiting electron is considered to be a current loop, what is the current \(I\)? (c) What is the magnetic moment of the atom due to the motion of the electron?

If a proton is exposed to an external magnetic field of 2 T that has a direction perpendicular to the axis of the proton's spin, what will be the torque on the proton? (a) 0; (b) 1.4 \(\times\) 10\(^{-26}\) N \(\cdot\) m; (c) 2.8 \(\times\) 10\(^{-26}\) N \(\cdot\) m; (d) 0.7 \(\times\) 10\(^{-26}\) N \(\cdot\) m.
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