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Ch 37: The Foundations of Modern Physics
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Non è quello che usi tu?Cambia libro di testo
Capitolo 37, Problema 41

A ²²²Rn atom (radon) in a 0.75 T magnetic field undergoes radioactive decay, emitting an alpha particle in a direction perpendicular to B→\(\overrightarrow{B}\). The alpha particle begins cyclotron motion with a radius of 45 cm. With what energy, in MeV, was the alpha particle emitted?

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Step 1: Identify the key physical principles involved in the problem. The alpha particle undergoes cyclotron motion in a magnetic field, which means the centripetal force required for circular motion is provided by the magnetic Lorentz force. The relationship between the radius of the motion, the velocity of the particle, and the magnetic field is given by the equation: r=mvqB, where r is the radius, m is the mass of the particle, v is its velocity, q is its charge, and B is the magnetic field strength.
Step 2: Rearrange the formula to solve for the velocity of the alpha particle. Using the given radius (r = 45 cm = 0.45 m), magnetic field strength (B = 0.75 T), and the charge of the alpha particle (q = 2e, where e = 1.6 × 10⁻¹⁹ C), the velocity can be expressed as: v=qBrm. Note that the mass of the alpha particle is approximately 6.64 × 10⁻²⁷ kg.
Step 3: Calculate the kinetic energy of the alpha particle using the formula for kinetic energy: K=1mv22. Substitute the mass of the alpha particle and the velocity obtained from Step 2 into this formula.
Step 4: Convert the kinetic energy from joules to MeV. Use the conversion factor: 1 MeV = 1.6 × 10⁻¹³ J. Divide the kinetic energy in joules by this factor to express the energy in MeV.
Step 5: Verify the units and ensure all values are consistent with the problem's requirements. The final energy value in MeV represents the energy with which the alpha particle was emitted.

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Cyclotron Motion

Cyclotron motion refers to the circular motion of a charged particle in a magnetic field. When a charged particle, such as an alpha particle, moves perpendicular to a magnetic field, it experiences a magnetic force that acts as a centripetal force, causing it to move in a circular path. The radius of this motion is determined by the particle's velocity, charge, and the strength of the magnetic field.
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Energy of a Charged Particle

The energy of a charged particle in cyclotron motion can be expressed in terms of its kinetic energy, which is related to its mass and velocity. For an alpha particle, this energy can be calculated using the formula E = (1/2)mv², where m is the mass and v is the velocity. Additionally, the energy can also be expressed in electronvolts (eV) or mega-electronvolts (MeV) for convenience in nuclear physics.
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Conservation of Charge

Magnetic Field Strength

Magnetic field strength, measured in teslas (T), influences the motion of charged particles. In this scenario, the 0.75 T magnetic field affects the radius of the alpha particle's cyclotron motion. The relationship between the magnetic field strength, the charge of the particle, and its velocity is crucial for determining the energy of the emitted alpha particle, as it dictates how tightly the particle will spiral in the magnetic field.
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