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?

Thank you Y a string 0.85 teslas during its radioactive decay, it releases an alpha particle. If alpha part moves in a direction that is perpendicular to the magnetic field B, which is, which creates a circular path, the diameter of 0.8 m, we're asked to determine the energy in mega electrons that the alpha particle possessed at the time it was emitted. There are multiple choice answers in units of mega electron volts are a 4.5 B 5.6 C 6.1 or D 7.3. So the key to solving this problem is we're calling two equations. So we have our centrifugal force which is equal to MB square divided by bar. And we're, we can recognize that we have a central force here because the particle is moving in a circular path. We also have our magnetic force which is given by the equation F of M is equal to QB multiplied by E because the magnetic force is the only force acted on the al particle, you can then set these equations equal to each other. So we have MV squared divided by R is equal to QVB. Now we're asked to find the energy in the problem, but we can solve for the speed first and then use our kinetic energy equation to solve for the energy. And so you can see one of those speeds cancel and then isolating speed we have V is equal to QVR divided by N. And from there, we can plug in our known values. We recognize that the charge for an awful particle or Q for alpha is equal to two multiplied by E that elementary charge, which we can recall is the constant 1.6 times 10 to the negative 19 cool lumps. And then our magnetic field magnitude is 0.85 teslas. And then our radius will be half of our diameter or 1.5 multiplied by 0.8 m divided by the mass. And all the part of all we can recall is a constant 6.64 times 10 to the negative 27 kg. And when we plug that in, we get a speed of 1.639 times 10 to the 7 m per second. And then our next step is going to be utilizing that kinetic energy equation A is equal to one half and B squared to solve for our energy. And so we have one half multiplied by the mass again, 6.64 times 10 is the negative 27 kg multiplied by that speed that we just saw for 1.639 times 10 is 7 m per second. And that quantity is squared. And so we get an energy of 8.9 times 10 to the negative 13 joules one last step. Because the answer ask for our energy in terms of mega electron volts, we can recall that conversion factor is one mega electron volt is equal to 1.6 times 10 to the negative 13 joules. And so we are left when 5.6 mega electrons and we look at our multiple choice answers, we see that, that aligns with answer choice B to B is the correct answer for this problem. That's all we have for this one. We'll see you in the next video.

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Problem 41

Textbook Question

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

Verified step by step guidance

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 = \frac{m v}{q B}

, 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 = \frac{q B r}{m}

. 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 = \frac{1 m v^{2}}{2}

. 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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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

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.

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.

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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