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singly charged uranium-238 ions are accelerated through a potential difference of \(2.00 \mathrm{kV}\) and enter a uniform magnetic field of \(1.20 \mathrm{T}\) directed perpendicular to their velocities. (a) Determine the radius of their circular path. (b) Repeat for uranium-235 ions. What If? How does the ratio of these path radii depend on the accelerating voltage and on the magnitude of the magnetic field?

Short Answer

Expert verified
The radii of the circular paths for the Uranium-238 and Uranium-235 ions can be calculated using the derived equation. The radii depend on the square root of the mass of the isotope, and inversely, on the strength of the magnetic field, but not on the accelerating voltage. The ratio of the radii is equal to the square root of the ratio of the masses of the isotopes.

Step by step solution

01

Understanding the Principle

An ion moving through a magnetic field experiences a force. The perpendicular force is given by \( F = qvB \) where F is the magnetic force, q is the charge of the ion, v is the velocity of the ion, and B is the magnetic field strength. When the ion moves in a circle due to this force, we also have \( F = \frac{mv^2}{r} \), where m is the mass of the ion and r is the radius of the circular path. Setting these two equal to each other allows us to solve for r.
02

Find the Velocity of the Ions

Because the ions are singly charged, there's a gain in kinetic energy equal to the loss in electric potential energy as the ions accelerate. So, \( \frac{1}{2} mv^2 = qV \), where V is the accelerating voltage. Solving for v, we get \( v = \sqrt{\frac{2qV}{m}} \)
03

Solve for the Radius

Insert the value of v from step 2 into the equation from step 1 to get radius : \( r = \frac{mv}{qB} = \frac{m\sqrt{\frac{2qV}{m}}}{qB} = \sqrt{\frac{2mV}{q}} \cdot \frac{1}{B} \). We can use this for both uranium-238 and uranium-235, remembering to adjust for the different masses.
04

Compute the Radii for Each Uranium Isotope

Substitute for each uranium isotope with the given values of V and B. The only difference is the mass. Use \( m = 238 \) and \( m = 235 \) separately and compute the radii.
05

Compute the Ratio of the Radii

The ratio of radii can be found as \( \frac{r_{238}}{r_{235}} \). From the radius equation in step 3, it can be seen that the radii are directly proportional to the square root of the mass \( m \), and inversely proportional to the magnetic field \( B \). As the voltage \( V \) is the same for both isotopes, it has no effect on the ratio.

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Key Concepts

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

Circular Motion
When an object moves in a circle at a constant speed, it experiences circular motion. This motion is governed by a force that acts towards the center of the circle, known as the centripetal force. In the case of ions in a magnetic field, such as uranium ions, this force is provided by the magnetic field itself. The magnetic force keeps the ions in a circular path.

The relationship between the magnetic force and the circular motion can be described by the equation for centripetal force: \ \( F = \frac{mv^2}{r} \), where \
  • \(m\) is the mass of the ion,
  • \(v\) is the velocity, and
  • \(r\) is the radius of the circular path.


For uranium ions specifically, this circular path helps differentiate isotopes based on their mass, as they will travel along different radii when subjected to the same magnetic force.
Ion Acceleration
Acceleration of ions occurs when they are subjected to an electric potential difference, which increases their kinetic energy. For singly charged ions, the change in kinetic energy equals the change in electric potential energy. This relationship is given by the formula: \ \( \frac{1}{2} mv^2 = qV \), where \
  • \(m\) is the mass of the ion,
  • \(v\) is the velocity of the ion,
  • \(q\) is the charge of the ion (for singly charged ions, typically the elementary charge), and
  • \(V\) is the potential difference.


Solving this equation for \(v\), we find the velocity of the ions post-acceleration: \ \( v = \sqrt{\frac{2qV}{m}} \). This velocity is crucial in determining how the ions move in a magnetic field and, consequently, their path's radius.
Uranium Isotopes
Uranium isotopes, such as uranium-238 and uranium-235, are variations of uranium atoms with different atomic masses but similar chemical properties. These differences in mass allow scientists to separate isotopes using their physical properties. For instance, when both isotopes are accelerated and subjected to a magnetic field, their paths will differ due to mass variance.

The separation relies on the principles of circular motion within a magnetic field. Since the radius of the path taken by an ion in the magnetic field is \ determined by \( \sqrt{\frac{2mV}{q}} \cdot \frac{1}{B} \), isotopes with different masses will naturally diverge due to the effect of \(m\) in the equation. This concept is not only vital in separating uranium isotopes but also in other fields such as mass spectrometry.
Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. For ions being accelerated, this energy is pivotal for achieving the necessary velocity to enter and be deflected by a magnetic field. Kinetic energy can be expressed as \( \frac{1}{2} mv^2 \).

For uranium ions moving through a magnetic field, their acquired kinetic energy from a potential difference is directly related to their subsequent motion. As these ions gain velocity through acceleration, their kinetic energy becomes the driving force that dictates the radius of their circular path in the magnetic field.

Understanding this energy transfer is crucial for determining how changes in mass affect their motion, especially when examining isotopes of different uranium masses. This insight helps us manipulate and control the motion for practical applications such as isotope separation.

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Most popular questions from this chapter

A cosmic-ray proton in interstellar space has an energy of \(10.0 \mathrm{MeV}\) and executes a circular orbit having a radius equal to that of Mercury's orbit around the Sun \((5.80 \times\) \(\left.10^{10} \mathrm{m}\right) .\) What is the magnetic field in that region of space?

A Hall-effect probe operates with a 120 -mA current. When the probe is placed in a uniform magnetic field of magnitude \(0.0800 \mathrm{T},\) it produces a Hall voltage of \(0.700 \mu \mathrm{V}\) (a) When it is measuring an unknown magnetic field, the Hall voltage is \(0.330 \mu \mathrm{V}\). What is the magnitude of the unknown field? (b) The thickness of the probe in the direction of \(\mathbf{B}\) is \(2.00 \mathrm{mm} .\) Find the density of the charge carriers, each of which has charge of magnitude \(e\)

A wire \(2.80 \mathrm{m}\) in length carries a current of \(5.00 \mathrm{A}\) in a region where a uniform magnetic field has a magnitude of \(0.390 \mathrm{T}\). Calculate the magnitude of the magnetic force on the wire assuming the angle between the magnetic field and the current is (a) \(60.0^{\circ},\) (b) \(90.0^{\circ},\) (c) \(120^{\circ}\)

(a) A proton moving in the \(+x\) direction with velocity \(\mathbf{v}=v_{i} \mathbf{i}\) experiences a magnetic force \(\mathbf{F}=F_{i} \hat{\mathbf{j}}\) in the \(+y\) direction. Explain what you can and cannot infer about \(\mathbf{B}\) from this information. (b) What If? In terms of \(F_{i},\) what would be the force on a proton in the same field moving with velocity \(\mathbf{v}=-v_{i} \hat{\mathbf{i}} ?\) (c) What would be the force on an electron in the same field moving with velocity \(\mathbf{v}=v_{i} \mathbf{i} ?\)

A current of \(17.0 \mathrm{mA}\) is maintained in a single circular loop of \(2.00 \mathrm{m}\) circumference. A magnetic field of \(0.800 \mathrm{T}\) is directed parallel to the plane of the loop. (a) Calculate the magnetic moment of the loop. (b) What is the magnitude of the torque exerted by the magnetic field on the loop?

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