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It is not possible to see very small objects, such as viruses, using an ordinary light microscope. An electron microscope can view such objects using an electron beam instead of a light beam. Electron microscopy has proved invaluable for investigations of viruses, cell membranes and subcellular structures, bacterial surfaces, visual receptors, chloroplasts, and the contractile properties of muscles. The 鈥渓enses鈥 of an electron microscope consist of electric and magnetic fields that control the electron beam. As an example of the manipulation of an electron beam, consider an electron traveling away from the origin along the x axis in the \(x y\) plane with initial velocity \(\mathbf{v}_{i}=v_{i} \mathbf{i} .\) As it passes through the region \(x=0\) to \(x=a,\) the electron experiences acceleration \(\mathbf{a}=a_{x} \hat{\mathbf{i}}+a_{y} \hat{\mathbf{j}}, \quad\) where \(a_{x}\) and \(a_{y}\) are constants. For the case \(v_{i}=1.80 \times 10^{7} \mathrm{m} / \mathrm{s}\) \(a_{x}=8.00 \times 10^{14} \mathrm{m} / \mathrm{s}^{2}\) and \(a_{y}=1.60 \times 10^{15} \mathrm{m} / \mathrm{s}^{2}\) , determine at \(x=d=0.0100 \mathrm{m}\) (a) the position of the electron, (b) the velocity of the electron, \((\mathrm{c})\) the speed of the electron, and (d) the direction of travel of the electron (i.e., the angle between its velocity and the \(x\) axis).

Short Answer

Expert verified
At x = 0.0100 m, (a) the position of the electron is 0.0100 m on the x-axis, (b) the velocity components are found using \(v_{fx} = v_{ix} + a_x t\) and \(v_{fy} = a_y t\), (c) the speed is \(\sqrt{v_{fx}^2 + v_{fy}^2}\), and (d) the direction of travel is \(\tan^{-1}(\frac{v_{fy}}{v_{fx}})\).

Step by step solution

01

Analyzing Position

To determine the position of the electron at x = d = 0.0100 m, we only need to consider the motion in the x-direction, as the electron starts at the origin and moves along the x-axis. Since acceleration in the x-direction is constant, the electron will have moved a linear distance of 0.0100 m from the origin.
02

Calculating Final Velocity in the x-direction

Using the kinematic equation for final velocity with constant acceleration: \(v_{fx} = v_{ix} + a_x t\). We need to find the time \(t\) it takes to reach \(x = d\). First, use the equation \(x = v_{ix}t + \frac{1}{2}a_x t^2\) to solve for time \(t\).
03

Solving for Time (t)

Given \(x = v_{ix}t + \frac{1}{2}a_x t^2\), we plug in \(x = 0.0100 m\), \(v_{ix} = 1.80 \times 10^7 m/s\), and \(a_x = 8.00 \times 10^{14} m/s^2\) to solve for \(t\). Since the initial velocity and acceleration are both in the same direction, we can solve this quadratic equation for \(t\).
04

Applying values to find time (t)

We have \(0.0100 = (1.80 \times 10^7)t + \frac{1}{2}(8.00 \times 10^{14})t^2\). Solving this quadratic equation gives us the time required for the electron to reach the position \(x = 0.0100 m\).
05

Calculate Final Velocity in the y-direction

Since the initial velocity in the y-direction is zero, we use the equation for velocity with constant acceleration: \(v_{fy} = v_{iy} + a_y t = a_y t\), to find the final velocity in y-direction.
06

Determining the Speed of the Electron

The speed is the magnitude of the velocity vector and can be found using the Pythagorean theorem for the two perpendicular velocity components: \(Speed = \sqrt{v_{fx}^2 + v_{fy}^2}\).
07

Finding the Direction of Travel

The direction of travel is the angle \(\theta\) between the velocity and the x-axis, which can be found using the inverse tangent function: \(\theta = \tan^{-1}(\frac{v_{fy}}{v_{fx}})\).

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Key Concepts

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

Electron Microscope Applications
Electron microscopes are an indispensable tool in modern science and technology. Unlike traditional light microscopes, which use photons to illuminate the sample, electron microscopes use a focused beam of electrons allowing for much higher resolution imaging. This powerful technology has a host of applications across various fields.

In biology, it has enabled researchers to visualize viruses and intricate cellular structures beyond the diffraction limit of light, providing a deeper understanding of cellular mechanisms. It assists in identifying new species and understanding the fine details of microorganisms. Electron microscopes are also widely used in materials science for studying the nanoscale structure of materials, which is crucial for developing new materials with desired properties.

Electron microscopy plays a pivotal role in semiconductor manufacturing, where it is employed to inspect the microscopic layers of chips for quality assurance. In forensic science, electron microscopy can reveal microscopic evidence that would be invisible under light microscopy, offering vital clues in investigations.

The versatility of electron microscopes extends even further, helping in the development of nanotechnology, the study of metallurgical samples, and the examination of chemical compositions with techniques like electron energy loss spectroscopy (EELS).
Manipulation of an Electron Beam
The core of electron microscope functionality lies in the precise manipulation of an electron beam. This is achieved using electromagnetic lenses that focus and direct the beam onto the sample. Unlike glass lenses used in light microscopes, electron microscope lenses control the path of electrons through the manipulation of electric and magnetic fields.

For an electron beam to be focused and steered properly, several factors are fine-tuned, such as the voltage applied to the electron gun, which is the source of electrons, and the current through the electromagnetic coils creating lenses. The electron beam is accelerated and decelerated using these electromagnetic fields to focus or defocus the beam, just like light is focused through optical lenses.

The manipulation of the beam鈥檚 path allows it to scan across the sample in a controlled manner, which is essential for techniques such as Scanning Electron Microscopy (SEM). In Transmission Electron Microscopy (TEM), the beam is precisely directed to pass through ultrathin samples, creating detailed internal images. Understanding this precise control in electron beam manipulation is essential to maintaining high-resolution images and also forms the basis for various advanced imaging techniques.
Kinematic Equations
Kinematics is a branch of classical mechanics that describes the motion of points, objects, and systems of bodies without considering the forces that cause motion. Kinematic equations, therefore, provide a mathematical description of the motion of particles when the acceleration is constant.

These equations are fundamental in predicting the future position and velocity of an object moving under constant acceleration. They can be summarized as follows:
  • The position-time equation: \( x = x_0 + v_0t + \frac{1}{2}at^2 \)
  • The velocity-time equation: \( v = v_0 + at \)
  • The equation that relates velocity, acceleration, and displacement: \( v^2 = v_0^2 + 2a(x - x_0) \)
  • The equation for displacement during a constant acceleration: \( \bar{v}t = x - x_0 \), where \( \bar{v} \) is the average velocity.
By using these equations, as seen in the electron microscope example, one can calculate the position, velocity, speed, and direction of an electron when subjected to known forces. This predictive ability is crucial not just in understanding electron beam manipulation within a microscope, but also in numerous engineering and physics applications where motion analysis is required.

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Most popular questions from this chapter

As their booster rockets separate, Space Shuttle astronauts typically feel accelerations up to \(3 g,\) where \(g=9.80 \mathrm{m} / \mathrm{s}^{2}\) . In their training, astronauts ride in a device where they experience such an acceleration as a centripetal acceleration. Specifically, the astronaut is fastened securely at the end of a mechanical arm that then turns at constant speed in a horizontal circle. Determine the rotation rate, in revolutions per second, required to give an astronaut a centripetal acceleration of 3.00g while in circular motion with radius 9.45 m.

Barry Bonds hits a home run so that the baseball just clears the top row of bleachers, 21.0 \(\mathrm{m}\) high, located 130 \(\mathrm{m}\) from home plate. The ball is hit at an angle of \(35.0^{\circ}\) to the horizontal, and air resistance is negligible. Find (a) the initial speed of the ball, (b) the time at which the ball reaches the cheap seats, and (c) the velocity components and the speed of the ball when it passes over the top row. Assume the ball is hit at a height of 1.00 m above the ground.

A soccer player kicks a rock horizontally off a 40.0-m high cliff into a pool of water. If the player hears the sound of the splash 3.00 s later, what was the initial speed given to the rock? Assume the speed of sound in air to be 343 m/s.

The pilot of an airplane notes that the compass indicates a heading due west. The airplane鈥檚 speed relative to the air is 150 km/h. If there is a wind of 30.0 km/h toward the north, find the velocity of the airplane relative to the ground.

A motorist drives south at 20.0 m/s for 3.00 min, then turns west and travels at 25.0 m/s for 2.00 min, and finally travels northwest at 30.0 m/s for 1.00 min. For this 6.00-min trip, find (a) the total vector displacement, (b) the average speed, and (c) the average velocity. Let the positive x axis point east.

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