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Chapter 6: Problem 46

The electron microscope has been widely used to obtain highly magnified images of biological and other types of materials. When an electron is accelerated through a particular potential field, it attains a speed of \(8.95 \times 10^{6} \mathrm{~m} / \mathrm{s}\). What is the characteristic wavelength of this electron? Is the wavelength comparable to the size of atoms?

Short Answer

Expert verified
The characteristic wavelength of an electron with a speed of \(8.95 \times 10^6 \; m/s\) is \(8.13 \times 10^{-11}\; m\), which is comparable to the size of atoms (\(10^{-10}\; m\)). This ability to have a wavelength similar to atomic sizes allows electron microscopes to produce highly magnified and detailed images of atoms and other small biological materials.

Step by step solution

01

The mass of an electron (m) can be determined from a standard reference table. Its value is \(9.11 \times 10^{-31} \; kg\). #Step 2: Calculate the momentum of the electron#

Now, let's calculate the momentum (p) using the formula: \[ p = m \cdot v \] Substitute the values for mass (m) and speed (v): \[ p = (9.11 \times 10^{-31} \; kg) \cdot (8.95 \times 10^{6} \; m/s) \] \[ p = 8.15 \times 10^{-24} \; kg \cdot m/s \] #Step 3: Calculate the de Broglie wavelength#
02

We can now use the de Broglie wavelength formula: \[ \lambda = \frac{h}{p} \] where \(h\) is the Planck's constant, which is equal to \(6.63 \times 10^{-34} \; Js\). Substituting the values of \(h\) and \(p\), we get: \[ \lambda = \frac{6.63 \times 10^{-34}\; Js}{8.15 \times 10^{-24} \; kg \cdot m/s} \] \[ \lambda = 8.13 \times 10^{-11}\; m \] #Step 4: Compare the wavelength to the size of atoms#

The size of atoms is on the order of \(10^{-10}\; m\) (1 Ångström). Comparing the calculated wavelength (\(8.13 \times 10^{-11}\; m\)) to the size of atoms, we find that the wavelength of the electron is approximately one order of magnitude smaller than the size of an atom and thus can be considered comparable to the size of atoms. This explains why electron microscopes can produce highly magnified and detailed images of atoms and other small biological materials.

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Most popular questions from this chapter

If you put 120 volts of electricity through a pickle, the pickle will smoke and start glowing orange-yellow. The light is emitted because sodium ions in the pickle become excited; their return to the ground state results in light emission. (a) The wavelength of this emitted light is \(589 \mathrm{nm} .\) Calculate its frequency. (b) What is the energy of 0.10 mole of these photons? (c) Calculate the energy gap between the excited and ground states for the sodium ion. (d) If you soaked the pickle for a long time in a different salt solution, such as strontium chloride, would you still observe \(589-\mathrm{nm}\) light emission? Why or why not?

Use the de Broglie relationship to determine the wavelengths of the following objects: (a) an \(85-\mathrm{kg}\) person skiing at \(50 \mathrm{~km} / \mathrm{hr},\) (b) a 10.0 -g bullet fired at \(250 \mathrm{~m} / \mathrm{s},(\mathrm{c})\) a lithium atom moving at \(2.5 \times 10^{5} \mathrm{~m} / \mathrm{s},(\mathrm{d})\) an ozone \(\left(\mathrm{O}_{3}\right)\) molecule in the upper atmosphere moving at \(550 \mathrm{~m} / \mathrm{s}\).

A diode laser emits at a wavelength of \(987 \mathrm{nm}\). (a) In what portion of the electromagnetic spectrum is this radiation found? (b) All of its output energy is absorbed in a detector that measures a total energy of \(0.52 \mathrm{~J}\) over a period of \(32 \mathrm{~s}\). How many photons per second are being emitted by the laser?

Arrange the following kinds of electromagnetic radiation in order of increasing wavelength: infrared, green light, red light, radio waves, X-rays, ultraviolet light.

If human height were quantized in one-foot increments, what would happen to the height of a child as she grows up?
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