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(a) In an electron microscope, what accelerating voltage is needed to produce electrons with wavelength 0.0600 nm? (b) If protons are used instead of electrons, what accelerating voltage is needed to produce protons with wavelength 0.0600 nm? (\(Hint\): In each case the initial kinetic energy is negligible.)

Short Answer

Expert verified
The accelerating voltage for electrons is approximately 104,000 V; for protons, it is 54.1 billion V.

Step by step solution

01

Understanding the De Broglie Wavelength

The de Broglie wavelength \( \lambda \) of a particle is given by the equation \( \lambda = \frac{h}{p} \), where \( h \) is Planck's constant and \( p \) is the momentum of the particle. For particles accelerated by a voltage \( V \), the momentum \( p \) can be expressed in terms of their kinetic energy, which derives from the electrical potential energy.
02

Relating Wavelength to Voltage for Electrons

For electrons, the de Broglie wavelength \( \lambda = 0.0600 \times 10^{-9} \, \text{m} \). The momentum \( p \) is related to kinetic energy by \( p^2 = 2meKE \), where \( KE = eV \). Therefore, \( \lambda = \frac{h}{\sqrt{2meV}} \), where \( m \) is the electron mass and \( e \) is the elemental charge.
03

Calculate Accelerating Voltage for Electrons

Rearrange the equation \( \lambda = \frac{h}{\sqrt{2meV}} \) to find \( V \). Substituting \( \lambda = 0.0600 \times 10^{-9} \, \text{m} \), \( m = 9.109 \times 10^{-31} \, \text{kg} \), \( e = 1.602 \times 10^{-19} \, \text{C} \), and \( h = 6.626 \times 10^{-34} \, \text{J} \cdot \text{s} \), solve for \( V \). This gives approximately \( V = 104 \times 10^{3} \, \text{V} \).
04

Relating Wavelength to Voltage for Protons

For protons, use the same relation: \( \lambda = \frac{h}{\sqrt{2mpV}} \), where \( mp \) is the proton mass \( 1.673 \times 10^{-27} \, \text{kg} \). Substitute \( \lambda = 0.0600 \times 10^{-9} \, \text{m} \), along with known \( h \) and \( e \), into this expression.
05

Calculate Accelerating Voltage for Protons

Using the rearranged formula and substituting the appropriate values for protons, solve for \( V \). This results in approximately \( V = 54.1 \times 10^{9} \, \text{V} \).

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Key Concepts

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

Electron Microscope
An electron microscope is a powerful tool that allows us to see images of tiny objects, such as cells and bacteria, at incredibly high magnifications. Unlike light microscopes that use light to illuminate specimens, electron microscopes use a beam of electrons. This allows for much higher resolution since electrons have much shorter wavelengths than visible light.
Electron microscopes work by accelerating electrons towards the sample we are examining. Once the electrons hit the sample, they interact with it, providing a detailed image of its structure. The shorter the wavelength of the electrons, the more detailed the image. This is why the concept of de Broglie wavelength is critical—because it relates to how we control the wavelength of these electrons using voltage.
  • Shorter wavelengths yield higher resolution images.
  • Electrons are accelerated using high voltages.
  • De Broglie wavelength helps us understand how to calculate the needed voltage for clarity.
Accelerating Voltage
Accelerating voltage is a crucial component in electron microscopy. It refers to the voltage used to speed up the electrons in the electron microscope. The greater the voltage, the faster the electrons are accelerated, reducing their wavelength in the process.
The relationship between voltage and electron speed is founded on the conversion of electric potential energy into kinetic energy. As electrons are released from a cathode, they are accelerated across a voltage difference. This process converts the potential energy (due to the voltage) directly into the electrons' kinetic energy.
  • Higher accelerating voltage reduces the effective wavelength of electrons.
  • This lower wavelength aids in achieving greater resolution in electron microscopes.
Understanding and calculating the right accelerating voltage is vital. It allows scientists to acquire clearer and more detailed images.
Kinetic Energy
Kinetic energy in the context of an electron microscope represents the energy that electrons gain as they are accelerated by the voltage applied. Initially, electrons have negligible kinetic energy, but upon acceleration, they gain significant kinetic energy due to their movement.
In our equations, the kinetic energy is related to the voltage by the simple relationship: \[ KE = eV \] where \( KE \) is the kinetic energy, \( e \) is the elementary charge, and \( V \) is the voltage. This makes kinetic energy directly proportional to the accelerating voltage.
  • As voltage increases, so does the kinetic energy of electrons.
  • Higher kinetic energy leads to shorter wavelengths and higher resolution images.
Planck's Constant
Planck's constant is a fundamental constant in physics, symbolized by \( h \). It is essential in the field of quantum mechanics, being a core component of the formula for the de Broglie wavelength.
Planck's constant allows us to calculate a particle's wavelength from its momentum. In the context of electron microscopes, this is crucial for determining how we can influence the wavelength through momentum changes driven by kinetic energy and voltage.
  • Planck's constant \( (h) \) is approximately \( 6.626 \times 10^{-34} \, \text{J} \cdot \text{s} \).
  • A key part of de Broglie's equation: \( \lambda = \frac{h}{p} \).
  • It helps in understanding the wave-particle duality of matter.
Understanding Planck's constant's role in the equations helps explain how we manipulate and interpret the behavior of electrons and other particles in electron microscopes.

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Most popular questions from this chapter

A scientist has devised a new method of isolating individual particles. He claims that this method enables him to detect simultaneously the position of a particle along an axis with a standard deviation of 0.12 nm and its momentum component along this axis with a standard deviation of 3.0 \(\times\) 10\(^{-25}\) kg \(\bullet\) m/s. Use the Heisenberg uncertainty principle to evaluate the validity of this claim.

In the Bohr model of the hydrogen atom, what is the de Broglie wavelength of the electron when it is in (a) the \(n\) = 1 level and (b) the \(n\) = 4 level? In both cases, compare the de Broglie wavelength to the circumference 2\(\pi{r_n}\) of the orbit.

Consider a particle with mass m moving in a potential \(U = {1\over2} kx^2\), as in a mass-spring system. The total energy of the particle is \(E = (p^2/2m) + 12 kx^2\). Assume that \(p\) and \(x\) are approximately related by the Heisenberg uncertainty principle, so \(px \approx h\). (a) Calculate the minimum possible value of the energy \(E\), and the value of \(x\) that gives this minimum E. This lowest possible energy, which is not zero, is called the \(zero-point \space energy\). (b) For the \(x\) calculated in part (a), what is the ratio of the kinetic to the potential energy of the particle?

(a) An atom initially in an energy level with \(E\) = -6.52 eV absorbs a photon that has wavelength 860 nm. What is the internal energy of the atom after it absorbs the photon? (b) An atom initially in an energy level with \(E\) = -2.68 eV emits a photon that has wavelength 420 nm. What is the internal energy of the atom after it emits the photon?

The brightest star in the sky is Sirius, the Dog Star. It is actually a binary system of two stars, the smaller one (Sirius B) being a white dwarf. Spectral analysis of Sirius B indicates that its surface temperature is 24,000 K and that it radiates energy at a total rate of 1.0 \(\times\) 10\(^{25}\) W. Assume that it behaves like an ideal blackbody. (a) What is the total radiated intensity of Sirius B? (b) What is the peak-intensity wavelength? Is this wavelength visible to humans? (c) What is the radius of Sirius B? Express your answer in kilometers and as a fraction of our sun's radius. (d) Which star radiates more \(total\) energy per second, the hot Sirius B or the (relatively) cool sun with a surface temperature of 5800 K? To find out, calculate the ratio of the total power radiated by our sun to the power radiated by Sirius B.

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