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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 \(\mathrm{nm}\) ? (Hint: In each case the initial kinetic energy is negligible.)

Short Answer

Expert verified
(a) Electron accelerating voltage is approximately 6210 V. (b) Proton accelerating voltage is approximately 3.41 × 10^8 V.

Step by step solution

01

Understand the De Broglie Wavelength Formula

The de Broglie wavelength formula is given by \[ \lambda = \frac{h}{p} \] where \( \lambda \) is the wavelength, \( h \) is Planck's constant, and \( p \) is the momentum of the particle. The momentum \( p \) can be expressed as \( p = \sqrt{2mE} \), where \( m \) is the mass and \( E \) is the kinetic energy of the particle.
02

Relate Kinetic Energy to Voltage

The kinetic energy \( E \) of a charged particle like an electron or proton in an electric field is related to the electric potential or voltage \( V \) by the equation\[ E = eV \]where \( e \) is the charge of the particle. Thus, we can substitute this into the momentum formula.
03

Solve for Voltage with Electrons

For electrons, substituting \( p = \sqrt{2mE} \) and \( E = eV \) into the de Broglie formula gives\[ \lambda = \frac{h}{\sqrt{2meV}} \]Rearranging for \( V \),\[ V = \frac{h^2}{2me\lambda^2} \]Substitute the constants: \( h = 6.63 \times 10^{-34} \) Js, \( m_e = 9.109 \times 10^{-31} \) kg, \( e = 1.602 \times 10^{-19} \) C, and \( \lambda = 0.0600 \times 10^{-9} \) m. Calculate \( V \).
04

Calculate Voltage for Electrons

Substitute the values into the formula:\[ V = \frac{(6.63 \times 10^{-34})^2}{2 \times 9.109 \times 10^{-31} \times 1.602 \times 10^{-19} \times (0.0600 \times 10^{-9})^2} \]Solving this gives \( V \approx 6.21 \times 10^3 \) V.
05

Solve for Voltage with Protons

For protons, the mass \( m_p \) is different, so use:\[ V = \frac{h^2}{2m_p e\lambda^2} \]Substitute \( m_p = 1.673 \times 10^{-27} \) kg and calculate \( V \) with the same \( \lambda \).
06

Calculate Voltage for Protons

Substitute the values into the formula:\[ V = \frac{(6.63 \times 10^{-34})^2}{2 \times 1.673 \times 10^{-27} \times 1.602 \times 10^{-19} \times (0.0600 \times 10^{-9})^2} \]Solving this gives \( V \approx 3.41 \times 10^8 \) 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 uses a beam of electrons to create an image of a specimen. It allows scientists to see at resolutions much higher than with traditional light microscopes. This is because electrons have much shorter wavelengths than visible light. The de Broglie wavelength of electrons used in an electron microscope can be adjusted by changing their velocity, which is controlled by an accelerating voltage. This enables us to probe at a much finer scale, revealing the intricate details of structures at a molecular or even atomic level.

These microscopes work using lenses that adjust the direction and focus of the electron beams. Since electrons have charge, magnetic lenses can be used effectively. The shorter wavelength of electrons compared to photons means that electron microscopes can resolve much finer details than optical microscopes. This advancement has been key in fields like biology, materials science, and nanotechnology.
  • Resolution is significantly higher than light microscopes.
  • Accelerating voltage controls the de Broglie wavelength of electrons.
Accelerating Voltage
The accelerating voltage in the context of electron microscopy is crucial because it directly influences the velocity and kinetic energy of the electrons, and hence their de Broglie wavelength. When electrons are accelerated through a potential difference, they gain kinetic energy, given by the formula \( E = eV \). In this formula, \( e \) is the electron charge and \( V \) is the accelerating voltage.

By manipulating the accelerating voltage, technicians can obtain the desired wavelength for the electrons. A higher voltage results in higher electron velocity, reducing the wavelength and allowing for greater resolution in the microscope image. To achieve a certain wavelength, one must adjust the voltage according to the de Broglie wavelength formula.
  • Voltage affects electron velocity and wavelength.
  • Higher voltage leads to shorter wavelengths and improved resolution.
Kinetic Energy
Kinetic energy, the energy possessed by a particle due to its motion, plays a fundamental role in determining the behavior of particles within an electron microscope. In this context, an electron gains kinetic energy as it is accelerated by the voltage set across the microscope's components.

The kinetic energy \( E \) is related to the voltage \( V \) by the formula \( E = eV \). It's noteworthy that the kinetic energy fundamentally influences the de Broglie wavelength since it determines the electron's velocity. A small change in kinetic energy, due to alteration in the voltage, can significantly influence the wavelength, thus impacting the resolution and quality of images produced. In an ideal setup, any initial kinetic energy is negligible compared to the energy gained from the accelerator, allowing precise control over the wavelength.
  • Kinetic energy depends on particle velocity and mass.
  • It’s influenced directly by the accelerating voltage in electron microscopes.
Planck's Constant
Planck's constant is a fundamental constant denoted by \( h \), which appears ubiquitously in quantum mechanics, setting the scale of quantum effects. In the context of an electron microscope and de Broglie wavelengths, it's particularly critical. Planck's constant relates a particle's momentum \( p \) to its wavelength \( \lambda \) through the de Broglie relation \( \lambda = \frac{h}{p} \).

This constant is essential for calculations involving wave-particle duality, allowing precise computation of the wavelengths of electrons or other particles in motion. As a tiny yet profound constant, it encapsulates the discrete nature of quantum phenomena and helps bridge the classical and quantum worlds. Its role in determining the wavelength of electrons underscores its pivotal role in advancing technologies based on quantum principles.
  • Planck’s constant forms the core of wave-particle duality concepts.
  • It plays a significant role in calculating particle wavelengths.

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Most popular questions from this chapter

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

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