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(a) In an electron microscope, what accelerating voltage is needed to produce electrons with wavelength \(0.0600 \mathrm{nm} ?\) (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
The voltage needed is approximately 10400 V for electrons and 5.71×10^7 V for protons.

Step by step solution

01

Identify the problem

The problem involves finding the accelerating voltage required to produce particles (electrons and protons) with a specified wavelength of 0.0600 nm using an electron microscope.
02

Use de Broglie Wavelength Formula

The de Broglie wavelength formula is \( \lambda = \frac{h}{p} \), 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 \( p = \sqrt{2mKE} \), where \( m \) is the mass and \( KE \) is the kinetic energy.
03

Express Kinetic Energy in Terms of Voltage

The kinetic energy of a charged particle is given by \( KE = eV \), where \( e \) is the elementary charge and \( V \) is the voltage. Thus, \( V = \frac{p^2}{2me} \). We substitute \( p = \frac{h}{\lambda} \) into this equation to find \( V \).
04

Calculate Voltage for Electrons

For electrons, known values are \( m_e = 9.11 \times 10^{-31} \) kg, \( e = 1.60 \times 10^{-19} \) C, and \( h = 6.63 \times 10^{-34} \) J·s. Substitute these along with the wavelength \( 0.0600 \) nm \( = 0.0600 \times 10^{-9} \) m into the voltage formula: \[ V_e = \frac{(\frac{6.63 \times 10^{-34}}{0.0600 \times 10^{-9}})^2}{2 \times 9.11 \times 10^{-31} \times 1.60 \times 10^{-19}} \approx 10400 \text{ V} \]
05

Calculate Voltage for Protons

For protons, use \( m_p = 1.67 \times 10^{-27} \) kg, \( e = 1.60 \times 10^{-19} \) C, and \( h = 6.63 \times 10^{-34} \) J·s, with same wavelength. Substitute into the formula:\[ V_p = \frac{(\frac{6.63 \times 10^{-34}}{0.0600 \times 10^{-9}})^2}{2 \times 1.67 \times 10^{-27} \times 1.60 \times 10^{-19}} \approx 5.71 \times 10^{7} \text{ V} \]

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Key Concepts

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

de Broglie Wavelength
In the world of quantum mechanics, the concept of de Broglie wavelength aids us in understanding the wave-particle duality of matter. **Louis de Broglie** proposed that particles such as electrons have a wavelength that can be calculated similarly to light waves. This wavelength is called the de Broglie wavelength. The formula for this is given by:\[ \lambda = \frac{h}{p} \] where:
  • \(\lambda\) is the de Broglie wavelength
  • \(h\) is Planck's constant
  • \(p\) is the momentum of the particle
Understanding the de Broglie wavelength is crucial. It helps predict the behavior of particles within an electron microscope. Since we can calculate this wavelength, it allows us to determine the needed accelerating voltage to reach a particular wavelength, such as the 0.0600 nm in this problem.
Accelerating Voltage
Accelerating voltage is fundamental when working with charged particles, like electrons and protons, in an electron microscope. This voltage contributes to the energy that propels these particles to a specific speed and consequently affects their wavelength. The relationship between energy and voltage is given by:\[ KE = eV \] where:
  • \(KE\) is the kinetic energy
  • \(e\) is the elementary charge
  • \(V\) is the accelerating voltage
The accelerating voltage determines how much kinetic energy the particles will gain. In essence, by adjusting the voltage, we control the velocity, momentum, and thus the de Broglie wavelength of the particle. By calculating the correct voltage, we can achieve the desired precise wavelengths necessary for high-resolution imaging in electron microscopes.
Kinetic Energy
Kinetic energy in the context of particle physics refers to the energy that a particle possesses due to its motion. For charged particles, like electrons in an electron microscope, the kinetic energy can be expressed in terms of the electric potential, or the accelerating voltage. The formula used is:\[ KE = \frac{p^2}{2m} \] and when expressed in terms of voltage:\[ KE = eV \] Here, kinetic energy is vital because it tells us how much energy the particle has as it moves through the electron microscope. It's essential to effectively calculate the kinetic energy to determine the right conditions for observation. The right amount of kinetic energy implies that particles have the sufficient force or speed to exhibit their wave properties, measured as their de Broglie wavelength.
Planck's Constant
Planck's constant is a pivotal figure in quantum mechanics. It serves as a conversion factor between the energy of a photon and the frequency of its electromagnetic wave. Denoted as \(h\), this constant is expressed as approximately \(6.63 \times 10^{-34} \) J·s. In calculations involving wave-particle duality, such as determining de Broglie wavelengths, Planck's constant is an integral component:\[ \lambda = \frac{h}{p} \] Considering the minute magnitude of Planck's constant, it plays a crucial role in quantifying how tiny wavelengths are required for observing particle behaviors, thus affecting settings in an electron microscope. Its value allows us to link the macroscopic perceptions of wavelength and wave functions with the microscopic quantum behaviors of electrons and other particles.

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Most popular questions from this chapter

A photon with wavelength of \(0.1100 \mathrm{nm}\) collides with a free electron that is initially at rest. After the collision, the photon's wavelength is \(0.1132 \mathrm{nm} .\) (a) What is the kinetic energy of the electron after the collision? What is its speed? (b) If the electron is suddenly stopped (for example, in a solid target), all of its kinetic energy is used to create a photon. What is the wavelength of this photon?

For a hydrogen atom in the ground state, determine (a) the circumference of the electron orbit, (b) the speed of the electron, (c) the total energy of the electron, and (d) the minimum energy required to remove the electron completely from the atom.

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

(a) What accelerating potential is needed to produce electrons of wavelength \(5.00 \mathrm{nm} ?\) (b) What would be the energy of photons having the same wavelength as these electrons? (c) What would be the wavelength of photons having the same energy as the electrons in part (a)?

In a parallel universe, the value of Planck's constant is \(0.0663 \mathrm{~J} \cdot \mathrm{s}\). Assume that the physical laws and all other physical constants are the same as in our universe. In this other universe, two physics students are playing catch with a baseball. They are \(50 \mathrm{~m}\) apart, and one throws a \(0.10 \mathrm{~kg}\) ball with a speed of \(5.0 \mathrm{~m} / \mathrm{s}\). (a) What is the uncertainty in the ball's horizontal momentum in a direction perpendicular to that in which it is being thrown if the student throwing the ball knows that it is located within a cube with volume \(1000 \mathrm{~cm}^{3}\) at the time she throws it? (b) By what horizontal distance could the ball miss the second student?

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