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With what potential must the electrons be accelerated in an electron microscope if its theoretical power of resolution is to equal that of microscope operating with X-rays of energy \(100 \mathrm{keV}\) ?

Short Answer

Expert verified
The accelerating potential is approximately 25,600 volts.

Step by step solution

01

Understand the Problem

The problem asks us to find the accelerating potential required for electrons in an electron microscope to have the same resolving power as a microscope using X-rays of a given energy, which is 100 keV.
02

Formula for Wavelength of X-rays

The wavelength of X-rays can be determined using the energy-wavelength relation \[ E = \frac{hc}{\lambda} \]where \(E\) is the energy of the X-ray, \(h\) is Planck's constant \( (6.626 \times 10^{-34} \, \text{Js}) \), and \(c\) is the speed of light \( (3.00 \times 10^8 \, \text{m/s}) \). Given that \( E = 100 \, \text{keV} = 100 \times 10^3 \times e\) (where \(e = 1.602 \times 10^{-19} \, \text{C}\)), calculate \(\lambda\).
03

Calculate Wavelength of X-rays

Convert 100 keV to joules: \[ E = 100 \times 10^3 \times 1.602 \times 10^{-19} = 1.602 \times 10^{-14} \, \text{J}\] Calculate the wavelength:\[ \lambda = \frac{hc}{E} = \frac{6.626 \times 10^{-34} \times 3.00 \times 10^8}{1.602 \times 10^{-14}} \approx 1.24 \times 10^{-11} \, \text{m} \]
04

Formula for Electron Wavelength

Use the de Broglie wavelength formula for electrons:\[ \lambda = \frac{h}{p} \]where \(p\) is the momentum of the electron. The momentum can be linked to the kinetic energy \( K \) by:\[ p = \sqrt{2m_eK} \]where \(m_e = 9.109 \times 10^{-31} \, \text{kg}\) is the electron mass.
05

Relate Kinetic Energy to Potential

The kinetic energy \( K \) of the electrons is given by:\[ K = eV \]where \( V \) is the accelerating potential and \( e = 1.602 \times 10^{-19} \, \text{C} \). Substitute this in the momentum expression:
06

Equate Wavelengths for Resolution Power

Equate the de Broglie wavelength to the X-ray wavelength found in Step 3:\[ \frac{h}{\sqrt{2m_e eV}} = 1.24 \times 10^{-11} \, \text{m} \]Solve for \( V \):\[ V = \frac{h^2}{2m_e e (1.24 \times 10^{-11})^2} \]
07

Calculate the Accelerating Potential

Substitute the constants and solve for \( V \):\[ V = \frac{(6.626 \times 10^{-34})^2}{2 \times 9.109 \times 10^{-31} \times 1.602 \times 10^{-19} \times (1.24 \times 10^{-11})^2} \approx 25,600 \, \text{volts} \]

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Key Concepts

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

Accelerating Potential
In electron microscopes, accelerating potential refers to the voltage applied to accelerate electrons. This voltage is crucial because it determines the electrons' kinetic energy. When electrons are accelerated, they gain energy proportional to the potential difference through which they travel. This is given by the formula:\[ K = eV \]where:
  • \( K \) is the kinetic energy of the electrons
  • \( e \) is the elementary charge \((1.602 \times 10^{-19} \text{ C})\)
  • \( V \) is the accelerating potential

In the context of electron microscopy, a higher accelerating potential leads to electrons with shorter wavelengths, which improves resolution. By managing this potential, scientists can manipulate the capability of an electron microscope, tailoring it to observe details at the atomic and molecular scales.
Resolution Power
Resolution power is a measure of an imaging system's ability to distinguish between two points that are close together. In microscopy, better resolution power means the microscope can reveal fine structural details of an object. For electron microscopes, this resolution is determined largely by the wavelength of the electrons. The smaller the wavelength, the better the resolution. For two different microscopes to have the same resolution power, their operational wavelengths must be the same. In our specific context, the electron microscope must achieve a wavelength equivalent to that of X-rays with a given energy to match their resolution power. This requirement leads to a close examination of both electron and photon behaviors under different conditions.
de Broglie Wavelength
The concept of de Broglie wavelength plays a pivotal role in understanding particle-wave duality, especially in electron microscopy. According to de Broglie's hypothesis, particles like electrons exhibit wave-like properties, with a wavelength determined by their momentum. This is captured in the formula:\[ \lambda = \frac{h}{p} \]where:
  • \( \lambda \) is the de Broglie wavelength
  • \( h \) is Planck's constant \((6.626 \times 10^{-34} \text{ Js})\)
  • \( p \) is the momentum of the particle

For electrons, momentum is linked to kinetic energy generated by the accelerating potential. By manipulating this potential, scientists can effectively alter the electron wavelength, and consequently, the resolving power of an electron microscope.
X-ray Energy
X-rays are a form of electromagnetic radiation with high energy and short wavelengths, which makes them suitable for imaging fine structures. The energy of X-rays can be directly correlated to their wavelength through the relation:\[ E = \frac{hc}{\lambda} \]where:
  • \( E \) is the energy of the X-ray
  • \( h \) is Planck's constant \((6.626 \times 10^{-34} \text{ Js})\)
  • \( c \) is the speed of light \((3.00 \times 10^8 \text{ m/s})\)
  • \( \lambda \) is the wavelength

When using X-rays in microscopy, the energy is typically high enough to give excellent resolution, but it requires special considerations for sample preparation and handling. By comparing this to electron wavelengths, scientists can engineer electron microscopes that rival X-ray resolution by tuning electron energies effectively.
Electron Wavelength
The wavelength of electrons is a central concept in electron microscopy as it dictates the resolution limit. Electrons, when accelerated to high speeds, exhibit wave-like properties and their wavelength is inversely proportional to their momentum.The de Broglie wavelength of electrons as accelerated by a potential \( V \) can be expressed as:\[ \lambda = \frac{h}{\sqrt{2m_e eV}} \]where:
  • \( \lambda \) is the electron wavelength
  • \( h \) is Planck's constant \((6.626 \times 10^{-34} \text{ Js})\)
  • \( m_e \) is the electron rest mass \((9.109 \times 10^{-31} \text{ kg})\)

By adjusting the accelerating potential \( V \), electron microscopes can reach resolutions comparable to those obtained with high-energy X-rays, pushing the boundaries of what can be visualized on small scales.

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Most popular questions from this chapter

Calculate the fractional change in the photon energy as a result of Compton scattering through an angle of \(90^{\circ}\) for photons of: a. X-rays with a wavelength of \(0.025 \mathrm{~nm}\); b. \(\quad y\) rays with an energy of \(1 \mathrm{MeV}\).

A sodium lamp emits yellow light with a wavelength of \(590 \mathrm{~nm}\). How many photons does a \(100 \mathrm{~W}\) sodium lamp emit each second?

When an electron penetrates the surface of a metal, its energy changes as a result of the metal's work function. a. Does an electron's energy increase or decrease when it penetrates a metal's surface? b. In the Davisson and Germer experiment, the electrons were accelerated through low potentials; \(54 \mathrm{~V}\) was a typical value. Calculate the velocity of the electrons before they entered the nickel crystal and after entering on the assumption that the work function of nickel is \(2.6 \mathrm{eV}\). c. Calculate the wavelengths of the electrons before and after they penetrate the surface.

Assuming that the human body has a surface area of 2 square meters and radiates like a black-body at a temperature of \(35^{\circ} \mathrm{C}\), calculate the rate at which it loses heat in surroundings that have a temperature of \(15^{\circ} \mathrm{C}\).

A \(100 \mathrm{~mW}\) argon laser emits a beam of light of wavelength \(488 \mathrm{~nm}\) in vacuo. a. How many photons does the laser emit each second b. What current will flow through a photoelectric cell whose cathode (emitting. plate) is illuminated by the laser's light beam if \(10 \%\) of the photons release an electron from it. c. What potential will be required between the cell's electrodes in order to stop the current, if the threshold frequency of the cathode is \(5.2 \cdot 10^{14} \mathrm{~Hz}\).

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