Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

To investigate the structure of extremely small objects, such as viruses, the wavelength of the probing wave should be about one-tenth the size of the object for sharp images. But as the wavelength gets shorter, the energy of a photon of light gets greater and could damage or destroy the object being studied. One alternative is to use electron matter waves instead of light. Viruses vary considerably in size, but 50 nm is not unusual. Suppose you want to study such a virus, using a wave of wavelength 5.00 nm. (a) If you use light of this wavelength, what would be the energy (in eV) of a single photon? (b) If you use an electron of this wavelength, what would be its kinetic energy (in eV)? Is it now clear why matter waves (such as in the electron microscope) are often preferable to electromagnetic waves for studying microscopic objects?

Short Answer

Expert verified
Light photon energy: 248 eV; electron energy: 0.375 eV. Electrons are preferred, causing less damage.

Step by step solution

01

Calculate photon energy using light

To find the energy of a photon, we use the equation \( E = \frac{hc}{\lambda} \), where \( h \) is Planck's constant \( 6.626 \times 10^{-34} \text{ J} \cdot \text{s} \), \( c \) is the speed of light \( 3.00 \times 10^8 \text{ m/s} \), and \( \lambda = 5.00 \times 10^{-9} \text{ m} \) is the wavelength. Calculate \( E \):\[E = \frac{6.626 \times 10^{-34} \times 3.00 \times 10^8}{5.00 \times 10^{-9}} = 3.97 \times 10^{-17} \text{ J}\]Convert the energy from Joules to electronvolts (1 eV = \( 1.602 \times 10^{-19} \text{ J} \)):\[E = \frac{3.97 \times 10^{-17}}{1.602 \times 10^{-19}} = 248 \text{ eV}\]
02

Calculate kinetic energy of an electron using matter waves

To find the kinetic energy of an electron with a given wavelength, use the de Broglie wavelength formula \( \lambda = \frac{h}{p} \) where \( p \) is momentum. Hence the momentum \( p = \frac{h}{\lambda} \). Since \( p = mv \) and kinetic energy (KE) is \( \frac{1}{2}mv^2 \), we can express \( KE \) as \( \frac{p^2}{2m} \):\[KE = \frac{h^2}{2m\lambda^2}\]Here, \( m = 9.11 \times 10^{-31} \text{ kg} \) (electron mass) and \( \lambda = 5.00 \times 10^{-9} \text{ m} \).\[KE = \frac{(6.626 \times 10^{-34})^2}{2 \times 9.11 \times 10^{-31} \times (5.00 \times 10^{-9})^2} = 6.02 \times 10^{-20} \text{ J}\]Convert to eV:\[KE = \frac{6.02 \times 10^{-20}}{1.602 \times 10^{-19}} = 0.375 \text{ eV}\]
03

Compare photon and electron energy

The energy of a photon with a 5 nm wavelength is 248 eV, while the kinetic energy of an electron with the same wavelength is 0.375 eV. This significant difference in energy levels explains why matter waves (like those in electron microscopes) are preferable, as they provide similar wavelengths for imaging while causing less potential damage.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

de Broglie wavelength
The de Broglie wavelength is a fundamental concept in physics that helps us understand the wave-particle duality of matter.
Louis de Broglie proposed that particles such as electrons exhibit both particle and wave properties, which can be described by a wavelength.
This wavelength is known as the de Broglie wavelength, and it is extremely useful in electron microscopy.
To calculate the de Broglie wavelength (\(\lambda\)) of a particle, you use the equation \(\lambda = \frac{h}{p}\), where \(h\) is Planck's constant, and \(p\) is the momentum of the particle. Momentum \(p\) is the product of mass (\(m\)) and velocity (\(v\)) of the particle.
  • Momentum is given by \(p = mv\)
  • Planck's constant is \(6.626 \times 10^{-34} \,\text{J}\cdot\text{s}\)
The de Broglie wavelength is particularly important when it comes to using electron microscopes.
Electrons with a smaller wavelength have higher resolution, allowing us to see extremely small details that are important for studying tiny objects like viruses.
photon energy
Photon energy is crucial when dealing with electromagnetic waves, such as light.
Photons, the basic unit of light, carry energy that can be determined using the wavelength of the light.
The energy \(E\) of a photon can be calculated using the formula \(E = \frac{hc}{\lambda}\).
  • Here, \(h\) is the Planck's constant \(6.626 \times 10^{-34} \,\text{J}\cdot\text{s}\)
  • \(c\) is the speed of light \(3.00 \times 10^8 \,\text{m/s}\)
  • \(\lambda\) is the wavelength of the photon
When comparing photons with different wavelengths, shorter wavelengths result in higher photon energy.
This is especially important in imaging, where high-energy photons can damage delicate structures, like viruses, which is why alternative methods, such as electron matter waves, are sometimes preferred.
kinetic energy of electrons
The kinetic energy of electrons is another essential aspect when we study tiny objects using electron beams.
Unlike photons, electrons have mass, and their kinetic energy is directly related to their momentum and speed.
The kinetic energy (\(KE\)) of an electron can be determined from its momentum (\(p\)) using the relation \(KE = \frac{p^2}{2m}\), where \(m\) is the mass of the electron.
  • Momentum can be expressed using the de Broglie relation: \(p = \frac{h}{\lambda}\)
  • The mass of an electron is \(9.11 \times 10^{-31} \,\text{kg}\)
This formula shows that by knowing the wavelength \(\lambda\) of an electron, one can determine its kinetic energy.
A key point in electron microscopy is that electrons at the same wavelength as photons have dramatically lower kinetic energies.
This lowers the risk of damaging the sample while still providing a high-resolution image, crucial for delicate samples like microorganisms.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

(a) Using the Bohr model, calculate the speed of the electron in a hydrogen atom in the \(n\) = 1, 2, and 3 levels. (b) Calculate the orbital period in each of these levels. (c) The average lifetime of the first excited level of a hydrogen atom is 1.0 \(\times\) 10\(^{-8}\) s. In the Bohr model, how many orbits does an electron in the \(n\) = 2 level complete before returning to the ground level?

Two stars, both of which behave like ideal blackbodies, radiate the same total energy per second. The cooler one has a surface temperature \(T\) and a diameter 3.0 times that of the hotter star. (a) What is the temperature of the hotter star in terms of \(T\) ? (b) What is the ratio of the peak-intensity wavelength of the hot star to the peak-intensity wavelength of the cool star?

A large number of neon atoms are in thermal equilibrium. What is the ratio of the number of atoms in a 5\(s\) state to the number in a 3\(p\) state at (a) 300 K; (b) 600 K; (c) 1200 K? The energies of these states, relative to the ground state, are E\(_{5s}\) = 20.66 eV and E\(_{3p}\) = 18.70 eV. (d) At any of these temperatures, the rate at which a neon gas will spontaneously emit 632.8-nm radiation is quite low. Explain why.

What is the de Broglie wavelength of a red blood cell, with mass 1.00 \(\times\) 10\(^{-11}\) g, that is moving with a speed of 0.400 cm/s? Do we need to be concerned with the wave nature of the blood cells when we describe the flow of blood in the body?

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.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free