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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=hcλ, where h is Planck's constant 6.626×1034 Js, c is the speed of light 3.00×108 m/s, and λ=5.00×109 m is the wavelength. Calculate E:E=6.626×1034×3.00×1085.00×109=3.97×1017 JConvert the energy from Joules to electronvolts (1 eV = 1.602×1019 J):E=3.97×10171.602×1019=248 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 λ=hp where p is momentum. Hence the momentum p=hλ. Since p=mv and kinetic energy (KE) is 12mv2, we can express KE as p22m:KE=h22mλ2Here, m=9.11×1031 kg (electron mass) and λ=5.00×109 m.KE=(6.626×1034)22×9.11×1031×(5.00×109)2=6.02×1020 JConvert to eV:KE=6.02×10201.602×1019=0.375 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.

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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 (λ) of a particle, you use the equation λ=hp, 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 6.626×1034Js
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=hcλ.
  • Here, h is the Planck's constant 6.626×1034Js
  • c is the speed of light 3.00×108m/s
  • λ 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=p22m, where m is the mass of the electron.
  • Momentum can be expressed using the de Broglie relation: p=hλ
  • The mass of an electron is 9.11×1031kg
This formula shows that by knowing the wavelength λ 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.

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Most popular questions from this chapter

Why is it easier to use helium ions rather than neutral helium atoms in such a microscope? (a) Helium atoms are not electrically charged, and only electrically charged particles have wave properties. (b) Helium atoms form molecules, which are too large to have wave properties. (c) Neutral helium atoms are more difficult to focus with electric and magnetic fields. (d) Helium atoms have much larger mass than helium ions do and thus are more difficult to accelerate.

For crystal diffraction experiments (discussed in Section 39.1), wavelengths on the order of 0.20 nm are often appropriate. Find the energy in electron volts for a particle with this wavelength if the particle is (a) a photon; (b) an electron; (c) an alpha particle (m = 6.64 × 1027 kg).

(a) What is the energy of a photon that has wavelength 0.10 μm ? (b) Through approximately what potential difference must electrons be accelerated so that they will exhibit wave nature in passing through a pinhole 0.10 μm in diameter? What is the speed of these electrons? (c) If protons rather than electrons were used, through what potential difference would protons have to be accelerated so they would exhibit wave nature in passing through this pinhole? What would be the speed of these protons?

(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 negative muon has a charge equal to that of an electron but a mass that is 207 times as great. Consider a hydrogenlike atom consisting of a proton and a muon. (a) What is the reduced mass of the atom? (b) What is the ground-level energy (in electron volts)? (c) What is the wavelength of the radiation emitted in the transition from the n = 2 level to the n = 1 level?

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