Physics for Scientists and Engineers with Modern Physics
10th Edition
ISBN: 9781337553292
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 39, Problem 26P
(a)
To determine
The magnitude of minimum kinetic energy required for the electrons.
(b)
To determine
The magnitude of minimum energy of photon.
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The resolving power of a microscope depends on the wavelength used. If you wanted to “see” an atom, a wavelength of approximately 1.00 × 10-11 m would be required. (a) If electrons are used (in an electron microscope), what minimum kinetic energy is required for the electrons? (b) What If? If photons are used, what minimum photon energy is needed to obtain the required resolution?
The resolving power of a microscope is proportional to the wavelength used. A resolution of 1.0 x 10-11 m (0.010 nm) would be required in order to “see” an atom. (a) If electrons were used (electron microscope), what minimum kinetic energy would be required of the electrons? (b) If photons were used, what minimum photon energy would be needed to obtain 1.0 x 10 -11 m resolution?
The resolving power of a microscope depends on the wavelength used. If you
wanted to "see" an atom, a wavelength of approximately 1.00 x 10-11 m would be
required.If electrons are used (in an electron microscope), what minimum kinetic
energy is required for the electrons?
Chapter 39 Solutions
Physics for Scientists and Engineers with Modern Physics
Ch. 39.1 - Prob. 39.1QQCh. 39.2 - Prob. 39.2QQCh. 39.2 - Prob. 39.3QQCh. 39.2 - Prob. 39.4QQCh. 39.3 - Prob. 39.5QQCh. 39.5 - Prob. 39.6QQCh. 39.6 - Prob. 39.7QQCh. 39 - Prob. 1PCh. 39 - Prob. 2PCh. 39 - Prob. 3P
Ch. 39 - Prob. 4PCh. 39 - Prob. 5PCh. 39 - Prob. 6PCh. 39 - Prob. 8PCh. 39 - Prob. 9PCh. 39 - Prob. 10PCh. 39 - Prob. 11PCh. 39 - Prob. 12PCh. 39 - Prob. 13PCh. 39 - Prob. 15PCh. 39 - Prob. 16PCh. 39 - Prob. 17PCh. 39 - Prob. 18PCh. 39 - Prob. 19PCh. 39 - Prob. 20PCh. 39 - Prob. 22PCh. 39 - Prob. 23PCh. 39 - Prob. 24PCh. 39 - Prob. 25PCh. 39 - Prob. 26PCh. 39 - Prob. 27PCh. 39 - Prob. 30PCh. 39 - Prob. 31PCh. 39 - Prob. 32PCh. 39 - Prob. 33PCh. 39 - Prob. 35PCh. 39 - Prob. 37PCh. 39 - Prob. 38PCh. 39 - Prob. 39PCh. 39 - Prob. 40APCh. 39 - Prob. 41APCh. 39 - Prob. 43APCh. 39 - Prob. 44APCh. 39 - Prob. 45APCh. 39 - Prob. 46APCh. 39 - Prob. 47CPCh. 39 - Prob. 48CPCh. 39 - Prob. 49CPCh. 39 - Prob. 50CP
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- At what velocity will an electron have a wavelength of 1.00 m?arrow_forwardAn electron microscope uses magnets to accelerate electrons with negligible initial speed through apotential difference of 50 million volts. Calculate the ratio of the resolution of this electron microscope to theresolution of a microscope that uses visible light with a wavelength of 5.00 × 10−7 m through a circularaperture with an aperture of 5.00 mm.arrow_forwardTo resolve an object in an electron microscope, the electrons' wavelength must be close to the diameter of the object. What kinetic energy must the electrons have in order to resolve a protein molecule that is 8.10 nm in diameter? Take the mass of an electron to be 9.11 ×× 1031 kg.arrow_forward
- An electron microscope passes 1.00-pm-wavelength electrons through a circular aperture 2.00 μm in diameter.What is the angle between two just-resolvable point sources for this microscope?arrow_forwardAn electron microscope passes 1.00-pm-wavelength electrons through a circular aperture 2.00 µm in diameter. What is the angle between two just-resolvable point sources for this microscope?arrow_forwardIn designing an experiment, you want a beam of photons and a beam of electrons with thesame wavelength of 0.281 nm, equal to the separation of the Na and Cl ions in a crystal ofNaCl. Find the energy of the photons and the kinetic energy of the electrons in electrovolts(eV).arrow_forward
- To study crystal diffraction we need wavelengths of about 0.5 x 10-10 m. What would be the corresponding kinetic energies in eV of (a) a photon, (b) an electron, and (c) a neutron?arrow_forwardFor crystal diffraction experiments, wavelengths on the order of 0.25 nm are often appropriate. Find the energy in electron volts for a particle with this wavelength if the particle is (a) a photon; and (b) an electron.arrow_forwardThe de Broglie wavelength of an electron has to do with spatial resolution of an electron microscope, which is often expressed in the unit of length Å (Angstrom). The 1 V potential difference causes an electron to gain kinetic energy EK of 1 electron Volt (eV). In the SI units, kinetic energy in eV must be converted to Joules. The conversion factor is 1 eV = 1.6 x 10-19 Joule. And, the formula for the wavelength is: λ = h / p = h / √(2 m EK) where m is electron mass. Calculate the de Broglie wavelength of an electron (in Å) when the electron is accelerated from rest through a potential difference of: a) 1 kV = 1,000 V (a low resolution setting of microscope), b) 10 kV = 10,000 V (intermediate resolution). c) 100 kV = 100,000 V (high resolution),arrow_forward
- An electron beam is shot through 2 thin slits with spacing 4.000 x 10-6 m and land on a detector 2.50 m away. The detector is 1.00 cm wide. You observe that the spacing of interference fringes is 8.25 x 10-6 m. (a) What is the wavelength of the electrons? (b) What is the momentum of the electrons? (c) What is the uncertainty in the momentum? (Hint: assume that the uncertainty in position is the width of the detector.)arrow_forwardRayleigh’s criterion is used to determine when two objects are barely resolved by a lens of diameter d. The angular separation must be greater than θR where θR = 1.22 λ/d In order to resolve two objects 4000 nm apart at a distance of 20 cm with a lens of diameter 5 cm, what energy (a) photons or (b) electrons should be used? Is this consistent with the uncertainty principle?arrow_forwardThe highest achievable resolving power of a microscope is limited only by the wavelength used; that is, the smallest item that can be distinguished has dimensions about equal to the wavelength. Suppose one wishes to “see” inside an atom. Assuming the atom to have a diameter of 100 pm, this means that one must be able to resolve a width of, say, 10 pm. (a) If an electron microscope is used, what minimum electron energy is required? (b) If a light microscope is used, what minimum photon energy is required? (c) Which microscope seems more practical? Why?arrow_forward
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