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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Question
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 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?
The resolving power of a microscope depends on the wavelength used. If you wanted to "see" an atom, a
resolution 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_forwardResolving ‘power’ of an electron microscope versus optical (photon) microscope: If a resolution of 1.0 x 10-11m (0.010nm) is required to ‘see’ an atom, (a) If electrons are used (e-microscope), what minimum kinetic energy of the electrons is required? Use deBroglie’s Hypothesis and KE = p2/2me and non-relativistic velocities: (b) If photons are used, what minimum KE (Eγ) is required to obtain 10-11m resolution?arrow_forwarda) A typical atomic nucleaus is about 5.00×10-¹5 m in radius. Use uncertanity principle to place a lower limit on the energy an electron must have if it is to be a part of nucleaus. b) Find phase velocites of 2 waves moving with velocities v₁= 7m/s and v2= 6m/s, having wavelengths A₁ = 15nm and A2= 22nm, (b) Investigate that combined waveform moves equal, faster/slower than either of its component waves?arrow_forward
- To 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_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_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_forward
- In 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_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_forwardTo 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_forward
- For 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_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_forward
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