COLLEGE PHYSICS,V.2
11th Edition
ISBN: 9781305965522
Author: SERWAY
Publisher: CENGAGE L
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Question
Chapter 27, Problem 33P
(a)
To determine
The minimum kinetic energy of the electrons.
(b)
To determine
The minimum energy of the 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 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),
Chapter 27 Solutions
COLLEGE PHYSICS,V.2
Ch. 27.5 - Prob. 27.1QQCh. 27.5 - Prob. 27.2QQCh. 27.5 - Prob. 27.3QQCh. 27.6 - Prob. 27.4QQCh. 27.6 - Prob. 27.5QQCh. 27 - Prob. 1CQCh. 27 - Prob. 2CQCh. 27 - Prob. 3CQCh. 27 - Prob. 4CQCh. 27 - Prob. 5CQ
Ch. 27 - Prob. 6CQCh. 27 - Prob. 7CQCh. 27 - Prob. 8CQCh. 27 - Prob. 9CQCh. 27 - Prob. 10CQCh. 27 - Prob. 11CQCh. 27 - Prob. 12CQCh. 27 - Prob. 13CQCh. 27 - Prob. 14CQCh. 27 - Prob. 15CQCh. 27 - Prob. 16CQCh. 27 - Prob. 1PCh. 27 - Prob. 2PCh. 27 - Prob. 3PCh. 27 - Prob. 4PCh. 27 - Prob. 5PCh. 27 - Prob. 6PCh. 27 - Prob. 7PCh. 27 - Prob. 8PCh. 27 - Prob. 9PCh. 27 - Prob. 10PCh. 27 - Prob. 11PCh. 27 - Prob. 12PCh. 27 - Prob. 13PCh. 27 - Prob. 14PCh. 27 - Prob. 15PCh. 27 - Prob. 16PCh. 27 - Prob. 17PCh. 27 - Prob. 18PCh. 27 - Prob. 19PCh. 27 - Prob. 20PCh. 27 - Prob. 21PCh. 27 - Prob. 22PCh. 27 - Prob. 23PCh. 27 - Prob. 24PCh. 27 - Prob. 25PCh. 27 - Prob. 26PCh. 27 - Prob. 27PCh. 27 - Prob. 28PCh. 27 - Prob. 29PCh. 27 - Prob. 30PCh. 27 - Prob. 31PCh. 27 - Prob. 32PCh. 27 - Prob. 33PCh. 27 - Prob. 34PCh. 27 - Prob. 35PCh. 27 - Prob. 36PCh. 27 - Prob. 37PCh. 27 - Prob. 38PCh. 27 - Prob. 39PCh. 27 - Prob. 40PCh. 27 - Prob. 41APCh. 27 - Prob. 42APCh. 27 - Prob. 43APCh. 27 - Prob. 44APCh. 27 - Prob. 45APCh. 27 - Prob. 46APCh. 27 - Prob. 47APCh. 27 - Prob. 48APCh. 27 - Prob. 49APCh. 27 - Prob. 50APCh. 27 - Prob. 51APCh. 27 - Prob. 52AP
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- At what velocity will an electron have a wavelength of 1.00 m?arrow_forwardDerive an expression for the ratio of X-ray photon frequency for two elements with atomic numbers Z1 and Z2.arrow_forwardAn X-ray tube has an applied voltage of 100 kV. (a) What is the most energetic X-ray photon it can produce? Express your answer in electron volts and joules. (b) Find the wavelength of such an X-ray.arrow_forward
- The work function for potassium is 2.26 eV. What is the cutoff frequency when this metal is used as photoelectrode? What is the stopping potential when for the emitted electrons when this photo electrode is exposed to radiation of frequency 1200 THz?arrow_forwardWhat is the de Brogue wavelength of a proton whose kinetic energy is 2.0 MeV? 10.0 MeV?arrow_forwardA 600-nm light falls on a photoelectric surface and electrons with the maximum kinetic energy of 0.17 eV are emitted. Determine (a) the work function and (b) the cutoff frequency of the surface. (c) What is the stopping potential when the surface is illuminated with light of wavelength 400 nm?arrow_forward
- Why are X-rays emitted only for electron transitions to inner shells? What type of photon is emitted for transitions between outer shells?arrow_forwardThe 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?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_forward
- To resolve an object in an electron microscope, the wavelength of an electrons must be close to the diameter of the object. What kinetic energy must they have in order to resolve a protein molecule that is 8.40 nm in diameter. Take the mass of an electron to be 9.11 x 10-31 kg.arrow_forwardThe smallest resolving power of a microscope is proportional to the magnitude of the wavelength used. a) 15 nm and b) How many energetic electrons are needed in the electron microscope for a resolution of 0.5nm?arrow_forwardWhat is the kinetic energy of each electron in a beam of electrons if the beam produces a diffraction pattern of a crystal which is similar to that of a beam of 1.00 eV neutrons? (knowing that electron mass is 9.11*10^-31 kg and neutron mass is 1.67*10^-26 kg). What are the specific equations that are being used in this problem?arrow_forward
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