COLLEGE PHYSICS,V.2
COLLEGE PHYSICS,V.2
11th Edition
ISBN: 9781305965522
Author: SERWAY
Publisher: CENGAGE L
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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 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),
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?
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