College Physics
College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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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 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 × 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),
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