General Physics, 2nd Edition
General Physics, 2nd Edition
2nd Edition
ISBN: 9780471522782
Author: Morton M. Sternheim
Publisher: WILEY
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Chapter 27, Problem 14E

(a)

To determine

The momentum of proton.

(b)

To determine

The kinetic energy of probe.

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The electron beam in a scanning electron microscope (SEM) can have an energy ranging from 0.2 keV to 40 keV. (Note:  an eV is an electronvolt, the measure of an amount of kinetic energy for a single electron.  1 eV = 1.60218×10−19 J and 1 J  = 1 kg m2 s−2).     What is the de Broglie wavelength of the electron beam in the SEM when the electrons are accelerated to kinetic energy of 6.6 keV?
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),
An electron moves with a speed of 3.5 x 10^5 m/s. What is its de Broglie wavelength? (b) A proton moves with the same speed. Determine its de Broglie wavelength.
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