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Modern Physics with de-Broglie and Electron Microscope

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1. Wavelength of particles
a. Consider a dust particle of diameter d =1 micron, mass m= 10^-15 kg, and speed v = 1 mm/s. Calculate the de Broglie wavelength of the particle and compare with the particle's size.
b. Now consider a thermal neutron, i.e, a neutron (m=1.67 x 10^-27 kg) with a speed corresponding to the average thermal energy at (absolute) temperature T, give by:
1/2 mv^2 = p^2 / 2m = 3/2 kBT,
where kB= 1.38 x 10^-23 J/K (joule/degree) is Boltzmann's constant. Find the wavelength of such a neutron at room temperature (T=300 K).

2. Wavelength and Resolution:
The smallest separation resolvable by a microscope is of the order of the wavelength used. In an electron microscope, what energy of electrons would be needed to resolve separations of:
a. 100 Amps
b. 5 Amps

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Answer:
The de Broglie.s wavelength  associated with a particle of mass m moving with velocity v is given by

where P is the momentum of the particle and h is plank constant and its value is 6.63*10-34.

Substituting m = 10-15 kg and v = 1 ...

Solution Summary

This solution discusses properties of de-Broglie wavelength and solves for particle size, wavelength of thermal neutron and the resolving power of an electron microscope.

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28. A wave packet describes a particle having momentum p. Starting with the relativistic relationship E^2 = p^2 c^2 + E0^2, show that the group velocity is Bc and the phase velocity is c/B (where B = v/c). How can the phase velocity physically be greater than c?

44. An electron microscope is designed to resolve objects as small as 0.14 nm. What energy electrons must be used in this instrument?

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