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Modern Physics with de-Broglie and Electron Microscope

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

See attachment for better formula representation.

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Solution Preview

Please see the attachment.

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.