Explore BrainMass

# De Broglie Wavelengths of Visible Particles

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

The De Broglie wavelengths of visible particles. Why does the wave nature of particles come as such a surprise to most people? If, as, De Broglie says, a wavelength can be associated with every moving particle, then why are we not forcibly made aware of this property in our everyday experience? In answering, calculate the de Broglie wavelength of each of the following "particles":

a)
an automobile of mass 2 metric tons (2000 kg) traveling at a speed of 50 mph (22 m/s)

b)
a marble of mass 10 g moving with a speed of 10 cm/sec

c)
a smoke particle of diameter 10^-5 cm (and a density of 2 g/cm3) being jostled about by air molecules at room temperature (27 c = 300 K). Assume that the particle has the same translational kinetic energy as the thermal average of the air molecules.

p^2/2m = 3kT/2

with k = Boltzmann's constant = 1.3*10^-16 erg/degree K

d)
speculative question: If the constants of nature were such that the de Broglie wavelength was of importance in decoding everyday experience, what forms would this experience take?

e)
interpretive question: Student A says that the wavelengths calculated in this exercise are utterly meaningless, since they are incapable of being verified. Student B maintains that, although they are miniscule, these wavelengths have an indisputable meaning. What criteria would you use in judging these competing claims?

https://brainmass.com/physics/kinetic-theory/de-broglie-wavelengths-visible-particles-525334

#### Solution Preview

See the attached file.
The de Broglie wavelength of a particle is given by λ = h/mv or λ = h/p where h is the Boltzmann constant and p is the linear momentum. Covert units where necessary.

a. λ = h/mv = 6.6 x 10-34 / 2000x22 = 1.5 x 10-38 m

b. λ = h/mv = 6.6 x 10-34 / 0.010x0.1 = 6.6 x 10-31 m

c. Mass = volume x density = 4/3 pi r^3 * density = 4/3 pi (0.5x10-7 )^3 * 2000 kg/m3 = 1.047 x 10-18 kg
P2/2m = 3kT/2 = 3 (1.38x10-23) 300/2 = 6.21 x 10-21 J
p2 = 6.21 x 10-21 x 2m = = 6.21 ...

#### Solution Summary

De Broglie's wavelengths of visible particles are examined in the solution.

\$2.19