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 marble of mass 10 g moving with a speed of 10 cm/sec
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
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?
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?
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 ...
De Broglie's wavelengths of visible particles are examined in the solution.
Solving for the De Broglie Wavelength
1. a) Calculate the De broglie wavelength of a golf ball that is 100.0 g and moving at 30.0 m/s.
b) Why does the golf ball look like a ball and not like a beam of light? (hint: where is the wavelength on the em spectrum).View Full Posting Details