please explain One of the smallest macroscopic particles we could imagine using

Question

please explain

One of the smallest macroscopic particles we could imagine using for an experiment would be a very small smoke or soot particle. These are almostequalto 1 mu m in diameter, too small to see with the naked eye and just barely at the limits of resolution of a microscope. A particle this size has mass m almostequalto 10^-18 kg. Estimate the de Broglie wavelength for a 1-mu m-diameter particle moving at the very slow speed of 1 mm/s. The particle's momentum is p = mv almostequalto 10^-21 kg middot m/s. The de Broglie wavelength of a particle with this momentum is lambda = h/p almostequalto 7 times 10^-14 m The wavelength is much, much smaller than the particle itself-much smaller than an individual atom! We don't expect to see this particle exhibiting wave-like behavior.

Explanation / Answer

formula for deBroglie wavelength is

lamda = h/p

p is momentum of particle it can be expressed as p = mv

lamda = h/ mv

here , h is plack's constant

lamda = 6.63 * 10^-34 / 10^-18 kg ( 1 * 10^-3 m/s)

=6.63 * 10^-13 m

rounding to one significant digit

lamda = 7 * 10^-13 m

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