Shows 1.0-μm diameter dust particles (m= .1.0 ×10^-15 kg) in...

shows $1.0-\mu \mathrm{m}$ diameter dust particles $(m=$ $\left.1.0 \times 10^{-15} \mathrm{kg}\right)$ in a vacuum chamber. The dust particles are released from rest above a $1.0-\mu$ m-diameter hole, fall through the hole (there's just barely room for the particles to go through), and land on a detector at distance $d$ below. a. If the particles were purely classical, they would all land in the same $1.0-\mu \mathrm{m}$ -diameter circle. But quantum effects don't allow this. If $d=1.0 \mathrm{m},$ by how much does the diameter of the circle in which most dust particles land exceed $1.0 \mu \mathrm{m} ?$ Is this increase in diameter likely to be detectable? b. Quantum effects would be noticeable if the detection-circle diameter increased by $10 \%$ to $1.1 \mu \mathrm{m}$. At what distance $d$ would the detector need to be placed to observe this increase in the diameter?

shows $1.0-\mu \mathrm{m}$ diameter dust particles $(m=$ $\left.1.0 \times 10^{-15} \mathrm{kg}\right)$ in a vacuum chamber. The dust particles are released from rest above
a $1.0-\mu$ m-diameter hole, fall through the hole (there's just barely room for the particles to go through), and land on a detector at distance $d$ below.
a. If the particles were purely classical, they would all land in the same $1.0-\mu \mathrm{m}$ -diameter circle. But quantum effects don't allow this. If $d=1.0 \mathrm{m},$ by how much does the diameter of the circle in which most dust particles land exceed $1.0 \mu \mathrm{m} ?$ Is this increase in diameter likely to be detectable?
b. Quantum effects would be noticeable if the detection-circle diameter increased by $10 \%$ to $1.1 \mu \mathrm{m}$. At what distance $d$ would the detector need to be placed to observe this increase in the diameter?

Key Concepts

Wave-Particle Duality

This concept refers to the idea that matter exhibits both particle-like and wave-like properties. In quantum mechanics, particles such as electrons, atoms, and even larger particles have associated wavelengths, leading to interference and diffraction effects that are not predicted by classical mechanics.

de Broglie Wavelength

The de Broglie wavelength is a fundamental concept in quantum mechanics that assigns a wavelength to a moving particle based on its momentum. It highlights that every particle with momentum has an associated wave, and its wavelength is inversely proportional to the momentum, influencing phenomena such as diffraction when the particle passes through a small aperture.

Quantum Diffraction Through an Aperture

Quantum diffraction occurs when particles with wave properties pass through an aperture that is comparable in size to their wavelength. The spreading out of the particle's wavefunction after passing through the aperture produces a pattern on a detector that is broader than the physical size of the aperture, contrasting with the predictions of classical mechanics.

Heisenberg’s Uncertainty Principle

This principle states that certain pairs of physical properties, such as position and momentum, cannot both be precisely known simultaneously. When a particle’s position is confined by a narrow aperture, the uncertainty in its momentum increases, leading to a greater spread in the particle's subsequent position, which is observed as an enlarged detection pattern.

Transcript

00:01 In this problem, we are given the particles mass, the width of the particles diameter, which is exactly the same width as the circle they go through, and then the distance between this circle and the detection plane, which is 1 meter.

00:25 With this, we want to know how much bigger is the circle that the particle form in the detection plane with regards to their diameter.

00:42 For this we perform the usual, which is we take into account that the eisenberg uncertainty principle tells us that this is bigger than age over two.

01:00 And the uncertainty and momentum is just the mass times uncertainty and velocity.

01:06 So we get velocity and certainty.

01:11 That's a over 2 times the mass times delta x.

01:18 This is 3 .32 times 10 to the minus 13 meters per second.

01:36 Okay.

01:37 So assuming a mean speed of 0 meters per second, then the range of speeds is from minus half of that to plus half of that, which is from minus 1 .66 to plus 1 .66 all the interval, the whole interval times 10 to the minus 13 meters per second.

02:14 Okay, so assuming these particles are in free fall, we can calculate the time they take to fall by knowing that the change in distance.

02:29 So in the y direction like this is a minus 1 meter, right? this is just the initial velocity, or should i say? so delta y is equal to, yeah, the initial velocity multiplied by the time.

03:02 It took minus one half acceleration of gravity multiplied by the time squared.

03:11 And if we consider the initial velocity to be zero, we just get one half of 9 .8 times t squared.

03:28 And this gives us a t of around.

03:35 So you need to take the square root and take the positive value.

03:38 You'll get 0 .45 seconds.

03:41 And assuming the maximum velocity of the particles is 1 .66 times 10 to the minus 13 meters per second.

03:53 The horizontal distance traveled in this time is simply so like this, x, dx, or should i say, to keep it.

04:12 All the same, i'll say delta x, no, that i have it above.

04:17 Okay, i'll call it.

04:23 Dx is delta v, the one we had.

04:31 So this in the x direction, multiplied by the time.

04:39 Time it took to fall, which means this is so 7 .5, so it's around 7 .5 times 10 to the minus 14 meters, which means that half the particles will move to the right and have a distance between zero and this and the other half to the left and have a zero, a distance between zero and minus this the x...

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