diffraction-effects-become-significant-when-the-width-of-an-aperture-is-comparable-to-the-wavelength

Question

Diffraction effects become significant when the width of an aperture is comparable to the wavelength of the waves being diffracted. (a) At what speed will the de Broglie wavelength of a $65-$ kg
student be equal to the $0.76-\mathrm{m}$ width of a doorway? (b) At this speed, how much time will it take the student to travel a distance
of 1.0 $\mathrm{mm}$ ? (For comparison, the age of the universe is approximately $4 	imes 10^{17} \mathrm{s.}$ .

Zulfiqar Ali · Accepted Answer

must&#x27;ve student equals 65 grands. Um moss off student is equal to 65 kilograms. And what work is 0.76 meters? We need to find out speed. Ah, land vehicles. It&#x27;s divided by B and B equals mass times lost e Therefore, lost equals. It&#x27;s divided by most times prevalent. Let&#x27;s plug in the values which is 6.63 Multiply the 10 to the dollar minus 34. Divided by mass is 65 kilograms multiplied by, uh, bad lint Lander. Um so oh, here lender is ah zero point 76 dividing the two numbers lost, he comes out to be ah 1.3 Multiply 10 to the power minus 35 meters for a second. So velocity is equal to 1.3 multiply. We tend to the power of minus 35 meter per second. We know that distance equals speed times, time and T equals speed divided by velocity. Well distance given his one millimeter divided by our speed is 1.3 multiply by 10 to the power minus 35 meter per second one multiplied by 10 people are minus three meters, divided by 1.3 multiply with tend to the bar minus 35. Dividing the two numbers. T equals 7.4 multiplied by damn dirty power, 31 seconds.

Ren Jie Tuieng · Answer

argument discussion that if you want to observe the friction effects as we passed through, it&#x27;ll be the the wavelength. There. You must have must be about one meters, because this about the order off magnitude off size off our door. Why don&#x27;t we? And therefore using this idea dick a debrief linked to be about one meters. You create this to a D. He built this formula, right? Deb always were saying this because each over the momentum can use this to find the momentum right and therefore the speak off. What we need to move it. Just hit your for m times under 60 kilograms reflex this one meter Sochi Fast boats 1.1 sent polmoney study Fife just a second. Well, at this particular speed, the time taken to move Just one step, which is your point. Meet us. Time taken would be gays. Your 0.8 divided by t lost he This case is one State minister five and this gives us seven. Treat Said posted four seconds where we divide this by 3.5156 Step out seven seconds. Put you right to get time in terms off years and this will be two point tree time. State of off to the seven. Yes, this is a very, very long time to take one step. And since we walked true Hey, go away! But it it&#x27;s a much faster rate much faster of ust get this b do not actually notice perfection effects.

Elan Stopnitzky · Answer

Okay, So in order to do so, Fracked through a door, you need a wavelength to broke me wavelength on the order of one meter. And this is related to the Mo mentum through P equals h divided by Landa and momentum is equal to M V. So if lambda is one is one meter, then we have is equal to 6.63 times 10 to the negative 34 divided by 60 kilograms which is Theus Umed Mass. And this comes out toe 1.1 times 10 to the negative, 35 meters per second. So one would have to be moving extremely slowly and were asked as well a tte this speed. How long would it take to take a step? Assuming a step, it is 0.8 meters long. Um, yeah, let&#x27;s convert this to units of years. So we have 60 times 60 times, 24 times 3 65 and this comes out to 2.3 times 10 to the 27 years. So the diffraction is really not a significant effect that human skills

Guilherme Barros · Answer

it&#x27;s exercise were given the information that if you had a wave ling lambda of 1 m, you would be able to do fracked, an exhibit wavelike properties. When passing through a door and in question, you have to calculate what is the speed of a person who has a mass of 60 kg so that she has, ah, a wavelength from 1 m. I remember that according to the Debra Lee formula, the momentum off any particle is age divided by its liberally wavelength lambda, and the momentum of massive particle is M times. The speed V so V is equal to age divided by M. Lunda. And if we sit through the numbers here, we have a H in 16 63 times. Stand to the monastery for this points constant. Divided by the mass, which is 60 kilograms times the wavelength of 1 m V is equal to 1.1 times 10 to the minus 35 m your second, so this should be the speed of someone who has a wavelength of 1 m. Been in Question V. We have to calculate how much time a person who&#x27;s moving with this speed would take in order to travel additions of 0.8 m, which is the approximate distance of one step. Well, remember that these equal to X, divided by T X, is a distance to use the time, meaning that t is equal to acts divided by V. Relax 0.8 m, while the speed V is 1.1 time standing a minus 35 meters per second. So the time is equal just to be 7.3 times 10 to the 34 seconds and makins transform this into ears. Basically, what we have to do is to divide the time by 336 100 seconds for our times 24 hours a day, times 365 days a year. When we get that, the number off years, he is 2.3 times 10 from 27. Now notice that the age of the universe is approximately 14 billion years, so this is 14 times 10 to the ninth years. So I noticed that the time it would take for someone who&#x27;s moving with that speed to ah complete one step would be much, much, much greater than the age of the universe. In fact, it would be 17 orders of magnitude reader. And for that reason, since people do not move that slow, it&#x27;s not possible to notice. People do fracking through doors, which includes or exercise.

Zachary Warner · Answer

we&#x27;re told that we have an electron, uh, which has a central, bright fringe of a diffraction pattern on a screen that&#x27;s identical with that of red light. So the distance between the screen, the slit is the same in each place and look and is large compared to the slit with how fast is the electron moving? Okay, so to do this, we&#x27;re gonna consider the, uh, the world you wavering wavelength. And we&#x27;re gonna say that lambda is equal to H over envy for the electron. Therefore, rearranging this to Sulphur V, we find that V is equal to H Over and Lambda, where h is Planck&#x27;s constant and M is just the mass of the electron. So plugging these values in, we find that this is equal to 1.1 times 10 to the three meters per second. Linkenbach said it is the solution to our question.

Problem

A particle has a mass $m$ and an electric charge …

Problem 70 Hard Difficulty

Answer

a) $1.34 \times 10^{-35} \mathrm{m} / \mathrm{s}$
b) $7.45 \times 10^{31} \mathrm{s}$

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a) $1.34 \times 10^{-35} \mathrm{m} / \mathrm{s}$b) $7.45 \times 10^{31} \mathrm{s}$

Physics

Liev B.

Marshall S.

Farnaz M.

Meghan M.

Wave Optics - Intro

Multiple Aperture Interference - Overview

James S. WalkerPhysics

Liev B.

Marshall S.

Farnaz M.

Meghan M.

a) $1.34 \times 10^{-35} \mathrm{m} / \mathrm{s}$
b) $7.45 \times 10^{31} \mathrm{s}$

James S. Walker

Physics