Imagine the following Young’s double-slit experiment using mat...

Question

Imagine the following Young’s double-slit experiment using matter rather than light: electrons are accelerated through a potential difference of 18 V, pass through two closely spaced slits separated by a distance d, and create an interference pattern. (a) Using Example 37–11 and Section 34–3 as guides, find the required value for d if the first-order interference fringe is to be produced at an angle of 10°. (b) Given the approximate size of atoms, would it be possible to construct the required two-slit set-up for this experiment?

Accepted Answer

Everyone. Let's take a look at this practice problem dealing with de Bru's wavelength. This problem says a young double slit experiment is set up using alpha particles. The particles are accelerated through a potential difference of 100 volts to achieve a first order interference fringe at an angle of 20 degrees. What is the necessary distance between the two slits? And we're given a hint that the mass of the alpha particle is 6.64 multiplied by 10 to the negative 27 kg. We give four possible choices as our answers. For choice A we have 2.1 peak of meters. For choice B, we have 3.0 pe of meter. For choice C we have 4.1 peak of meters. And for choice D, we have 5.5 peak of meters. Now we're asked to calculate these uh separation between the slits. And since we're looking at fringes here, we're gonna be looking at the maximum of our interference pattern. So we call your formula for finding the maxima of a double slit interference pattern. And so that formula is D multiplied by sign the is equal to M multiplied by LAMBDA, where D here is the separation between slits, the is the angular location of one of the Maxima on the screen. M is the order number for that maximum. And lambda is the wavelength. Now I need to solve this four D since that represents the separation between the two slits. So I'll divide both sides by sine theta. And so I'll get D equal to M LAMBDA divided by sine theta. Now I was given the order number and the problem which is one since we're looking at the 1st, 1st order interference fringe, and we're given the angle um for theta which is 28 degrees. But I was not given any information directly about the wavelength. However, I am dealing with particles here. So I'm going to want to use our droit wavelength or the wavelength in our interference formula. So recall your formula for the wavelength that is P is equal to H divided by lambda, where P here is the momentum of the particle H is planks constant and lambda is the right way of length. So I need to solve this for Lambda by multiplying both sides by lambda and dividing both sides by P. And so when I do that, I'll get lambda is equal to H divided by P. Now, I was not given any information about the momentum in the problem, but I was given the potential difference through which the particle is accelerated. And since I have a potential difference with a charge particle in it. That means I have some sort of potential energy and that potential energy gets converted into kinetic energy. And I can write my kinetic energy in terms of the momentum. So that's the route that we're going to do take in order to calculate our momentum in terms of this potential difference. So we're actually going to look at energy conservation, we call for energy conservation that we have our initial energy, which in this case is just our initial potential energy, which we'll call U I and that's gonna be equal to our final energy, which in this case is going to be all kinetic energy will just have KF So K here represents our final kinetic energy. Now recall your um formula for the potential energy of a charge particle digging in a, an electric potential. And that formula is Q multiplied by V where Q is the charge of the particle and V is the electric potential. And that's gonna be equal to our kinetic energy, which we can write in terms of our momentum. So recall that we'll have P squared divided by two M alpha where P here is our momentum and M alpha is the mass of the alpha particle. Now, four alpha particles here, they're made up of two protons and two neutrons. That means their charge is going to be a positive two E. So we can replace Q in our equation with two E will have two, E multiplied by V is equal to P squared divided by two M alpha. So at this point I just need to solve for P by multiplying both sides by two M alpha and then taking the square root. And so I have P is equal to the square root of four M alpha multiplied by E multiplied by V. So I can take that expression and plug it back in to our formula or the deploy wavelength. And so we'll have lambda is equal to H divided by the square root of the quantity of four M alpha multiplied by E multiplied by V. And now we can take that um formula for lambda and plug it back into our slit separation formula. And so we'll have D is equal to M multiplied by H divided by fine, the multiplied by the square root of four M alpha multiplied by E multiplied by V. And so we actually have values for everything on the right hand side. So we can just plug those in and calculate our D. We have D is equal to or M we're looking at the first order. So that's just gonna be one H is Planck's constant, which has value of 6.63 multiplied by 10 to the negative 34 dual seconds. And that's gonna be divided by the quantity of sign of the, and theta is equal to 20 degrees. And that's gonna be multiplied by the square root of the quantity of four multiplied by the mass of the alpha particle, which we were given. And that is six point 64 multiplied by 10 to the negative 27 kg. That's gonna be multiplied by the charge of the proton, which is 1.6 multiplied by 10 to the negative 19 cools. And then that is going to be multiplied by our potential difference, which is 100 volts. So I can take all those values and plug them into my calculator. And I'll get D is equal to 3.0 multiplied by 10 to the negative 12 m. And here just kept two significant figures. But if I look at my answer choices, they're all in PICO meters, they'll need to multiply this by the conversion factor of one om Meer divided by one multiplied by 10 to the negative 12 m. So that means that D gonna be equal to 3.0 peak of meters. And so that's the answer that I'm looking for and that corresponds to answer choice B. So just a quick recap of what we've done here, we started off by using our formula for finding the location of the maxima for a double slit interference pattern. But in order to use that formula, we needed to calculate the wavelength and the wavelength in this case was our droit wavelength. So we use our formula for the deroy wavelength in order to write the droit wavelength in terms of the momentum. And we then had to use our conservation of energy to write our momentum in terms of the potential difference that we were given in the problem. And once we combine all those together into our formula for the split separation, we were able to calculate a numerical value for that slip split separation. So I hope that this has been useful and I'll see you in the next video.

Key Concepts

Wave-Particle Duality

Interference Pattern

Potential Difference and Electron Acceleration

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