Skip to main content
Ch 36: Diffraction

Chapter 35, Problem 36

If the planes of a crystal are 3.50 Å (1 Å = 10-10 m = 1 Ångstrom unit) apart, (a) what wavelength of electromagnetic waves is needed so that the first strong interference maximum in the Bragg reflection occurs when the waves strike the planes at an angle of 22.0°, and in what part of the electromagnetic spectrum do these waves lie?

Verified Solution
Video duration:
4m
This video solution was recommended by our tutors as helpful for the problem above.
253
views
Was this helpful?

Video transcript

Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let's read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. An emerging class of two dimensional material termed mxene has ac lattice parameter. A planar spacing of 9.23 years. If the electromagnetic ray is made incident on the planes at an angle of 30 degrees, determine its wavelength for the first interference maximum, also determine the region in which the electromagnetic ray falls in terms of its wavelength. Awesome. OK. So we're given some multiple choice answers. Let's read them off to see what our final answer might be. A is LAMBDA equals 450 nanometers lies in the this region. B is lambda equals 1.1 0.51 nanometers lies in the IR infrared region. C is lambda equals 0.108 nanometers lies in the X ray region and D is lambda equals 0.923 nanometers lies in the X ray region. OK. So first off, let us recall Bragg's Law equation. So Bragg's Law equation states that two multiplied by the interp planar spacing multiplied by sine theta is equal to the nth order multiplied by a wavelength. So N is some integer hole number for the order. OK. So let's write down all of our known and unknown variables. So D equals 9.23 years, but we need to convert to meters. So when we convert that to meters, we should get 9.23 multiplied by 10 to the power of negative 10 m. We know that the in this case equals 30 this is given to us in the problem, we do not know what the wavelength is. We're trying to find the wavelength for the interference maximum. And we know that the end value is equal to one because we're dealing with a first order condition. OK. So let's plug in all of our known variables to solve for the wavelength. So since we know that N equals one, so it'll just be one multiplied by lambda the wavelength. So we know that the wavelength is equal to two multiplied by D which was 9.23 multiplied by 10 to the power of negative 10 meters multiplied by sign of 30 degrees. OK. So we plug into a calculator, we should get that Lambda is equal to 9.23 multiplied by 10 to the power of negative 10 m. But when we convert this two nanometers, which all we have to do is just multiply by 10 to the power of nine, we should get 0.923 nanometers. So that is our lambda. So lambda equals the wavelength equals 0.923 nanometers. OK. So when we refer to electromagnetic spectrum chart or looking up our nanometer value, it determines that it lies within the X ray region. So OK. So it lies in the X ray region hooray, we did it, we finished solving this problem and this one was a quick one. OK. So that means that our multiple choice answer, the correct one is D wavelength equals. So lambda equals 0.923 nanometers and it lies in the X ray region. Thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.
Related Practice
Textbook Question
In a two-slit interference pattern, the intensity at the peak of the central maximum is I0. (a) At a point in the pattern where the phase difference between the waves from the two slits is 60.0°, what is the intensity?
270
views
Textbook Question
Two slits spaced 0.260 mm apart are 0.900 m from a screen and illuminated by coherent light of wavelength 660 nm. The intensity at the center of the central maximum 1u = 0°2 is I0. What is the distance on the screen from the center of the central maximum (a) to the first minimum
294
views
Textbook Question
Two slits spaced 0.260 mm apart are 0.900 m from a screen and illuminated by coherent light of wavelength 660 nm. The intensity at the center of the central maximum 1u = 0°2 is I0. What is the distance on the screen from the center of the central maximum (b) to the point where the intensity has fallen to I0>2?
327
views
Textbook Question
Parallel rays of green mercury light with a wavelength of 546 nm pass through a slit covering a lens with a focal length of 60.0 cm. In the focal plane of the lens, the distance from the central maximum to the first minimum is 8.65 mm. What is the width of the slit?
256
views
Textbook Question
Light of wavelength 585 nm falls on a slit 0.0666 mm wide. (a) On a very large and distant screen, how many totally dark fringes (indicating complete cancellation) will there be, including both sides of the central bright spot? Solve this problem without calculating all the angles! (Hint: What is the largest that sin u can be? What does this tell you is the largest that m can be?) (b) At what angle will the dark fringe that is most distant from the central bright fringe occur?
335
views
Textbook Question
Diffraction occurs for all types of waves, including sound waves. High-frequency sound from a distant source with wavelength 9.00 cm passes through a slit 12.0 cm wide. A microphone is placed 8.00 m directly in front of the center of the slit, corresponding to point O in Fig. 36.5a . The microphone is then moved in a direction perpendicular to the line from the center of the slit to point O. At what distances from O will the intensity detected by the microphone be zero?
390
views