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Chapter 4: Problem 76

On a certain crystal, a first-order X-ray diffraction maximum is observed at an angle of \(27.1^{\circ}\) relative to its surface, using an X-ray source of unknown wavelength. Additionally, when illuminated with a different, this time of known wavelength \(0.137 \mathrm{nm},\) a second-order maximum is detected at \(37.3^{\circ} .\) Determine (a) the spacing between the reflecting planes, and (b) the unknown wavelength.

Short Answer

Expert verified
The spacing between the reflecting planes is approximately 0.235 nm, and the unknown wavelength is approximately 0.207 nm.

Step by step solution

01

Use Bragg's Law with the second-order maximum data

Using Bragg's Law and the given angle and wavelength for the second-order maximum, we can find the spacing between the reflecting planes (d). \(n \lambda = 2d \sin{\theta}\) We are given n = 2, λ = 0.137 nm, and θ = 37.3°. Plug in the values and solve for d: \(2 \times 0.137 \mathrm{nm} = 2d \sin{37.3^{\circ}}\)
02

Solve for the spacing between the reflecting planes (d)

Now, we solve the equation for d by dividing both sides by 2sin(37.3°): \(d = \frac{2 \times 0.137 \mathrm{nm}}{2\sin{37.3^{\circ}}}\) \(d = 0.137 \mathrm{nm} \div \sin{37.3^{\circ}}\) \(d \approx 0.235 \mathrm{nm}\) Thus, we find that the spacing between the reflecting planes is approximately 0.235 nm.
03

Use Bragg's Law with the first-order maximum data

Now, we will use Bragg's Law with the given angle for the first-order maximum and the calculated value of d to find the unknown wavelength (λ). \(n \lambda = 2d \sin{\theta}\) We have n = 1, d = 0.235 nm, and θ = 27.1°. Plug in the values and solve for λ: \(1 \times \lambda = 2 \times 0.235 \mathrm{nm} \sin{27.1^{\circ}}\)
04

Solve for the unknown wavelength (λ)

Now, we solve the equation for λ by dividing both sides by sin(27.1°): \(\lambda = 2 \times 0.235 \mathrm{nm} \div \sin{27.1^{\circ}}\) \(\lambda \approx 0.207 \mathrm{nm}\) Thus, we find that the unknown wavelength is approximately 0.207 nm. In conclusion, the spacing between the reflecting planes is approximately 0.235 nm, and the unknown wavelength is approximately 0.207 nm.

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Most popular questions from this chapter

On a bright clear day, you are at the top of a mountain and looking at a city \(12 \mathrm{km}\) away. There are two tall towers \(20.0 \mathrm{m}\) apart in the city. Can your eye resolve the two towers if the diameter of the pupil is 4.0 mm? If not, what should be the minimum magnification power of the telescope needed to resolve the two towers? In your calculations use \(550 \mathrm{nm}\) for the wavelength of the light.

Two slits of width \(2 \mu \mathrm{m},\) each in an opaque material, are separated by a center-to-center distance of \(6 \mu \mathrm{m}\). A monochromatic light of wavelength \(450 \mathrm{nm}\) is incident on the double- slit. One finds a combined interference and diffraction pattern on the screen. (a) How many peaks of the interference will be observed in the central maximum of the diffraction pattem? (b) How many peaks of the interference will be observed if the slit width is doubled while keeping the distance between the slits same? (c) How many peaks of interference will be observed if the slits are separated by twice the distance, that is, \(12 \mu \mathrm{m}\), while keeping the widths of the slits same? (d) What will happen in (a) if instead of 450-nm light another light of wavelength \(680 \mathrm{nm}\) is used? (e) What is the value of the ratio of the intensity of the central peak to the intensity of the next bright peak in (a)? (f) Does this ratio depend on the wavelength of the light? (g) Does this ratio depend on the width or separation of the slits?

Consider a single-slit diffraction pattem for \(\lambda=589 \mathrm{nm},\) projected on a screen that is \(1.00 \mathrm{m}\) from a slit of width \(0.25 \mathrm{mm}\). How far from the center of the pattern are the centers of the first and second dark fringes?

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