Q71P In Fig. 36-48, let a beam of x-r... [FREE SOLUTION]

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Chapter 36: Q71P (page 1113)

In Fig. 36-48, let a beam of x-rays of wavelength 0.125 nm be incident on an NaCl crystal at angle 胃 = 45.0掳 to the top face of the crystal and a family of reflecting planes. Let the reflecting planes have separation d = 0.252 nm. The crystal is turned through angle 蠒 around an axis perpendicular to the plane of the page until these reflecting planes give diffraction maxima. What are the (a) smaller and (b) larger value of 蠒 if the crystal is turned clockwise and the (c) smaller and (d) larger value of 蠒 if it is turned counter-clockwise

Short Answer

Expert verified

(a) $蠒 = 15.26 掳$ in clockwise for $m = 2$ .

(b) $蠒 = 30.64 掳$ in clockwise for $m = 1$ .

(d) $蠒 = 37.8 掳$ in counter-clockwise for $m = 4$ .

Step by step solution

Bragg’s law

If the incident angle of the wave $(胃)$ , measured from the surfaces of the reflecting plane and the radiation's wavelength $(位)$ , fulfil Bragg's equation, diffraction maxima (caused by constructive interference) is achieved. The Bragg鈥檚 equation is given as-

$2 诲蝉颈苍胃 = 尘位 (m = 1, 2, 3 . . .)$

Where $d$ is the interplanar distance and is the order of the maximum intensity.

Given Data

The wavelength of light used: $位 = 0.125 nm$

Plane separation: $d = 0.252 nm$

Angle of incidence: $胃 = 45 掳$

Calculate the value of ϕ for each subpart.

If the crystal is rotated by an angle $蠒$ , the new angle of incident is $胃卤蠒$ depending on the clockwise or counter-clockwise rotation.

Using the values given in the question,

For $m = 1$ , in clockwise direction, the angle $胃 - 蠒$ is

role="math" localid="1663051508257" $胃 - 蠒 = \sin^{- 1} \frac{m 位}{2 d} 45 掳 - 蠒 = \sin^{- 1} \frac{(0.125 n m) (1)}{2 (0.252 n m)} 蠒 = 45 掳 - \sin^{- 1} 0.248 蠒 = 30.64 掳$

For $m = 2$ , in clockwise direction, the angle $胃 - 蠒$ is

$胃 - 蠒 = \sin^{- 1} \frac{m 位}{2 d} 45 掳 - 蠒 = \sin^{- 1} \frac{(0.125 n m) (2)}{2 (0.252 n m)} 蠒 = 45 掳 - \sin^{- 1} 0.496 = 15.26 掳$

For $m = 3$ , in counter-clockwise direction, the angle $胃 + 蠒$ is

$胃 + 蠒 = \sin^{- 1} \frac{m 位}{2 d} 45 掳 + 蠒 = \sin^{- 1} \frac{(0.125 n m) (3)}{2 (0.252 n m)} 蠒 = \sin^{- 1} 0.744 - 45 掳 = 3.1 掳$

For $m = 4$ , in counter-clockwise direction, the angle $胃 + 蠒$ is

$胃 + 蠒 = \sin^{- 1} \frac{m 位}{2 d} 45 掳 + 蠒 = \sin^{- 1} \frac{(0.125 n m) (4)}{2 (0.252 n m)} 蠒 = \sin^{- 1} 0.992 - 45 掳 = 37.8 掳$

The smaller and larger values of $蠒$ for clockwise rotation are $蠒 = 15.26 掳$ and $蠒 = 30.64 掳$ respectively. The smaller and larger values of $蠒$ for counter-clockwise rotation are $蠒 = 3.1 掳$ and $蠒 = 37.8 掳$ respectively.