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  • In Bragg's experiment of diffraction by crystals using x-rays of wavelength  <img alt="\mathrm{ x \AA }" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%20x%20%5CAA%20%7D

In Bragg's experiment of diffraction by crystals using x-rays of wavelength  \mathrm{ x \AA }, the glancing angle of the first order spectrum is \mathrm{ 8^{\circ}}. If the lattice spacing is \mathrm{ 2.82 A^{\circ}}, then using \mathrm{ \sin 5^{\circ}=0.1392,} we can have the value of  x as:

Option: 1

78.51 \mathrm{hm}


Option: 2

7.851 \mathrm{hm}


Option: 3

0.7851 \mathrm{~nm}


Option: 4

0.07 \mathrm{~nm}


Answers (1)

best_answer

\mathrm{ Given, \sin 8^{\circ}=0.1392 } \\ {Lattice\, \, \, spacing a=2.82 \mathrm{~A}=2.82 \times 10^{-10} \mathrm{~m}} \\ {Use \, \, \, Bragg's \, \, equation - }

\mathrm{ \begin{aligned} 2 a \sin \theta & =n \lambda \\ 2 \times 2.82 \times 0.1392 \times 10^{-10} & =n \lambda \end{aligned} }
for n=1
\mathrm{ \begin{aligned} & \therefore \quad \lambda=5.64 \times 0.1392 \times 10^{-10} \\ & \lambda=0.7551 \times 10^{-10}=7.551 \times 10^{-9} \mathrm{~m} \\ & =7.55 \mathrm{~nm} \text { Ans. }\end{aligned}}

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manish

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