If the lattice constant of a crystal is 3 tim | vedclass

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(B) Given: Lattice constant $d = 3 	imes 10^{-8} \, cm$,glancing angle $	heta = 30^{\circ}$,and order of diffraction $n = 1$.Using Bragg's Law: $2d

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$3$

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(B) Given: Lattice constant $d = 3 	imes 10^{-8} \, cm$,glancing angle $	heta = 30^{\circ}$,and order of diffraction $n = 1$.Using Bragg's Law: $2d \sin 	heta = n\lambda$.Substituting the values: $\lambda = \frac{2 	imes (3 	imes 10^{-8} \, cm) 	imes \sin(30^{\circ})}{1}$.Since $\sin(30^{\circ}) = 0.5$,we get $\lambda = 2 	imes 3 	imes 10^{-8} 	imes 0.5 = 3 	imes 10^{-8} \, cm$.Thus,the value is $3 	imes 10^{-8} \, cm$.

If the lattice constant of a crystal is $3 \times 10^{-8} \, cm$ and the glancing angle for the first-order diffraction of $X$-rays is $30^{\circ}$,then the value of $\lambda$ will be ..... $\times 10^{-8} \, cm$.

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