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(D) According to Bragg's Law,the condition for diffraction is $2d \sin 	heta = n \lambda$.To find the maximum wavelength $\lambda_{\max}$,we rearrang

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$20$

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(D) According to Bragg's Law,the condition for diffraction is $2d \sin 	heta = n \lambda$.To find the maximum wavelength $\lambda_{\max}$,we rearrange the formula: $\lambda = \frac{2d \sin 	heta}{n}$.For $\lambda$ to be maximum,$n$ must be the minimum $(n = 1)$ and $\sin 	heta$ must be the maximum $(\sin 	heta = 1)$.Thus,$\lambda_{\max} = 2d$.Given $d = 10^{-7} \ cm = 10^{-9} \ m = 10 \ \mathring{A}$.Substituting the values: $\lambda_{\max} = 2 	imes 10 \ \mathring{A} = 20 \ \mathring{A}$.

The maximum distance between the inter-atomic planes of a crystal is $10^{-7} \ cm$. The maximum wavelength of $X$-rays that can be studied by this crystal is ........... $\mathring{A}$.

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