(A) Given: Angle of diffraction $(2 \theta) = 90^{\circ}$,so $\theta = 45^{\circ}$.Distance between two planes,$d = 2.28 \ \mathring{A}$.Order of diff

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(A) Given: Angle of diffraction $(2 \theta) = 90^{\circ}$,so $\theta = 45^{\circ}$.Distance between two planes,$d = 2.28 \ \mathring{A}$.Order of diffraction,$n = 2$.Bragg's equation is given by $n \lambda = 2 d \sin \theta$.Substituting the values: $2 \times \lambda = 2 \times 2.28 \times \sin 45^{\circ}$.Since $\sin 45^{\circ} = \frac{1}{\sqrt{2}} \approx 0.707$,we have $2 \lambda = 2 \times 2.28 \times 0.707$.$\lambda = 2.28 \times 0.707 = 1.612 \ \mathring{A}$.

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For a crystal,the angle of diffraction $(2 \theta)$ is $90^{\circ}$ and the second order line has a $d$ value of $2.28 \ \mathring{A}$. The wavelength (in $\mathring{A}$) of $X$-rays used for Bragg's diffraction is

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A
$1.612$
B
$2.00$
C
$2.28$
D
$4.00$

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For a crystal,the angle of diffraction $(2 \theta)$ is $90^{\circ}$ and the second order line has a $d$ value of $2.28 \ \mathring{A}$. The wavelength (in $\mathring{A}$) of $X$-rays used for Bragg's diffraction is

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