The diffraction pattern of a crystalline solid gave a peak at $2 \theta = 60^{\circ}$. What is the distance (in $cm$) between the layers which gave this peak? ($\lambda$ of $X$-rays is $1.54 \mathring{A}$) ($\sin 30^{\circ} = 0.5$,$\sin 60^{\circ} = 0.866$; $n = 1$)
- A$8.89 \times 10^{-9}$
- B$8.89 \times 10^{-1}$
- C$1.54 \times 10^{-8}$
- D$1.54$
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