Q20 / 160 · AP EAMCET 2024 (18 May Shift 1) · Chemistry
The diffraction pattern of 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 $1.54 Å)\left(\sin 30^{\circ}=0.5, \sin 60^{\circ}=0.866\right.$; $n=1$ )
- A$8.89 \times 10^{-9}$
- B$8.89 \times 10^{-1}$
- Correct answerC$1.54 \times 10^{-8}$
- D$1.54$
Solution
According to Bragg's law
$\mathrm{n} \lambda=2 \mathrm{~d} \sin \theta$
Given, $\mathrm{n}=$ order $=1$
$\begin{aligned} & \lambda=\text { wavelength of } \mathrm{x} \text {-rays }=1.54 \mathrm{~A}^{\circ} \\ & \theta=\text { angle of incidence }=30^{\circ} \\ & d=\text { interplaner distance }\end{aligned}$
(distance between the layers)
$\begin{aligned} & \mathrm{n} \lambda=2 \mathrm{~d} \sin \theta \\ & 1.54 Å=2 \times \mathrm{d} \times \sin 30^{\circ}\end{aligned}$
$\begin{aligned} & 1.54 \times 10^{-8} \mathrm{~cm}=2 \times \mathrm{d} \times 0.5=\left[1 Å=10^{-8} \mathrm{~cm}\right] \\ & \mathrm{d}=1.54 \times 10^{-8} \mathrm{~cm}\end{aligned}$
Hence, the correct answer is option: 'C'.