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X-ray diffraction pattern of a crystal is given below. It undergoes first order reflection at a Bragg angle of 30 degrees from a crystal.

What is the value of the interplanar spacing (d) of the crystal, given that the X-ray wavelength used is 0.15 nm?

Option: 1

0.10 nm

 


Option: 2

0.15 nm


Option: 3

0.20 nm

 


Option: 4

0.25 nm


Answers (1)

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The given X-ray diffraction pattern is due to the constructive interference of X-rays diffracted by the crystal planes. According to Bragg's law, the condition for constructive interference is given by 2 d \sin \theta=n \lambda , where d is the interplanar spacing,  \theta is the angle of incidence,\lambda is the wavelength of the X-rays, and n is an integer. 

As it undergoes first order reflection at a Bragg angle of 30 degrees from a crystal.

Therefore, we can calculate the interplanar spacing (d) using the formula as:

\begin{aligned} & 2 d \sin \theta=n \lambda \\ & 2 d \sin 30^{\circ}=1(0.15 \mathrm{~nm}) \\ & d=\frac{0.15 \mathrm{~nm}}{2 \sin 30^{\circ}} \\ & d=\frac{0.15 \mathrm{~nm}}{1} \\ & \mathrm{~d}=0.15 \mathrm{~nm} \end{aligned}

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