Question

Calculate the distance between the planes of ions if the smallest angle of reflection is x-ray diffraction is ${5.97}^0$ and the wavelength of incident ray is 58.6 pm? $(sin{5.97}^0$ = 0.104)

• A. 282 pm
• B. 382 pm
• C. 156 pm
• D. 78 pm

Answer 1

The correct option is A

282 pm

Do you remember Bragg’s equation?
It was, $n\lambda =2dsin\theta$

Here it has to be noted that whenever order of reflection is not given in the question, we will take it as 1.

Here, we know the angle $\left(\theta \right)$ and wavelength of x – ray.

$\left(\theta \right)={5.97}^0$

$\lambda$ = 58.6 pm

so, $d=\frac{n\lambda }{2sin\theta }$

$=\frac{58.6\times {10}^{-12}}{2\times sin\left({5.97}^0\right)}=\frac{58.6\times {10}^{-12}}{2\times 0.104}=281.7\times {10}^{-12}m=281.7\times pm\cong 282pmAns.$

So correct option is A.