Using X-rays of wavelength 154.1 pm and starting from the glancing angle, the reflection from silver crystal was found to occur at theta=22.20^@. Calculate the spacing between the plane of Ag atoms that gave rise to the above reflection. (sin 22.20^@=0.3778)

Using X-rays of wavelength 154.1 pm and starting from the glancing ang

1.	Using X-rays of wavelength 154.1 pm and starting from the glancing angle, the reflection from silver crystal was found to occur at theta=22.20^@. Calculate the spacing between the plane of Ag atoms that gave rise to the above reflection. (sin 22.20^@=0.3778)
Answer» Solution :Here, `lambda`=154.1 pm , n=1, `THETA=22.20 ^@` using Bragg's EQUATION , VIZ. , `2 d sin theta= nlambda` `d=(nlambda)(2 sin theta)=(1xx154.1)/(2xx0.3778)` pm =204 pm