`CsBr` crystallizes in a body-centred cubic unit lattice with an edge length of `4.287 Å`. Calculate the angles at which the second-order reflection maxima may be expected for `(2, 0, 0)`, `(1, 1, 0)`, planes when `X`-rays of `gamma = 0.50 Å` are used.

Explanation

For `bc c` lattice, `d_(200) = a//2`
So, for second-orderreflection, `2gamma = 2 xx (a)/(2) sin theta_(1)`
or `sin theta_(1) = (2gamma)/(a)`
i.e., `sin theta_(1) = (2 xx 0.50)/(4.287)` and `theta_(1) = 13^(@)29'`
`d_(110) = (a)/(sqrt2)`
So, `2 gamma = 2 xx (a)/(sqrt2) sin theta_(2) = (sqrt2gamma)/(a)`.
`sin theta_(2) = (sqrt2 xx 0.50)/(4.284) implies theta_(2) = 9^(@)30'`
`d_(111) = (a)/(2sqrt3)`, so `2gamma = 2 xx (a)/(2sqrt3) sin theta_(3) implies sin theta_(3) = (2sqrt3gamma)/(a)`
i.e., `sin theta_(3) = (2 xx sqrt3 xx 0.50)/(4.287)` and `theta_(3) = 23^(@)49'`