Question

Find the distance (345) planes in a cubic lattice having length 7.7A

Answer 1

Answer:

Explanation:

Interplanar distance for Cubic system is given by,

(\frac{1}{d_{hkl}})^{2}=\frac{h^{2}+k^{2}+l^{2}}{a^{2}}

where d_{hkl} is interplanar distance with miller indices h,k,l and 'a' is lattice length or edge length

In question given that,

h=3 , k=4 , l=5

a= 7.7 A

(\frac{1}{d_{hkl}})^{2}=\frac{h^{2}+k^{2}+l^{2}}{a^{2}}  Put the above values in this equation

\frac{1}{d_{hkl}}=\frac{\sqrt{h^{2}+k^{2}+l^{2}}}{a}

       or

d_{hkl}=\frac{a}{\sqrt{h^{2}+k^{2}+l^{2}}}

d_{hkl}=\frac{a}{\sqrt{3^{2}+4^{2}+5^{2}}}

d_{hkl}=\frac{a}{\sqrt{9+16+25}}

d_{hkl}=\frac{a}{\sqrt{50}}

d_{hkl}=\frac{7.7}{7.07}=1.089 A

So, the distance between (345) plane is 1.089 A.