Question

An element forms a simple cubic cell.If the radius of an atom of an element is 1.53A calculate the internuclear distance between neighboring atom in the unit cell​

Answer 1

$\sf\small\red{Given:-}</p><p>$$\rm Element = 1.53 \: A$$\sf\small\red{Find:-}</p><p>$calculate the internuclear distance between neighboring atom in the unit cell.$\sf\small\red{Solution:-}$

$\rm\rightarrow \: 4r = 2 \sqrt{2}a$

$\rm\rightarrow \: a = \frac{4r}{2 \sqrt{2} }$

$\rm\rightarrow \: a = \frac{4 \times 1.53}{2 \times 1.41}$

$\rm\rightarrow \: a = \frac{6.12}{2.82}$

$\dashrightarrow\rm\pink {2. 17}$