To solve this problem, we need to find the distance between the body-centered atom and one corner atom in a cube.$\newline$Given:$\newline$• The cube has a side length $a. \newline$• The body-centered atom is located at the center of the cube.$\newline$• The corner atom is located at one of the vertices of the cube.$\newline$Let's analyze the geometry of the cube:$\newline$• The coordinates of the corner atom can be taken as $(0, 0, 0). \newline$• The coordinates of the body-centered atom are $\left(\frac{a}{2}, \frac{a}{2}, \frac{a}{2}\right). \newline$The distance $d$ between these two points is given by the distance formula:$\newline d = \sqrt{\left(\frac{a}{2} - 0\right)^2 + \left(\frac{a}{2} - 0\right)^2 + \left(\frac{a}{2} - 0\right)^2} \newline$Simplifying the expression:$\newline d = \sqrt{\left(\frac{a}{2}\right)^2 + \left(\frac{a}{2}\right)^2 + \left(\frac{a}{2}\right)^2} \newline d = \sqrt{3 \times \left(\frac{a}{2}\right)^2} \newline d = \sqrt{3} \times \frac{a}{2} \newline d = \frac{\sqrt{3}}{2} a \newline$Therefore, the distance between the body-centered atom and one corner atom in the cube is $\frac{\sqrt{3}}{2} a. \newline$This corresponds to Option 4.