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Niobium is an element whose density is 8.55 g/cm3 and atomic weight is 93. It crystallizes in body-body-centered structure. With this information, calculate the atomic radius of niobium atom.The quest
Niobium is an element whose density is 8.55 g/cm3 and atomic weight is 93. It crystallizes in body-body-centered structure. With this information, calculate the atomic radius of niobium atom.
The question requires that the Edge of length of cell a, be calculated. We know that the density of the element is given as p = 8.55 g /cm3, while the number of atoms in unit cell of BCC lattice Z is known to be 2. The Avogadro number NA is a fixed amount known to be 6.022x1023. The edge of length of a cell, a, can be calculated using a series of formula.
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******* = *NoZ=2M= **** *** g/cm3Using the formula ******* * ************* * *6022*1023 = ********** ****** ********** cm3a = 3306*10-8cmRadius *** ********* ** ********** ***** *** ************** =sqrt ******** **** *** ***** ** ** 3306*10-8cmHence * * *********************** *
