Density of Li is `0.53"g cm"^(-3)`. The edge length of Li is `3.5Å`. Find the number of Li atoms in a unit cell `(N_(0)=06.023xx10^(23),M=6.94)`.

Density of Lithium atom is 0.53gcm^-3 . The edge length of Li is 3.5A^@ . Find out the number of Li atom in a unit cell.

Density of lithium atom is 0.53 g//cm^(3) . The edge length of Li is 3.5 A^.The number of lithium atoms in a unit cell will be.. . (Atomic mass of lithium is 6.94)

Density of lithium atom is 0.53 g//cm^(3) . The edge length of Li is 3.5 Å .The number of lithium atoms in a unit cell will be……….. . (Atomic mass of lithium is 6.94)

The density of KBr is 2.75"gm cm"^(-3) . The edge length of unit cell is 654 pm. Predict the type of cubic lattice to which unit cell of KBr belongs. (N_(0)=6xx023xx10^(23)"mol"^(-1)) . Atomic mass of K = 39, Br = 80.

An element occurs in body centered cubic structure. Its density is 8.0 g//cm^(3) . If the cell edge is 250 pm, calculate the atomic mass of an atom of this element. (N_(A)=6.023xx10^(23))

The unit cell of an element of atomic mass 108 u and density 10.5 g cm^(-3) is cube with edge length 409 pm. Find the type of unit cell of the crystal. (Given : Avogadro's constant = 6.023 xx 10^(23) mol)