Question

an element having atomic nmass 60.2 has a face center cubic unit cell the edgelemgth of the unit cell is 100 pm find out density of the element​

Answer 1

answer- d≈.62×10^7

I hope it might help u ❤

Answer 2

Answer:

Given = atomic mass = 60.2

             FCC (length)= 100pm= 100 x 10⁻¹⁰

   

to find = density

solution =  

density of unit cell = mass of unit cell / volume of unit cell

mass of unit cell = no.of atoms in the unit cell x mass of each atom

mass of unit cell                     = 4 x 60 /6.023 x 10²³

no. of atoms in FCC unit cell = 8 x 1/8 ++ 6 x 1/2 = 4

volume of the unit cell           = (100 x 10⁻¹⁰)³= 10 x 10⁻²⁴

so put this value in above formula ,

                                                =4 x60 / 6.023 x 10²³ x  10 x 10⁻²⁴

                                                =240 / 6.023 x 10²³ x  10 x 10⁻²⁴

0.25g/cm³

so the final density will be 0.25g/cm³