Question

An element has BCC structure with cell edge $ 288{{ pm}} $ . Density of the element is $ 7.2{{ gm/cc}} $ . How many atoms are present in $ 208{{ gm}} $ of elements.

Answer 1

Hint: To answer this question, you must follow a number of steps. We are given the mass of the element. To find the number of atoms in the given amount of the solid substance, we must find the number of unit cells present in that amount of the solid and for which we further need to find the volume occupied by the given mass of solid. We shall find the volume of the mass given and use it to get the number of unit cells by dividing it by the volume of one unit cell whose edge is given. Then, we can find the number of atoms.

Formula used: $ {{\rho }} = \dfrac{{{m}}}{{{V}}} $ or $ {{V}} = \dfrac{{{m}}}{{{\rho }}} $
Where $ \rho $ depicts the density of the given element in its solid state.
 $ m $ depicts the given mass of the given solid element
And $ V $ depicts the volume occupied by the given mass of the element in the solid state.

Complete step by step solution:
The volume occupied by $ 208{{ gm}} $ of the element is $ {{V}} = \dfrac{{{m}}}{{{\rho }}} = \dfrac{{208}}{{7.2}} = 28.89{{ c}}{{{m}}^3} $
We are given the edge length of the unit cell as $ 288{{ pm}} $ and so we can calculate the volume of one unit cell as $ {\left( {288{{ pm}}} \right)^3} = {\left( {288 \times {{10}^{ - 10}}{{ cm}}} \right)^3} = 2.389 \times {10^{ - 23}}{{ c}}{{{m}}^3} $
So, we can find the number of unit cells as $ = \dfrac{{{{\text{Total Volume}}}}}{{{{\text{Volume of unit cell}}}}} = \dfrac{{28.89}}{{2.389 \times {{10}^{ - 23}}}} = 12.09 \times {10^{23}} $
The number of atoms present in one unit cell $ = 2 $
So the number of atoms present in $ 208{{ gm}} $ of the element is $ N = 2 \times 12.09 \times {10^{23}} $
 $ \Rightarrow N = {24.18^{23}} $ .

Note:
We know that in a body centred cubic unit cell, the particles are present at the corners of the cube and the body- center of the cube. So, we can write that,
 $ {n_c} = $ number of particles in the corners of the bcc unit cell $ = 8 $
 $ {n_f} = $ number of particles on the six faces of the bcc unit cell $ = 0 $
 $ {n_i} = $ number of particles at the body center of the fcc unit cell $ = 1 $
 $ {n_e} = $ number of particles at the edge centres of the fcc unit cell $ = 0 $
Thus, the total number of particles in the body centred unit cell is
 $ n = \dfrac{{{n_c}}}{8} + \dfrac{{{n_f}}}{2} + \dfrac{{{n_i}}}{1} + \dfrac{{{n_e}}}{4} $
Substituting the values, we get
 $ n = \dfrac{8}{8} + \dfrac{0}{2} + \dfrac{1}{1} + \dfrac{0}{4}{{ }} $
 $ n = 1 + 0 + 1 + 0{{ }} $
Thus, $ n = 2 $ .