Question

Silver (atomic weight = 108 g$mo{{l}^{-1}}$) has a density of 10.5 g$c{{m}^{-3}}$. The number of silver atoms on a surface of area ${{10}^{-12}}{{m}^{2}}$ can be expressed in scientific notation as $y\times {{10}^{x}}$. The value of ‘x’ is:
(A).7
(B).8
(C).6
(D).9

Answer 1

This question is talking about only silver (Ag) metal, so I can understand that this question is from the solid state chapter. Crystal has an FCC unit cell.

Complete step by step solution:
Face centered unit cell of Ag crystal.
Given that, atomic weight of Ag = 108 g$mo{{l}^{-1}}$
Density = 10.5 g $c{{m}^{-3}}$
Surface area = ${{10}^{-12}}{{m}^{2}}$
So, from the value of density
Density = $\dfrac{4\times 108}{6.023\times {{10}^{23}}\times {{a}^{3}}}$ (here a is the dimension of unit cell and ${{a}^{3}}$is the volume of the unit cell)
10.5 = $\dfrac{4\times 108}{6.023\times {{10}^{23}}\times {{a}^{3}}}$
By rearranging the equation and calculating cube root:
${{a}^{3}}$= 6.83 x ${{10}^{-23}}$,
‘a’ = 4 x ${{10}^{-8}}$cm
‘a’ = 4 x ${{10}^{-10}}$ m
So, the surface area of unit cell:
${{a}^{2}}$= ${{(4\times {{10}^{-10}})}^{2}}$
So, the number of unit cells on ${{10}^{-12}}{{m}^{2}}$ surface area:
= $\dfrac{{{10}^{-12}}}{1.6\times {{10}^{-19}}}$
= $6.25\times {{10}^{6}}$
And we know that there are effectively two atoms on the surface of unit cell
So, the number of atoms on ${{10}^{-12}}{{m}^{2}}$ surface area = 2 x number of unit cells
= 2 x $6.25\times {{10}^{6}}$
= 1.25 x ${{10}^{7}}$
On comparing with $y\times {{10}^{x}}$
We get the value of ‘x = 7’
So, the correct answer is ‘A’.

Note:
Here you should know the structure of the unit cell of FCC and position atoms in it and the number of atoms in its unit cell. You should know the formula of density of the unit cell.