Solveeit Logo
Login

Question

Question: Potassium has a bcc structure with nearest neighbour distance \( 4.52{{\text{A}}^{\text{o}}} \) , it...

Potassium has a bcc structure with nearest neighbour distance 4.52Ao4.52{{\text{A}}^{\text{o}}} , its atomic weight is 39. Its density (in kg m−1{{\text{m}}^{ - 1}} ) will be:
(A) 454
(B) 804
(C) 852
(D) 908

Explanation

Solution

Density is equal to mass per volume. Density is also known as volumetric mass density or specific mass. Density of any material varies with temperature and pressure. To solve this question, we will use the concept of solid state and density. We shall calculate the edge length and use it in the formula given.

Formula used: Density = M×ZNA×a3\dfrac{{M \times Z}}{{{N_A} \times {a^3}}}
Distance between two atoms = a32\dfrac{{a\sqrt 3 }}{2}
Here M is the molecular mass of potassium in kg, Z is number of atoms per unit cell, NA{N_A} is Avogadro’s number that is 6.022×10236.022 \times {10^{23}} and ‘a’ is the length of edge of unit cell.

Complete step by step solution:
Unit cell is the smallest repeating unit of crystal lattice. Unit cell is the building block of a crystal.
A Body-centred cubic (bcc) unit cell has atoms at each corner of the cube and an atom at the centre of the structure.
In bcc the distance between two nearest atoms is given by a32\dfrac{{a\sqrt 3 }}{2} .
a32\dfrac{{a\sqrt 3 }}{2} = 4.52A∘4.52{A^ \circ }
Therefore, a = 4.52×23\dfrac{{4.52 \times 2}}{{\sqrt 3 }}
Density = 39×2×10−36.022×1023×(2×4.52×10−103)3\dfrac{{39 \times 2 \times {{10}^{ - 3}}}}{{6.022 \times {{10}^{23}} \times {{(\dfrac{{2 \times 4.52 \times {{10}^{ - 10}}}}{{\sqrt 3 }})}^3}}}
Here we have substituted Z as 2 as the number of atoms in bcc are two.
Here, the molecular weight of potassium is given as 39 grams so to convert it into kilogram that is the S.I unit we divide it by 1000 that is multiplied by 10−3{10^{ - 3}} .
Density = 908 kgm−3{\text{kg}}{{\text{m}}^{ - 3}}
So, the correct answer is option D.

Note:
In body centred cubic the atoms are present at:
In Body–centred cubic unit cell every corner has an atom.
There is one atom present at the centre of the cube.
According to this body- centred cubic, the atom at the body centres completely belongs to the unit cell in which it is present.