Solveeit Logo
Login

Question

Question: If the radius of gold \( \left( Au \right) \) and platinum \( \left( Pt \right) \) atoms in pm is \(...

If the radius of gold (Au)\left( Au \right) and platinum (Pt)\left( Pt \right) atoms in pm is 1002100\sqrt{2} ​ and 90290\sqrt{2} respectively, find nAu/nPt{{n}_{Au}}/{{n}_{Pt}} or nAunPt\dfrac{{{n}_{Au}}}{{{n}_{Pt}}}
(A) 0.520.52
(B) 0.680.68
(C) 0.730.73
(D) 0.740.74

Explanation

Solution

We have to calculate the numbers of atoms present in the unit cell of FCC atom and then, we have to calculate the mass of the unit cell of FCC type using molar mass of gold and Avogadro number. From the density and mass of the unit cell, we have to calculate the volume of the unit cell. From the volume of the unit cell, we can calculate the edge of the unit cell. The radius of the unit cell is calculated using the edge of the unit cell. The formula for number of atoms per unit volume is given by V=43πr3V=\dfrac{4}{3}\pi {{r}^{3}}

Complete step by step solution:
The given data here we have are radius of gold and platinum;
rPt=902{{r}_{Pt}}=90\sqrt{2} and rAu1002{{r}_{Au}}100\sqrt{2}
Now that we know the formula for number of atoms per unit volume is inversely proportional to the volume of one atom, we get the ratio of radius of gold to platinum;
nAunPt=VAuVPt=43πrPt343πrAu3\dfrac{{{n}_{Au}}}{{{n}_{Pt}}}=\dfrac{{{V}_{Au}}}{{{V}_{Pt}}}=\dfrac{\dfrac{4}{3}\pi {{r}_{Pt}}^{3}}{\dfrac{4}{3}\pi {{r}_{Au}}^{3}}
Now by further solving we get;
nAunPt=VAuVPt=rPt3rAu3\dfrac{{{n}_{Au}}}{{{n}_{Pt}}}=\dfrac{{{V}_{Au}}}{{{V}_{Pt}}}=\dfrac{{{r}_{Pt}}^{3}}{{{r}_{Au}}^{3}}
Now we have to substitute the data for radius of gold and platinum;
nAunPt=VAuVPt=rPt3rAu3=(902)3(1002)3\dfrac{{{n}_{Au}}}{{{n}_{Pt}}}=\dfrac{{{V}_{Au}}}{{{V}_{Pt}}}=\dfrac{{{r}_{Pt}}^{3}}{{{r}_{Au}}^{3}}=\dfrac{{{\left( 90\sqrt{2} \right)}^{3}}}{{{\left( 100\sqrt{2} \right)}^{3}}}
By solving we get
nAunPt=0.73\dfrac{{{n}_{Au}}}{{{n}_{Pt}}}=0.73
Therefore, the correct answer is option C i.e. nAunPt\dfrac{{{n}_{Au}}}{{{n}_{Pt}}} is 0. 0.73.0.73. .

Note:
An example of a compound that has face-centered cubic lattice is sodium chloride. Lithium fluoride lithium chloride, Sodium fluoride and potassium chloride etc. are examples of compounds that contain face centered cubic structures. Simple cubic cell, and body centered cubic cell are the other two crystal structures.