Fisika Study Center

The 4th IPhO took place in Moscow, Soviet Union in 1970. Here is a sample problem from the 4th IPhO Theoretical Problems.

Problem 2
A unit cell of a crystal of natrium chloride (common salt- NaCl) is a cube with the edge length a = 5.6⋅10^-10 m (Fig.2). The black circles in the figure stand for the position of natrium atoms whereas the white ones are chlorine atoms. The entire crystal of common salt turns out to be a repetition of such unit cells. The relative atomic mass of natrium is 23 and that of chlorine is 35,5. The density of the common salt ρ = 2.22⋅10³ kg/m³ . Find the mass of a hydrogen atom.

Solution
Let’s calculate the quantities of natrium atoms (n₁) and chlorine atoms (n₂) embedded in a single NaCl unit crystal cell (Fig.2).
One atom of natrium occupies the middle of the cell and it entirely belongs to the cell. 12 atoms of natrium hold the edges of a large cube and they belong to three more cells so as 1/4 part of each belongs to the first cell.
Thus we have
n₁ = 1+12⋅1/4 = 4 atoms of natrium per unit cell.
In one cell there are 6 atoms of chlorine placed on the side of the cube and 8 placed in the vertices. Each atom from a side belongs to another cell and the atom in the vertex - to seven others. Then for one cell we have
n₂= 6⋅1/2 + 8⋅ 1/8 = 4 atoms of chlorine.
Thus 4 atoms of natriun and 4 atoms of chlorine belong to one unit cell of NaCl crystal.
The mass m of such a cell is equal
m = 4(m_rNa + m_rCl) (amu),
where m_rNa and m_rCl are relative atomic masses of natrium and clorine. Since the mass of hydrogen atom mH is approximately equal to one atomic mass unit: mH = 1.008 amu ≈ 1 amu then the mass of an unit cell of NaCl is
m = 4(m_rNa + m_rCl) mm_H .
On the other hand, it is equal m = ρa³ , hence



Sources/Literature:
-The Olympic home page www.jyu.fi/ipho
-Problems of the IV International Olympiad, Moscow, 1970 The publication is prepared by Prof. S. Kozel & Prof. V.Orlov (Moscow Institute of Physics and Technology)