Bohr's Model: Hydrogen Atom and Basic Crystal Structure

1. ENE 311 Lecture 3

2. Bohr’s model • Niels Bohr came out with a model for hydrogen atom from emission spectra experiments. • The simplest Bohr’s model is that the atom has a positively charged nucleus and negatively charged electrons. • The total charge of all electrons is equal to that of the nucleus. Therefore, the whole atom is neutral.

3. Bohr’s model Bohr made some statements about electrons as • Electrons exist in certain stable and orbit circularly around the nucleus without radiating.

4. Bohr’s model 2. The electron could shift to a higher or lower level of energy by gaining or losing energy equal to the difference in the energy levels.

5. Bohr’s model 3. The angular momentum p of electron in an orbit is

6. Bohr’s model • Hydrogen atom is the best model to study Schrödinger’s equationin 3 dimensions. • It consists of an electron and a proton. The potential energy of electron at distance r from the proton is where ε0 = permittivity of free space = 8.85 x 10-12 F/m

7. Hydrogen atom We consider only time-independent Schrödinger’s equation since we are now interested in the probability of electron to be found at a distance r from the nucleus and its allowed energy levels.

8. Hydrogen atom Use spherical coordinate (r,,), ψ only depends on r.

9. Hydrogen atom A solution of this equation is Ψ = exp(-c0r) The energy of hydrogen can be calculated as

10. Hydrogen atom • Bohr further said that a stationary orbit is determined by the quantization of energy represented by 4 quantum numbers (n, l, m, s). • Any combination of quantum number specifies one allow state or energy level which is “Pauli exclusion principle”.

11. Hydrogen atom For hydrogen atom, quantized energy is *For n = 1, this is called a “ground state”.

12. Basic Crystal Structure Solids may be classified into 3 groups: • Amorphous • Polycrystalline • Crystalline

13. Basic Crystal Structure • Amorphous: Atoms are randomly arranged (formlessness).

14. Basic Crystal Structure 2. Polycrystalline: Atoms have many small regions. Each region has a well organized structure but differs from its neighboring regions.

15. Basic Crystal Structure 3. Crystalline: Atoms are arranged in an orderly array.

16. Basic Crystal Structure • A collection of point periodically arranged is called a lattice. • Lattice contains a volume called a unit cell which is representative of the entire lattice. • By repeating the unit cell throughout the crystal, one can generate the entire lattice.

17. Unit Cell • Unit cell has an atom at each corner shared with adjacent cells. • This unit cell is characterized by integral multiples of vectors such as a, b, and c called basis vectors. • Basis vectors are not normal to each other and not necessarily equal in length.

18. Unit Cell Every equivalent lattice point in the crystal can be expressed by • Two-dimensional case: R = ma + nb • Three-dimensional case: R = ma + nb +pc where m, n, and p are integers

19. Unit Cell The smallest unit cell that can be repeated to form the lattice is called a primitive cell shown below.

20. Unit Cell Ex. 2-D lattice r = 3a+2b

21. Basic Crystal Structure • There are three basic cubic-crystal structures: (a) simple cubic (b) body-centered cubic, and (c) face-centered cubic.

22. Simple Cubic (SC) • Each corner of cubic lattice is occupied by an atom which is shared equally by eight adjacent unit cells. • The total number of atoms is equal to

23. Simple Cubic (SC) For the case of maximum packing (atoms in the corners touching each other), about 52%of the SC unit cell volume is filled with hard spheres, and about 48% of the volume is empty.

24. Body-centered cubic (BCC) • An additional atom is located at the center of the cube. • The total number of atoms is equal to

25. Body-centered cubic (BCC) Ex. Ba, Ce, Cr, Mo, Nb, K

26. Face-centered cubic (FCC) • One atom is added at each of six cubic faces in addition to the eight corner atoms. • The total number of atoms is equal to Ex. Al, Ag, Au, Cu, Ne

27. The Diamond Structure • This structure is like the FCC crystal family. • It results from the interpenetration of two FCC lattices with one displaced from the other by one-quarter of the distance along a body diagonal of the cube or a displacement of .

28. The Diamond Structure • If a corner atom has one nearest neighbor in the body diagonal, then it will have no nearest neighbor in the reverse direction. • Therefore, there are eight atoms in one unit cell. • Examples for this structure are Si and Ge.

29. Zinc blende structure • For the Zinc blende structure, this results from the diamond structure with mixed atoms such that one FCC sublattice has column III (or V) atoms and the other has Column V (or III) atoms. • This is a typical structure of III-V compounds. Ex. GaAs, GaP, ZnS, CdS

30. Fourteen Bravais lattices

31. Lattice systems

32. Crystal Planes and Miller Indices • The crystal properties along different planes are different. • The device characteristics (not only mechanical properties but also electrical properties) are dependent on the crystal orientation. • The way to define crystal planes is to use Miller indices. • The Miller indices are useful to specify the orientation of crystal planes and directions.

33. Crystal Planes and Miller Indices Miller indices can be obtained using this following procedure: • Determine the intercepts of the plane with crystal axes (three Cartesian coordinates). • Take the reciprocals of the numbers from 1. • Multiply the fractions by the least common multiple of the intercepts. • Label the plane in parentheses (hkl) as the Miller indices for a single plane.

34. Crystal Planes and Miller Indices Ex.Find the Miller indices of the below plane.

35. Crystal Planes and Miller Indices • Intercepts are a, 3a, 2a: (132) • Take the reciprocals of these intercepts: 1,1/3,1/2 • Multiply the fractions by the least common multiple, which in this case is 6, we get 6,2,3. • Label the plane as (623)

36. Crystal Planes and Miller Indices The Miller indices of important planes in a cubic crystal.

37. Crystal Planes and Miller Indices There are some other conventions as: • For a plane that intercepts the negative x-axis, for example, • {hkl}: For planes of equivalent symmetry depending only on the orientation of the axes, e.g. {100} for (100), (010), (001), , ,

38. Crystal Planes and Miller Indices 3. [hkl]: For a crystal direction, such as [010] for the y-axis. By definition, the [010] direction is normal to (010) plane, and the [111] direction is perpendicular to the (111) plane.

39. Crystal Planes and Miller Indices 4. <hkl>: For a full set of equivalent directions, e.g. <100> for [100], [010], [001], , ,

40. Density of crystal • Crystal with a lattice constant “a” and “n” atoms within unit cell, then the weight of unit cell is equal to the weight of atoms per unit cell. • The density of the crystal will be expressed as

41. Density of crystal where = density of crystal [g/cm3] M = atomic weight [g/mole] NA = Avogadro’s number = 6.022 x 1023 atoms/mole = number of atoms per unit volume

42. Density of crystal Ex. At 300 K the lattice constant for Si is 5.43 Å. Calculate the number of Si atoms per cubic centimeter and the density of Si at room temperature. Note: Atomic weight of Si = 28.09 g/mole

43. Density of crystal Ex. At 300 K the lattice constant for Si is 5.43 Å. Calculate the number of Si atoms per cubic centimeter and the density of Si at room temperature. Note: Atomic weight of Si = 28.09 g/mole Soln There are eight atoms per unit cell. Therefore, there are 8/a3 = 8/(5.43 x 10-8)3 = 5 x 1022 atoms/cm3

44. Density of crystal Ex. Find the density of Cu in [g/cm3]. Cu has an atomic radius of 1.278 Å and atomic weight of 63.5 g/mole.

45. Density of crystal Ex. Find the density of Cu in [g/cm3]. Cu has an atomic radius of 1.278 Å and atomic weight of 63.5 g/mole. • Soln Cu has an FCC structure. Å Number of atoms/unit cell is 4.