The plot of Figure 1.1 is very instructive for many quantum phenomena since it s
ID: 883963 • Letter: T
Question
The plot of Figure 1.1 is very instructive for many quantum phenomena since it shows the dominance of surface effects for nanomaterials such as nanocrystals in the size (diameter) range of 5 to 100 nm. Assuming the crystals are roughly spherical, calculate the number of (Pb + Sb) atoms for diameters of 20 nm, 40 nm, 60 nm, and 80 nm. Useful data: density of crysalline PbS = 7.6 g/cm3. Avogadro’s Number = 6.02 x 1023, atomic mass of Pb = 207.2 g/mole, atomic mass of sulfur = 32.06 g/mole. 1 nm = 10-7 cm.
Explanation / Answer
average mass of one molecule = (207.2+32.06) /6.02x10^23= 39.744x10^-23g/molecule.
volume of sphere = 4/3 pir^3= 4/3x22/7 xr^3=volume of the unit cell:
r1=10nm
V1= 4x22x1000x10^-21/7x3= 4.19x10^-18cm^3
V2= 88x8000x10^-21/7x3=33.52x10^-18cm^3
V3= 88x27000x10^-21/21=113.14x10^-18cm^3
V4= 88x64000x10^-21/21= 268.2x10^-18 cm^3
density of crysalline PbS = 7.6 g/cm^3
molecules in 1 cm3: =7.6/39 .744 x10^-23 = 1.9122x10^22 molecules
no of molecule in V1= 1.9122x10^22 x4.19x10^-18= 8x10^4
no of atoms= 2x8x10^4= 16x10^4
no of molecule in V2= 1.9122x10^22x 33.52x10^-18= 64x10^4
no of atoms= 812x10^4
no of molecule in V3=1.91x10^22 x113.14x10^-18=216x10^4
no of atoms= 432x10^4
no of molecule in V4= 1.9122x10^22 x268.2x10^-18 = 10^4 x513
no of atoms= 1026x10^4
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.