3. a) For a nanocrystalline sample containing particles of unknown size, you measure a band gap energy of 3.65x10 9 Use the plot and the equation of the best-fit line (below) to calculate the particle size (in nm) of this unknown sample. Be sure to show your calculation. (Note: "E" in the plot represents 10^) b) What is the band gap of the bulk material? Trend in Bandgap Energy with Particle Size 5.00E-19 y3D4,29E-19x+3.04E 19 450E 19 4.00E 19 3.50E-19 3.00E-19 250E 19 0.000 0.100 0.200 0300 (1/Size) (nn) (r) A

3. a) For a nanocrystalline sample containing particles of unknown size, you measure a band gap energy of 3.65x10 9 Use the plot and the equation of the best-fit line (below) to calculate the particle size (in nm) of this unknown sample. Be sure to show your calculation. (Note: "E" in the plot represents 10^) b) What is the band gap of the bulk material? Trend in Bandgap Energy with Particle Size 5.00E-19 y3D4,29E-19x+3.04E 19 450E 19 4.00E 19 3.50E-19 3.00E-19 250E 19 0.000 0.100 0.200 0300 (1/Size) (nn) (r) A

Principles of Modern Chemistry

8th Edition

ISBN:9781305079113

Author:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler

Publisher:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler

Chapter22: Inorganic Materials

Section: Chapter Questions

Problem 35P

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8.97 The doping of semiconductors can be done with enough precision to tune the size of the band gap in the material. Generally, in order to have a larger band gap, the dopant should be smaller than the main material. If you are a materials engineer and need a semiconductor that has lower conductivity thin pure silicon, what clement or elements could you use as your dopant? (You do not want either an n- or a p- type material) Explain your reasoning.
When you shine light of band gap energy or higher on asemiconductor and promote electrons from the valenceband to the conduction band, do you expect the conductivityof the semiconductor to (a) remain unchanged,(b) increase, or (c) decrease?
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The band gap for the elemental semiconductor silicon is 1.11 eV. That energy corresponds to what wavelength and region of the electromagnetic spectrum? (1eV = 1.602 x 10-19J; 1 nm = 1 x 10-9) 1.778 x 10E(-19) J; Ultra-violet 1117 nm; Ultra-violet 1.117 x 10E(-6) nm; Infrared 1117 nm; Infrared
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