The number of electrons excited to the conduction band per cubic centimeter in a semiconductor can be estimated from the following equation: ne = (4.8 × 10'5 cm¯³ K¯¥2) T/2 e¯E{2RT) K¯32) T³² e¯E«M2RT) п where T is the temperature in kelvins and Eg the band gap in joules per mole. The band gap of diamond at 300 K is 8.7 × 10-19 J. How many electrons are thermally excited to the conduction band at this temperature in a 1.00-cm diamond crystal?

The number of electrons excited to the conduction band per cubic centimeter in a semiconductor can be estimated from the following equation: ne = (4.8 × 10'5 cm¯³ K¯¥2) T/2 e¯E{2RT) K¯32) T³² e¯E«M2RT) п where T is the temperature in kelvins and Eg the band gap in joules per mole. The band gap of diamond at 300 K is 8.7 × 10-19 J. How many electrons are thermally excited to the conduction band at this temperature in a 1.00-cm diamond crystal?

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

See similar textbooks

Similar questions

How many molecular orbitals are present in the conduction band of a lithium crystal with a mass of 11.2 g?
Melting gold with other metals makes 10-karat gold. Which term best describes 10-karat gold?
Calculate the number of molecular orbitals present in the valence band of a lithium crystal with a mass of 2.52 g.
Define the term, Band-Gap?
When an X-ray beam with a wavelength of 126.5 pm is incident on a metal crystal, it produces a peak in the X-ray spectrum at 31.2°. Assuming n = 1, determine the separation between layers of atoms (in pm) that gives rise to this peak. Report your answer to 1 decimal place.
A crystalline sample has a diffraction peak at 17.2°. What is the corresponding d space in pm? Assume n = 1 and A = 71.1 pm. Report your answer to the ones place.
The figure to the right shows sunlight intensity as a function of wavelength. Answer the following questions about materials that could act as the absorber (p-type semiconductor) in a heterojunction solar cell. a.What advantage would there be in using a 1000-nm wavelength bandgap absorber vs. one that had a 400-nm wavelength bandgap? (HINT: the bandgap is the minimum energy required to promote electrons.) b. One reason the electron-volt (eV) is often used as the bandgap unit in solar cell research is that the bandgap in eV is also equal to the maximum voltage that can be generated by the solar cell. With this in mind, what advantage would there be in using a 400-nm wavelength bandgap absorber vs. one that had a 1000-nm wavelength bandgap? c.In light of your answers to (a) and (b), can there be one “best” bandgap to use in a solar cell?
Niobium crystallizes in BCC structure and has a density of 8.55 g/cm^3. Calculate its atomic radius if its atomic mass is 93 g/mol. *
What colour are SnC12.2H2O crystals?
QI. Calculate the linear density of directions [011] for BCC, and FCC crystal structures?
Silicon has a band gap of 1.1 eV at room temperature.(a) What wavelength of light would a photon of this energycorrespond to? (b) Draw a vertical line at this wavelength inthe figure shown, which shows the light output of the Sunas a function of wavelength. Does silicon absorb all, none,or a portion of the visible light that comes from the Sun?(c) You can estimate the portion of the overall solar spectrumthat silicon absorbs by considering the area under thecurve. If you call the area under the entire curve “100%,”what approximate percentage of the area under the curve isabsorbed by silicon?
The semiconductor CdSe has a band gap of 1.74 eV. Whatwavelength of light would be emitted from an LED madefrom CdSe? What region of the electromagnetic spectrum isthis?