University Physics with Modern Physics (14th Edition)
14th Edition
ISBN: 9780321973610
Author: Hugh D. Young, Roger A. Freedman
Publisher: PEARSON
expand_more
expand_more
format_list_bulleted
Question
Chapter 42, Problem 42.17E
(a)
To determine
The energy gap between the valance and
(b)
To determine
To explain: why the pure silicon is opaque.
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
A germanium diode is used to detect radiation with energy of 1.6 MeV. To do so,
the anode (positive terminal) is exposed to the incoming radiation. Assume that the
energy is absorbed entirely at the point of entry or the anode. With mobilities of 1,150
cm?/V.s for holes and 3,750 cm?/V.s for electrons, calculate the current through the
diode with reverse bias: V = 24 V and d = 12 mm. Neglect the effects of electrodes and
4.
of the n and p layers and assume a single radiation event, that is a single particle or a
short burst of radiation. (Charge of an electron = 1.602 x 10-19 C; d is the distance
between anode and cathode).
The forbidden energy bandgap of AIP is
2.43 eV. Determine the wavelength (in
nm) of an incident photon that can
interact with a valence electron and
elevate the electron into the conduction
band.
Oλ = 511 nm
Oλ = 419 nm
O λ = 882 nm
λ = 575 nm
The gap between valence and conduction bands in diamond is 5.47 eV.What is the maximum wavelength of a photon that can excite an electron from the top of the valence band into the conduction band? In what region of the electromagnetic spectrum does this photon lie?
Chapter 42 Solutions
University Physics with Modern Physics (14th Edition)
Ch. 42.1 - If electrons obeyed the exclusion principle but...Ch. 42.2 - Prob. 42.2TYUCh. 42.3 - Prob. 42.3TYUCh. 42.4 - One type of thermometer works by measuring the...Ch. 42.5 - Prob. 42.5TYUCh. 42.6 - Prob. 42.6TYUCh. 42.7 - Suppose a negative charge is placed on the gate of...Ch. 42 - Van der Waals bonds occur in many molecules, but...Ch. 42 - Prob. 42.2DQCh. 42 - The H2+ molecule consists of two hydrogen nuclei...
Ch. 42 - The moment of inertia for an axis through the...Ch. 42 - Prob. 42.5DQCh. 42 - Prob. 42.6DQCh. 42 - Prob. 42.7DQCh. 42 - The air you are breathing contains primarily...Ch. 42 - Prob. 42.9DQCh. 42 - Prob. 42.10DQCh. 42 - What factors determine whether a material is a...Ch. 42 - Prob. 42.12DQCh. 42 - Prob. 42.13DQCh. 42 - Prob. 42.14DQCh. 42 - Prob. 42.15DQCh. 42 - Prob. 42.16DQCh. 42 - Prob. 42.17DQCh. 42 - Prob. 42.18DQCh. 42 - Prob. 42.19DQCh. 42 - Prob. 42.20DQCh. 42 - Prob. 42.21DQCh. 42 - Prob. 42.22DQCh. 42 - Prob. 42.23DQCh. 42 - Prob. 42.24DQCh. 42 - If the energy of the H2 covalent bond is 4.48 eV,...Ch. 42 - An Ionic Bond, (a) Calculate the electric...Ch. 42 - Prob. 42.3ECh. 42 - Prob. 42.4ECh. 42 - Prob. 42.5ECh. 42 - Prob. 42.6ECh. 42 - Prob. 42.7ECh. 42 - Two atoms of cesium (Cs) can form a Cs2 molecule....Ch. 42 - Prob. 42.9ECh. 42 - Prob. 42.10ECh. 42 - A lithium atom has mass 1.17 1026 kg, and a...Ch. 42 - Prob. 42.12ECh. 42 - When a hypothetical diatomic molecule having atoms...Ch. 42 - The vibrational and rotational energies of the CO...Ch. 42 - Prob. 42.15ECh. 42 - Prob. 42.16ECh. 42 - Prob. 42.17ECh. 42 - Prob. 42.18ECh. 42 - Prob. 42.19ECh. 42 - Prob. 42.20ECh. 42 - Prob. 42.21ECh. 42 - Prob. 42.22ECh. 42 - Prob. 42.23ECh. 42 - Prob. 42.24ECh. 42 - Prob. 42.25ECh. 42 - Prob. 42.26ECh. 42 - Prob. 42.27ECh. 42 - Prob. 42.28ECh. 42 - Prob. 42.29ECh. 42 - Prob. 42.30ECh. 42 - Prob. 42.31ECh. 42 - Prob. 42.32ECh. 42 - Prob. 42.33PCh. 42 - Prob. 42.34PCh. 42 - Prob. 42.35PCh. 42 - The binding energy of a potassium chloride...Ch. 42 - (a) For the sodium chloride molecule (NaCl)...Ch. 42 - Prob. 42.38PCh. 42 - Prob. 42.39PCh. 42 - Prob. 42.40PCh. 42 - Prob. 42.41PCh. 42 - Prob. 42.42PCh. 42 - Prob. 42.43PCh. 42 - Prob. 42.44PCh. 42 - Prob. 42.45PCh. 42 - Prob. 42.46PCh. 42 - Prob. 42.47PCh. 42 - Prob. 42.48PCh. 42 - Prob. 42.49PCh. 42 - Prob. 42.50PCh. 42 - Prob. 42.51PCh. 42 - Prob. 42.52PCh. 42 - Prob. 42.53CPCh. 42 - Prob. 42.54CPCh. 42 - Prob. 42.55CPCh. 42 - Prob. 42.56PPCh. 42 - Prob. 42.57PPCh. 42 - Prob. 42.58PP
Knowledge Booster
Similar questions
- The two nuclei in the carbon monoxide (CO) molecules are 0.1128 nm apart. The mass of the carbon atom is 1.993x10-26 kg. The mass of the oxygen atom is 2.656x10-26 kg. Spectroscopic measurements show that adjacent vibrational energy levels for the CO molecule are 0.269 eV. What is the effective spring constant of the CO molecule? (Give your answer in N/m.)arrow_forward1 A bar of n-type germanium 10mmx1mmx1mm. The electron density in the bar is 7×10²1 m³ and B=0.2 Wb/m². If one millivolt is applied across the long ends of the bar, determine the current through the bar, the Hall coefficient and the Hall voltage. Assume n = 0.39 m²/v.s. Ans: 43.6 μA; 8.9×10-4 m³/C; 7.76 μV.arrow_forwardThe gap between valence and conduction bands in diamond is 5.47 eV. (a) What is the maximum wavelength of a photon that can excite an electron from the top of the valence band into the conduction band? In what region of the electromagnetic spectrum does this photon lie? (b) Explain why pure diamond is transparent and colorless. (c) Most gem diamonds have a yellow color. Explain how impurities in the diamond can cause this color.arrow_forward
- The number of silicon atoms per m3 is 5 × 1028. This is doped simultaneously with 5 × 1022 atoms per m3 of Arsenic and 5 × 1020 per m3 atoms of Indium. Calculate the number of electrons and holes.arrow_forwardProblem 5. Assume that at T=300 K, the electron mobility in a silicon sample is 1300 cm²/Vs. If an electric field of 100 V/cm is applied what is the excess energy of the electrons? How does this excess energy compare with the thermal energy? If you assume that the mobility is unchanged how does the same comparison work out at a field of 5 V/cm. (NOTE: Excess energy is equal to ½ m₂* v² where Va is the drift velocity.)arrow_forwardThe number of silicon atoms per m3 is 5 × 1028. This is doped simultaneously with 5 × 1022 atoms per m3 of Arsenic and 5 × 1020 per m3 atoms of Indium. Calculate the number of electrons and holes. Given that ni = 1.5 × 1016 m-3. Is the material n-type or p-type?arrow_forward
- The two nuclei in the carbon monoxide (CO) molecules are 0.1128 nm apart. The mass of the carbon atom is 1.993x10-26 kg. The mass of the oxygen atom is 2.656x10-26 kg. What is the first excited rotational energy level for the CO molecule? (Give the your answer in meV.)arrow_forwardSilicon atoms with a concentration of 7x 1010 cm are added to gallium arsenide GaAs at T = 400 K. Assume that the silicon atoms act as fully ionized dopant atoms and that 15% of the concentration added replaces gallium atoms to free electrons and 85% replaces arsenic to create holes. Use the following parameters for GaAs at T = 300 K: N. = 4.7 x 1017cm-3 and N, = 7 x 1018 cm-3. The bandgap is E, = 1.42 eV and it is constant over the temperature range. The intrinsic concentration?arrow_forwardHowever, the molecule we can encounter everyday continuously vibrates and interact with the surrounding causing its bond vector to vary slightly. According to a new spectroscopy analysis, the adjacent bond vectors was found to be A = 1.01i + 0.84j + 0.9k B = 0.99i + -0.92j + -1.04k What is the angle (in degrees) between the bonds based on this new data?arrow_forward
- Calculate the London penetration depth of the electron in the copper metal. The concentration of free electrons in the copper is 8.5x10²³ /m³. (a) 1.8x10 m (b) 0.2x10* m (c) 2.6x10* m (d) 3.4x10 marrow_forwardX. Determine the resistivity of a germanium crystal at 300 K that has the simultaneous presence of (i) donor impurity of 1 in 10' and (ii) acceptor impurity of 1 in 10%. Given data: Atomic concentration in Germanium is 4.4 x 1022 atoms/cm³, n₁ = 2.5 × 10¹3 /cm³, He = 3800 cm² /Vs and h= 180 cm² /V s.arrow_forwardA pn junction diode and a Schottky diode have equal cross-sectional areas and have forward-bias currents of 0.5 mA. The reverse-saturation current of the Schottky diode is The difference in forward-bias voltages between the two diodes is 0.30 V. Determine the reverse-saturation current of the pn junction diode.arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
University Physics Volume 3PhysicsISBN:9781938168185Author:William Moebs, Jeff SannyPublisher:OpenStax
University Physics Volume 3
Physics
ISBN:9781938168185
Author:William Moebs, Jeff Sanny
Publisher:OpenStax