/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 18 Three photodiodes \(D_{1}, D_{2}... [FREE SOLUTION] | 91影视

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Chapter 26: Problem 18

Three photodiodes \(D_{1}, D_{2}\) and \(D_{3}\) are made of semiconductors having band gaps of \(2.5 \mathrm{eV}, 2 \mathrm{eV}\) and \(3 \mathrm{eV}\), respectively. Which one will be able to detect light of wavelength of \(6000 \AA\) A \(?\) (a) \(1.05 \mathrm{eV}\) (b) \(0.05 \mathrm{eV}\) (c) \(0.25 \mathrm{eV}\) (d) \(2.06 \mathrm{eV}\)

Short Answer

Expert verified
Photodiode \(D_2\) will detect the light.

Step by step solution

01

Understand the Problem

The problem requires us to determine which photodiode can detect light of a given wavelength by comparing the energy of the light with the band gaps of the photodiodes.
02

Calculate the Energy of the Photon

We need to calculate the energy of a photon with a wavelength of 6000 脜. First, convert the wavelength from 脜ngstr枚ms to meters: \(6000 \ \text{脜} = 6000 \times 10^{-10} \ \text{m}\). Use the formula \( E = \frac{hc}{\lambda} \), where \( h = 6.626 \times 10^{-34} \ \text{Js} \) is Planck's constant and \( c = 3 \times 10^8 \ \text{m/s} \) is the speed of light.
03

Perform the Calculation

Substitute the values into the formula:\[E = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{6000 \times 10^{-10}}\]Calculating the above gives:\[ E \approx 3.31 \times 10^{-19} \ \text{J} \]Convert this energy from joules to electronvolts using: \(1 \ \text{eV} = 1.602 \times 10^{-19} \ \text{J}\):\[E \approx \frac{3.31 \times 10^{-19}}{1.602 \times 10^{-19}} \approx 2.06 \ \text{eV}\]
04

Compare with Band Gaps

We compare the energy of the photon to the band gaps of the three photodiodes:- \(D_1\): Band gap = 2.5 eV- \(D_2\): Band gap = 2 eV- \(D_3\): Band gap = 3 eVA photodiode can detect light when the photon's energy is equal to or greater than its band gap. Photon energy is 2.06 eV.
05

Determine the Detecting Photodiode

The photon energy (2.06 eV) is greater than the band gap of \(D_2\) (2 eV) but less than that of \(D_1\) (2.5 eV) and \(D_3\) (3 eV). Thus, only photodiode \(D_2\) can detect this light.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Band Gap Energy
In semiconductors, band gap energy is a crucial concept. It is the energy difference between the valence band and the conduction band. Essentially, it's the minimum energy required to move an electron from a bound state in the valence band to a free state in the conduction band. The band gap energy determines the semiconductor's ability to conduct electricity and its optical properties. For a photodiode to detect light, the energy of the incoming photons must be equal to or exceed the band gap energy.
  • A small band gap allows more photons to be absorbed, as less energy is needed to excite an electron across the band gap.
  • A large band gap material may only respond to photons with higher energies.
In this exercise, we're told that photodiodes have different band gap energies at 2.5 eV, 2 eV, and 3 eV. The photodiode with the smallest band gap that is greater than or equal to the photon energy will detect the light.
Photon Energy Calculation
The calculation of photon energy is central to the operation of photodiodes. Photons carry energy, and this energy can be calculated using the formula:\[ E = \frac{hc}{\lambda} \]where:
  • \( h \) is Planck's constant \( (6.626 \times 10^{-34} \text{ Js}) \)
  • \( c \) is the speed of light \( (3 \times 10^8 \text{ m/s}) \)
  • \( \lambda \) is the wavelength of the photon
By substituting the wavelength value into this formula, one can find out how much energy each photon carries. In our case, for a wavelength of 6000 脜, the photon energy is calculated to be about 2.06 eV. This value is then used to determine if the photodiode can detect the incoming light based on its band gap energy. If the photon energy is equal to or higher than the band gap energy, the photodiode will detect the light.
Wavelength Conversion
Understanding how to convert wavelengths into manageable calculations is vital. Wavelengths can be given in various units, but they need to be consistent for energy calculations using the formula \( E = \frac{hc}{\lambda} \).
  • Commonly, wavelengths are expressed in nanometers (nm) or 脜ngstr枚ms (脜), where \(1 \text{ 脜} = 10^{-10} \text{ meters}\).
  • For precise calculations, conversion to meters is necessary: a wavelength of 6000 脜 becomes \( 6000 \times 10^{-10} \) meters.
Making accurate conversions ensures the calculated photon energy is also accurate. Using meter-based units gives consistency with the units used for Planck's constant and the speed of light, both of which are in SI units.
Planck's Constant
Planck's constant \( h \) is one of the fundamental constants in physics, pivotal for calculations in quantum mechanics. It is a proportionality constant that connects the energy of photons and their frequency. In the formula used for photon energy calculation, \( E = \frac{hc}{\lambda} \), Planck's constant \( h \) (\(6.626 \times 10^{-34} \text{ Js}\)) helps transform the product of photon frequency \( u \) and wavelength \( \lambda \) into energy \( E \).
  • It suggests that energy and frequency are directly proportional.
  • Planck's constant is critical in the quantization of energy.
By utilizing this constant in calculations, the precise energy that photons carry at specific wavelengths can be determined, enabling us to assess which semiconductors can respond to specific wavelengths based on their band gap.

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Most popular questions from this chapter

For transistor action, which of the following statements are correct? (a) Base, emitter and collector regions should have similar size and doping concentrations (b) The base region must be very thin and lightly doped (c) The emitter junction is forward biased and collector junction is reverse biased (d) Both the emitter junction as well as the collector junction are forward biased

A silicon specimen is made into a \(p\)-type semiconductor by dopping, on an average, one indium atom per \(5 \times 10^{7}\) silicon atoms. If the number density of atoms in the silicon specimen is \(5 \times 10^{28}\) atoms \(\mathrm{m}^{-3}\), then the number of acceptor atoms in silicon per cubic centimetre will be (a) \(2.5 \times 10^{30}\) atoms \(\mathrm{cm}^{-3}\) (b) \(2.5 \times 10^{35}\) atoms \(\mathrm{cm}^{-3}\) (c) \(1.0 \times 10^{13}\) atoms \(\mathrm{cm}^{-3}\) (d) \(1.0 \times 10^{15}\) atoms \(\mathrm{cm}^{-3}\)

An \(n\)-type and a \(p\)-type silicon semiconductor can be obtained by doping pure silicon with (a) sodium and magnesium (b) phosphorus and boron respectively (c) boron and phosphorus respectively (d) indium and sodium respectively

In a \(p-n\) junction diode (a) the current in the reverse biased condition is generally very small (b) the current in the reverse biased condition is small but the forward biased current is independent of the bias voltage (c) the reverse biased current is strongly dependent on the applied bias voltage (d) the forward biased current is very small in small in comparison to reverse biased current

The correct relation between the two current gains \(\alpha\) and \(\beta\) in a transistor is (a) \(\beta=\frac{\alpha}{1+\alpha}\) (b) \(\alpha=\frac{\beta}{1-\beta}\) (c) \(\alpha=\frac{\beta}{1+\beta}\) (d) \(\alpha=\frac{1+\beta}{\beta}\)
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