/*! 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 34 You are using light of wavelengt... [FREE SOLUTION] | 91影视

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Chapter 28: Problem 34

You are using light of wavelength \(1400 \mathrm{~nm} \mathrm{in}\) a photodetector. Which of the following semiconductor is to be used having band gap as? \(\left.C h=6.63 \times 10^{-34} J s ; c=3 \times 10^{9} \mathrm{~ms}^{-1}\right)\) (a) \(1 \mathrm{eV}\) (b) \(2 \mathrm{eV}\) (c) \(3 \mathrm{cV}\) (d) \(3.5 \mathrm{cV}\)

Short Answer

Expert verified
The semiconductor with a band gap of 1 eV (option a) is suitable.

Step by step solution

01

Convert Wavelength to Nanometers

To determine which semiconductor should be used, first, convert the wavelength from nanometers (nm) to meters (m). The given wavelength is \(1400 \, nm\). Since \(1 \, nm = 10^{-9} \, m\), the wavelength in meters is \(1400 \, nm = 1400 \times 10^{-9} \, m = 1.4 \times 10^{-6} \, m\).
02

Calculate the Energy of the Photon

Use the formula \(E = \frac{hc}{\lambda}\) to find the energy of a photon with the given wavelength. Here, \(h = 6.63 \times 10^{-34} \, Js\), \(c = 3 \times 10^{8} \, m/s\), and \(\lambda = 1.4 \times 10^{-6} \, m\). Calculate \(E\): \[E = \frac{(6.63 \times 10^{-34})(3 \times 10^{8})}{1.4 \times 10^{-6}} \approx 1.42 \times 10^{-19} \, J\]
03

Convert Energy to Electron Volts

Convert the energy from joules to electron volts to compare with the band gaps provided in the options. Use the conversion \(1 \, eV = 1.6 \times 10^{-19} \, J\):\[E = \frac{1.42 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 0.89 \, eV\]
04

Determine the Appropriate Semiconductor

Compare the calculated energy of the photon (approximately \(0.89 \, eV\)) with the band gap values of the semiconductors given in the options. The energy must be equal to or greater than the band gap energy for electron excitation to occur. Since the photon energy is \(0.89 \, eV\), the semiconductor with a band gap of \(1 \, eV\) or less can potentially absorb this photon. The closest option available in the choices is option (a) \(1 \, eV\).

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Key Concepts

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

Photon Energy Calculation
When discussing the photoelectric effect, calculating the energy of a photon is fundamental. This energy is crucial because it determines if a photon can excite an electron within a semiconductor. To find the photon energy, you use the formula:\[E = \frac{hc}{\lambda}\]where:
  • \(E\) is the energy of the photon in joules.
  • \(h = 6.63 \times 10^{-34} \, Js\) is Planck's constant.
  • \(c = 3 \times 10^{8} \, m/s\) is the speed of light in a vacuum.
  • \(\lambda\) is the wavelength of the light in meters.
To dive deeper, consider that each electron-volt (eV) is an amount of energy equal to \(1.6 \times 10^{-19} \, J\). So, when calculating photon energy, if you end up with joules, you often need to convert them to electron-volts for practical comparisons with band gap energies of semiconductors. This conversion helps in identifying if a material can absorb a specific photon.
Band Gap Energy
Band gap energy in semiconductors is a pivotal concept in understanding how materials absorb light. It signifies the minimal energy required to excite an electron from the valence band to the conduction band within the material.
A photon must have equal or greater energy than the band gap to induce electron excitation. In our exercise, we compare photon energy with various band gaps, values specified in electron-volts. These comparisons guide in selecting appropriate semiconductors that can effectively interact with these photons.
This interaction determines a semiconductor's suitability for various applications, like solar cells or photodetectors. A correct match between photon energy and band gap is key to ensuring that the material can absorb and activate its electrons, contributing to efficient energy use in devices.
Wavelength to Energy Conversion
Wavelength to energy conversion is integral when analyzing the photoelectric effect. Photons carry energy inversely proportional to their wavelength, meaning longer wavelengths contain less energy.
By using the relationship \(E = \frac{hc}{\lambda}\), where \(\lambda\) represents the wavelength, we can translate this into beating frequency terms to understand energy presence in a given scenario.
For practical use, such as within a photodetector scenario, it's crucial to convert nm wavelengths to meters to plug them into the formula accurately. Fully understanding this conversion provides insight into if a material's band gap matches the photon energy, ensuring that photon absorption and conductivity occur naturally and effectively within the semiconductor. Converting wavelength to energy thus forms a bridge between theoretical calculations and real-world applications.

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Most popular questions from this chapter

Three waves \(A, B\) and \(C\) of frequencies \(1600 \mathrm{kHz}\), \(5 \mathrm{MHz}\) and \(60 \mathrm{MHz}\), respectively are to be transmitted from one place to another. Which of the following is the most appropriate mode of communication? [NCERT Exemplar] (a) \(A\) is transmitted via space wave while \(B\) and \(C\) are transmitted via sky wave (b) \(A\) is transmitted via ground wave, \(B\) via sky wave and \(C\) via space wave (c) \(B\) and \(C\) are transmitted via ground wave while \(A\) is transmitted via sky wave (d) \(B\) is transmitted via ground wave while \(A\) and \(C\) are transmitted via space wave

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A diode AM detector with the output circuit consisting of \(R=1 \mathrm{k} \Omega\) and \(C=1 \mu \mathrm{F}\) would be more suitable for detecting a carrier signal of (a) \(10 \mathrm{kH}_{2}\) (b) \(1 \mathrm{kH}_{2}\) [c) \(0.5 \mathrm{kHz}\) [d) \(0.1 \mathrm{kHz}\)

An optical fibre made of glass with a core of refractive index of \(1.55\) and include with another glass with a refractive index of 1.51. Launching takes place from air. What is the value of eritical angle for core-clad boundary? (a) \(65^{\prime}\) (b) \(72^{=}\) (c) \(77^{*}\) (d) \(82=\)
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