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Chapter 19: Problem 2632

A P-N photodiode is made of a material with a band gap of 2.0ev. The minimum frequency of the radiation that can be absorbed by the material is nearly (Take hc \(=1240 \mathrm{eVnm}\) ) (A) \(5 \times 10^{14} \mathrm{~Hz}\) (B) \(20 \times 10^{14} \mathrm{~Hz}\) (C) \(1 \times 10^{14} \mathrm{~Hz}\) (D) \(10 \times 10^{14} \mathrm{~Hz}\)

Short Answer

Expert verified
The minimum frequency of the radiation that can be absorbed by the material is nearly (A) \( 5 \times 10^{14} \mathrm{~Hz} \).

Step by step solution

01

Find the minimum wavelength

We know the band gap energy of the P-N photodiode is 2.0 eV. First, we will find the minimum wavelength (\( \lambda_{min} \)) that can be absorbed by the material using the energy formula. Energy = (hc) / \( \lambda \) 2.0 eV = (1240 eVnm) / \( \lambda_{min} \) Now, let's solve for \( \lambda_{min} \).
02

Solve for the minimum wavelength

Rearrange the formula to solve for the minimum wavelength: \( \lambda_{min} \) = (1240 eVnm) / (2.0 eV) \( \lambda_{min} \) = 620 nm Now we have the minimum wavelength for the radiation that can be absorbed by the material.
03

Determine the minimum frequency

Next, we need to use this minimum wavelength to find the minimum frequency (υ_min) of the radiation. The relation between frequency, speed of light (c), and wavelength is: υ_min = c / \( \lambda_{min} \) In order to find the minimum frequency, we need to convert the wavelength to meters. We know: \( 1 \mathrm{~nm} = 10^{-9} \mathrm{~m} \) So, the minimum wavelength in meters is: \( \lambda_{min} \) = 620 nm × \( 10^{-9} \mathrm{~m/nm} \) = \( 6.20 \times 10^{-7} \mathrm{~m} \) Now, the speed of light (c) is: c = \( 3 \times 10^8 \mathrm{~m/s} \) Let's plug these values into the formula:
04

Calculate the minimum frequency

Using the formula and the values we have found: υ_min = \( \frac{3 \times 10^{8} \mathrm{~m/s}}{6.20 \times 10^{-7} \mathrm{~m}} \) υ_min = \( 4.84 \times 10^{14} \mathrm{~Hz} \) The minimum frequency of the radiation that can be absorbed by the material is nearly \( 4.84 \times 10^{14} \mathrm{~Hz} \). Comparing this value to the available options, we can round it up to: υ_min ≈ \( 5 \times 10^{14} \mathrm{~Hz} \)
05

Choose the correct answer

By comparing the calculated value of the minimum frequency with the options given, we can determine that the correct answer is: (A) \( 5 \times 10^{14} \mathrm{~Hz} \)

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