Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 28: Problem 49

A photoconductor (see Conceptual Question 13 ) allows charge to flow freely when photons of wavelength \(640 \mathrm{nm}\) or less are incident on it. What is the band gap for this photoconductor?

Short Answer

Expert verified
Answer: The band gap energy for the photoconductor is \(3.1 \times 10^{-19} \, \text{J}\).

Step by step solution

01

Identify the given information

The given information is that the photoconductor allows charge to flow when the photon's wavelength is 640 nm or less. We will need to convert this wavelength into energy to find the band gap.
02

Apply the Planck's constant

In order to find the energy of a photon, we can use the Planck's constant (\(h\)) and the speed of light (\(c\)). The formula to calculate the energy (\(E\)) of a photon is: \(E = \frac{hc}{\lambda}\) Where, \(E\) is the energy \(h = 6.63 \times 10^{-34} \, \text{Js}\) (Planck's constant) \(c = 3 \times 10^{8} \, \text{m/s}\) (speed of light) \(\lambda = 640 \, \text{nm}\) (wavelength) We first need to convert the wavelength from nm to meters. To do that, we divide the wavelength value by \(10^9\).
03

Convert the wavelength

\(\lambda = \frac{640}{10^9} \, \text{m} = 6.4 \times 10^{-7} \, \text{m}\) Now we can substitute the values for \(h\), \(c\), and \(\lambda\) in the energy formula.
04

Calculate the energy of a photon

\(E = \frac{(6.63 \times 10^{-34} \, \text{Js})(3 \times 10^{8} \, \text{m/s})}{6.4 \times 10^{-7} \, \text{m}}\) \(E = 3.1 \times 10^{-19} \, \text{J}\) The energy of a photon with a wavelength of 640 nm is \(3.1 \times 10^{-19} \, \text{J}\).
05

Determine the band gap energy

Now we know the energy required to overcome the band gap is equal to the energy of the photon. Therefore, the band gap energy is: \(E_{\text{band gap}} = 3.1 \times 10^{-19} \, \text{J}\) The band gap for the photoconductor is \(3.1 \times 10^{-19} \, \text{J}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The particle in a box model is often used to make rough estimates of ground- state energies. Suppose that you have a neutron confined to a one-dimensional box of length equal to a nuclear diameter (say \(10^{-14} \mathrm{m}\) ). What is the ground-state energy of the confined neutron?
The particle in a box model is often used to make rough estimates of energy level spacings. For a metal wire \(10 \mathrm{cm}\) long, treat a conduction electron as a particle confined to a one-dimensional box of length \(10 \mathrm{cm} .\) (a) Sketch the wave function \(\psi\) as a function of position for the electron in this box for the ground state and each of the first three excited states. (b) Estimate the spacing between energy levels of the conduction electrons by finding the energy spacing between the ground state and the first excited state.
The phenomenon of Brownian motion is the random motion of microscopically small particles as they are buffeted by the still smaller molecules of a fluid in which they are suspended. For a particle of mass $1.0 \times 10^{-16} \mathrm{kg},\( the fluctuations in velocity are of the order of \)0.010 \mathrm{m} / \mathrm{s} .$ For comparison, how large is the change in this particle's velocity when the particle absorbs a photon of light with a wavelength of \(660 \mathrm{nm}\) such as might be used in observing its motion under a microscope?
The beam emerging from a ruby laser passes through a circular aperture $5.0 \mathrm{mm}$ in diameter. (a) If the spread of the beam is limited only by diffraction, what is the angular spread of the beam? (b) If the beam is aimed at the Moon, how large a spot would be illuminated on the Moon's surface?
The neutrons produced in fission reactors have a wide range of kinetic energies. After the neutrons make several collisions with atoms, they give up their excess kinetic energy and are left with the same average kinetic energy as the atoms, which is \(\frac{3}{2} k_{\mathrm{B}} T .\) If the temperature of the reactor core is \(T=400.0 \mathrm{K},\) find (a) the average kinetic energy of the thermal neutrons, and (b) the de Broglie wavelength of a neutron with this kinetic energy.
See all solutions

Recommended explanations on Physics Textbooks

Work Energy and Power

Read Explanation

Kinematics Physics

Read Explanation

Modern Physics

Read Explanation

Fluids

Read Explanation

Geometrical and Physical Optics

Read Explanation

Physical Quantities And Units

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free