Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 10: Q64E (page 470)

Question: The photons emitted by an LED arise from the energy given up in electron-hole recombinations across the energy gap. How large should the energy gap be to give photons at the red end of the visible spectrum (700nm) ?

Short Answer

Expert verified

Answer

The energy gap to give photons at the red end of the visible spectrum is 1.8 eV .

Step by step solution

01

Given data 

Wavelength of emitted photons, λ=700nm.

02

Concept of photon energy 

Energy corresponding to wavelength of photon is,E=hcλ .

Where, h is Planck's constant =6.64×10-34J·s.

C Is velocity of light=3×108m/s .

03

Step 3:Determine the energy gap to give photons at the red end of the visible spectrum

Substitute numerical values in photon energy equation.

E=6.64×10-34J·s×3×108m/s700×10-9m=0.0284×10-17J=1.8eV

The energy gap to give photons at the red end of the visible spectrum is 1.8eV

.

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

From the diagrams in Figure 7.15and the qualitative behavior of the wave functions they represent, argue that a combination of the2pZ(i.e.,ml=0) and the negative of the 2swould produce a function that sticks out preferentially in the positive Zdirection. This is known as a hybridspstate.

By the “vector” technique of example 10.1 , show that the angles between all lobes of the hybridsp3states are 109.5°..

Question: When electrons cross from the n-type to the p-type to equalize the Fermi energy on both sides in an unbiased diode they leave the n-type side with an excess of positive charge and give the p-type side an excess of negative. Charge layers oppose one another on either side of the depletion zone, producing. in essence, a capacitor which harbors the so-called built-in electric

field. The crossing of the electrons to equalize the Fermi energy produces the dogleg in the bands of roughly Egap , and the corresponding potential differerence

is Egap /e. The depletion zone in a typical diode is wide, and the band gap is 1.0 eV. How large is the built-in electric field?

The resistivity of the silver is 1.6×10-8 Ω⋅mat room temperature of (300 K), while that of silicon is about10 Ω⋅m

(a) Show that this disparity follows, at least to a rough order of magnitude from the approximate 1 eV band gap in silicon.

(b) What would you expect for the room temperature resistivity of diamond, which has a band gap of about 5 eV.

Section 10.2 discusses - bonds and σ- bonds for p - states and π-bonds for s-states but not σ- bonds for π - states. Why not?
See all solutions

Recommended explanations on Physics Textbooks

Measurements

Read Explanation

Radiation

Read Explanation

Nuclear Physics

Read Explanation

Electrostatics

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Famous Physicists

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