/*! 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} Q67E Question: An LED is connected in... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Question: An LED is connected in series with a resistor and a power supply of variable voltage. The supply voltage starts from —5 V and increases linearly with time to a maximum of +5 V. The figure shows the supply voltage the voltage across the diode, and the current in the Circuit as functions of time. The LED does not glow for the whole time shown. (a) Explain the plots, and identify the period when the LED glows. (b) Approximately what color is the light produced by the LED?

Short Answer

Expert verified

Answer

(a) The band gap of the LED is 2.5e V .

(b) The light produced by the LED is cyan color light.

Step by step solution

01

Given data

The given diagram is shown in figure 1.

Figure 1

02

Concept of the energy of a photon

The expression to determine the wavelength of the light is, λ=hcV .

03

Determine the explanation for the plots and identify the period when the LED glows

(a)

The LED only glows when the current flows in the circuit and the time period is 15s<t<20s.

From the above graph the voltage required by the LED to turn on is, V= 2.5 e V .

Therefore, the band gap of the LED is V = 2.5 e V .

04

Determine the color of the light produced by the LED

(b)

Calculate the wavelength of the light.

λ=hcV=1240 eV.nm2.5 eV=496 nm

The above value shows that the light is of cyan color.

Therefore, the light produced by the LED is cyan color light.

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.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

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

Assuming an interatomic spacing of 0.15 nm, obtain a rough value for the width (in eV) of then=2 band in a one-dimensional crystal.

Section 10.2 gives the energy and approximate proton separation of the H2+ molecule. What is the energy of the electron alone?

The bond length of theN2molecule is 0.11nm, and its effect the spring constant is 2.3×103N/m .

(a) From the size other energy jumps for rotation and vibration, determine whether either of these modes of energy storage should be active at 300K .

(b) According to the equipartition theorem, the heat capacity of a diatomic molecule storing energy in rotations but not vibrations should be52R(3 translational +2rotational degrees of freedom). If it is also storing energy in vibrations. it should be72R(adding 2 vibrational degrees). Nitrogen's molar heat capacity is 20.8J/mol.K at 300K. Does this agree with your findings in part (a)?

Question: The interatomic potential energy in a diatomic molecule (Figure 10.16) has many features: a minimum energy, an equilibrium separation a curvature and so on. (a) Upon what features do rotational energy levels depend? (b) Upon what features do the vibration levels depend?

Brass is a metal consisting principally of copper alloyed with a smaller amount of zinc, whose atoms do not alternate in a regular pattern in the crystal lattice but are somewhat randomly scattered about. The resistivity of brass is higher than that of either copper or zinc at room temperature, and it drops much slower as the temperature is lowered. What do these behaviors tell us about electrical conductivity in general?

See all solutions

Recommended explanations on Physics Textbooks

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.