Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 21: Problem 1

Visible light having a wavelength of \(5 \times 10^{-7} \mathrm{~m}\) appears green. Compute the frequency and energy of a photon of this light.

Short Answer

Expert verified
Answer: The frequency of the green light is \(6.0 \times 10^{14}\) Hz and the energy of a photon of this light is \(3.978 \times 10^{-19}\) J.

Step by step solution

01

Identify the given information

We are given the wavelength of the green light, which is \(5 \times 10^{-7} \mathrm{~m}\).
02

Calculate the frequency

First, we will find the frequency of the green light using the speed of light formula: $$ c = \lambda \times f $$ Where: \(c\) is the speed of light (\(3.0 \times 10^8 \mathrm{~m/s}\)), \(\lambda\) is the wavelength, and \(f\) is the frequency. We have the wavelength and we know the speed of light, so we can solve for the frequency: $$ f = \frac{c}{\lambda} $$ Substitute the given values: $$ f = \frac{3.0 \times 10^8 \mathrm{~m/s}}{5 \times 10^{-7} \mathrm{~m}} $$
03

Evaluate the frequency

Calculate the frequency: $$ f = \frac{3.0 \times 10^8 \mathrm{~m/s}}{5 \times 10^{-7} \mathrm{~m}} = 6.0 \times 10^{14} \mathrm{~Hz} $$ So the frequency of the green light is \(6.0 \times 10^{14} \mathrm{~Hz}\).
04

Calculate the energy

Next, we will find the energy of a photon of this light using the Planck's constant formula: $$ E = h \times f $$ Where: \(E\) is the energy of a photon, \(h\) is the Planck's constant (\(6.63 \times 10^{-34} \mathrm{~Js}\)), and \(f\) is the frequency. We found the frequency in step 3 and we know the Planck's constant, so we can substitute these values: $$ E = (6.63 \times 10^{-34} \mathrm{~Js}) \times (6.0 \times 10^{14} \mathrm{~Hz}) $$
05

Evaluate the energy

Calculate the energy of a photon: $$ E = (6.63 \times 10^{-34} \mathrm{~Js}) \times (6.0 \times 10^{14} \mathrm{~Hz}) = 3.978 \times 10^{-19} \mathrm{~J} $$ So the energy of a photon of this green light is \(3.978 \times 10^{-19} \mathrm{~J}\). In conclusion, the frequency of the green light is \(6.0 \times 10^{14} \mathrm{~Hz}\) and the energy of a photon of this light is \(3.978 \times 10^{-19} \mathrm{~J}\).

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

Briefly describe the three absorption mechanisms in nonmetallic materials.

Compute the velocity of light in diamond, which has a dielectric constant \(\epsilon_{r}\) of \(5.5\) (at frequencies within the visible range) and a magnetic susceptibility of \(-2.17 \times 10^{-5}\)

Briefly explain why amorphous polymers are transparent, whereas predominantly crystalline polymers appear opaque or, at best, translucent.

Using the data in Table \(21.1\), estimate the dielectric constants for silica glass (fused silica), soda-lime glass, polytetrafluoroethylene, polyethylene, and polystyrene, and compare these values with those cited in the following table. Briefly explain any discrepancies. \begin{tabular}{lc} \hline Material & Dielectric Constant (1 MHz) \\ \hline Silica glass & \(3.8\) \\ \hline Soda-lime glass & \(6.9\) \\ \hline Polytetrafluoroethylene & \(2.1\) \\ \hline Polyethylene & \(2.3\) \\ \hline Polystyrene & \(2.6\) \\ \hline \end{tabular}

Briefly explain the operation of a photographic lightmeter.
See all solutions

Recommended explanations on Physics Textbooks

Electric Charge Field and Potential

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Electrostatics

Read Explanation

Measurements

Read Explanation

Oscillations

Read Explanation

Nuclear Physics

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