/*! 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} Problem 21 The retina of a human eye can de... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 21

The retina of a human eye can detect light when radiant energy incident on it is at least \(4.0 \times 10^{-17} \mathrm{~J}\). For light of 585 -nm wavelength, how many photons does this energy correspond to?

Short Answer

Expert verified
Approximately 118 photons.

Step by step solution

01

Write the Energy-Photon Relationship Formula

The energy of one photon can be calculated using the formula \( E = \frac{hc}{\lambda} \), where \( E \) is the energy of a photon, \( h \) is Planck's constant \( (6.63 \times 10^{-34} \text{ J s}) \), \( c \) is the speed of light \( (3.00 \times 10^{8} \text{ m/s}) \), and \( \lambda \) is the wavelength.
02

Convert Wavelength to Meters

Convert the given wavelength from nanometers to meters: \( 585 \text{ nm} = 585 \times 10^{-9} \text{ m} \).
03

Calculate Energy of a Single Photon

Use the energy formula to find the energy of one photon: \[ E = \frac{(6.63 \times 10^{-34} \, \text{J s}) \times (3.00 \times 10^{8} \, \text{m/s})}{585 \times 10^{-9} \, \text{m}} \approx 3.40 \times 10^{-19} \text{ J} \] per photon.
04

Find Number of Photons

To find the number of photons, divide the total energy by the energy per photon: \[ n = \frac{4.0 \times 10^{-17} \, \text{J}}{3.40 \times 10^{-19} \, \text{J/photon}} \approx 118 \] photons.

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 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Planck's Constant
Planck's constant is a crucial quantity in quantum mechanics, symbolized by the letter "h".
This fundamental concept shows the relationship between energy and frequency for particles like photons.
Planck's constant (denoted as \(h\)) has a value of \(6.63 \times 10^{-34} \text{ J s}\).
  • It's used to express the size of quanta or "packets" of energy.
  • It's essential for calculating the energy of a photon.
In an equation, like the one used to find a photon's energy \(E = \frac{hc}{\lambda}\), Planck's constant interacts with other constants to determine energy.
Each photon's energy is calculated by multiplying Planck's constant by the speed of light, and then dividing by the wavelength.
This equation is central to many calculations in quantum physics, helping us understand how light and its particles behave.
Speed of Light
The speed of light, represented by \(c\), is a key constant in physics.
It plays a vital role in the formulas used to calculate photon energy.
The speed of light is precisely \(3.00 \times 10^{8} \text{ m/s}\).
  • Light travels at this ultimate speed in a vacuum.
  • This value becomes a part of calculations that solve for crucial information about photons.
When we calculate the energy of a photon, the speed of light is multiplied by Planck's constant in the formula \(E = \frac{hc}{\lambda}\).
This shows that light's speed is directly related to photon energy, impacting how we interpret light in both theoretical and practical applications.
Wavelength Conversion
Wavelength conversion is an important aspect to understand when dealing with calculations involving light.
The wavelength is the distance between successive crests of a wave and is commonly measured in nanometers (nm).
  • 1 nanometer equals \(10^{-9}\) meters.
  • Converting wavelength to meters is essential for calculations within SI units.
In the problem we explored, the wavelength of light given was 585 nm, which needed conversion to meters:
\(585 \text{ nm} = 585 \times 10^{-9} \text{ m}\).
This conversion allows us to apply it in the energy formula, \(E = \frac{hc}{\lambda}\), effectively and without error.
A proper understanding of wavelength conversion is key to performing accurate scientific calculations across various areas of physics.

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

In a photoelectric experiment a student uses a light source whose frequency is greater than that needed to eject electrons from a certain metal. However, after continuously shining the light on the same area of the metal for a long period of time the student notices that the maximum kinetic energy of ejected electrons begins to decrease, even though the frequency of the light is held constant. How would you account for this behavior?

Discuss the similarities and differences between a \(1 s\) and a \(2 s\) orbital.

Certain sunglasses have small crystals of silver chloride (AgCl) incorporated in the lenses. When the lenses are exposed to light of the appropriate wavelength, the following reaction occurs: $$ \mathrm{AgCl} \longrightarrow \mathrm{Ag}+\mathrm{Cl} $$ The Ag atoms formed produce a uniform grey color that reduces the glare. If \(\Delta H\) for the preceding reaction is \(248 \mathrm{~kJ} / \mathrm{mol}\), calculate the maximum wavelength of light that can induce this process.

A photoelectric experiment was performed by separately shining a laser at \(450 \mathrm{nm}\) (blue light) and a laser at \(560 \mathrm{nm}\) (yellow light) on a clean metal surface and measuring the number and kinetic energy of the ejected electrons. Which light would generate more electrons? Which light would eject electrons with greater kinetic energy? Assume that the same amount of energy is delivered to the metal surface by each laser and that the frequencies of the laser lights exceed the threshold frequency.

Calculate the wavelength (in \(\mathrm{nm}\) ) of a photon emitted by a hydrogen atom when its electron drops from the \(n=7\) state to the \(n=2\) state.
See all solutions

Recommended explanations on Chemistry Textbooks

Physical Chemistry

Read Explanation

Kinetics

Read Explanation

Nuclear Chemistry

Read Explanation

Chemistry Branches

Read Explanation

The Earths Atmosphere

Read Explanation

Organic Chemistry

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.