/*! 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 24 The number of photons emitted by... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The number of photons emitted by a \(60 \mathrm{~W}\) bulb per second, if \(10 \%\) of the electrical energy supplied to an incandescent light bulb is radiated as visible light, is: (a) \(1.8 \times 10^{19}\) (b) \(1.8 \times 10^{16}\) (c) \(1.8 \times 10^{11}\) (d) \(1.8 \times 10^{21}\)

Short Answer

Expert verified
The number of photons emitted per second is approximately \(1.8 \times 10^{19}\), so the answer is (a).

Step by step solution

01

Calculate the Energy Radiated as Visible Light

Given that only 10% of electrical energy is radiated as visible light by the bulb, we first need to determine how much energy is used for this purpose. The bulb has a power of 60 W, which means it consumes 60 joules of energy per second. Since only 10% of this energy is radiated as visible light, we calculate:\[\text{Energy for visible light} = 0.1 \times 60 \text{ J/s} = 6 \text{ J/s}\]
02

Calculate Energy of a Single Photon

To calculate the number of photons, we must first know the energy of a single photon of visible light. The energy of a photon is determined by the formula: \(E = \frac{hc}{\lambda}\), where \(h\) is Planck's constant \(6.63 \times 10^{-34} \text{ Jâ‹…s}\), \(c\) is the speed of light \(3 \times 10^8 \text{ m/s}\), and \(\lambda\) is the wavelength. Assuming an average wavelength of visible light is 550 nm \(= 550 \times 10^{-9} \text{ m}\), we calculate the energy of one photon:\[E = \frac{(6.63 \times 10^{-34}) \times (3 \times 10^8)}{550 \times 10^{-9}} = 3.61 \times 10^{-19} \text{ J/photon}\]
03

Calculate the Number of Photons Emitted per Second

With the energy radiated as visible light and the energy per photon known, we can now find the number of photons emitted per second by dividing the total energy by the energy of one photon.\[\text{Number of photons} = \frac{6 \text{ J/s}}{3.61 \times 10^{-19} \text{ J/photon}} \approx 1.66 \times 10^{19} \text{ photons/s}\]
04

Round the Answer to the Nearest Provided Option

The calculated number of photons is approximately \(1.66 \times 10^{19}\), close to option (a), which is \(1.8 \times 10^{19}\). This rounding considers significant figures and measurement uncertainties. Therefore, the closest answer matches option (a).

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!

Key Concepts

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

Visible Light
Visible light is the portion of the electromagnetic spectrum that the human eye can see. It encompasses a range of wavelengths from about 400 nm to 700 nm. This is the light that illuminates our world, making it colorful and vivid.
Different wavelengths within this range correspond to different colors. For instance, shorter wavelengths near 400 nm appear as violet, while longer wavelengths around 700 nm look red. The middle of this spectrum, around 550 nm, is perceived as green, which is often used as an average in calculations.
Visible light plays a crucial role in various applications:
  • Sight: Allows humans and animals to see.
  • Communication: Used in technologies such as fiber optics.
  • Energy: Sources like sunlight are exploited for solar power systems.
Understanding visible light involves not just recognizing it as the light we see, but appreciating its role and significance in science and technology.
Planck's Constant
Planck's constant is a fundamental constant in physics, denoted by the symbol \(h\). It is essential in the quantum mechanics field, relating to the energy of photons and their frequency. The value of Planck's constant is approximately \(6.63 \times 10^{-34}\) Jâ‹…s.
This constant is crucial when dealing with calculations about quantum phenomena, particularly those involving photons—particles of light. The relationship between photon energy and its frequency \(f\) is given by the equation:
\[ E = hf \]
Where:
  • \(E\) is the energy of the photon.
  • \(f\) is the frequency of the light wave.
Understanding Planck's constant helps in:
  • Exploring quantum physics and its implications.
  • Calculating energies in photon emission processes.
  • Building technologies that depend on light interaction, such as lasers and LEDs.
By using Planck's constant, we bridge classical and quantum concepts, creating comprehensive insights into energy transfer processes at the micro level.
Wavelength of Light
The wavelength of light refers to the distance between successive crests of a wave. It is a key property of waves that defines the color of light and relates to other properties such as energy and frequency.
Wavelength is typically measured in nanometers (nm), where 1 nm = \(10^{-9}\) meters. The visible light spectrum, ranging from approximately 400 nm to 700 nm, includes all colors perceivable by the human eye.
The wavelength \(\lambda\) affects the energy \(E\) of a photon based on the equation:\[ E = \frac{hc}{\lambda} \] Where:
  • \(h\) is Planck's constant.
  • \(c\) is the speed of light \(3 \times 10^8 \text{ m/s}\).
From these relations, we understand:
  • Shorter wavelengths imply higher energy photons (e.g., blue light).
  • Longer wavelengths mean lower energy photons (e.g., red light).
The wavelength of light plays a vital role in optics, communication technologies, and energy systems. Recognizing how wavelength governs the behavior and characteristics of light helps in various scientific and practical applications.

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 an oil drop experiment, the following charges (in arbitrary units) were found on a series of oil droplets : \(2.30 \times 10^{-15}, \quad 6.90 \times 10^{-15}, \quad 1.38 \times 10^{-14}, \quad 5.75 \times 10^{-15}\), \(1.955 \times 10^{-14}\). The charge on electron (in the same unit) should be : (a) \(2.30 \times 10^{-15}\) (b) \(1.15 \times 10^{-15}\) (c) \(1.38 \times 10^{-14}\) (d) \(1.955 \times 10^{-14}\)

Two sotrces \(A\) and \(B\) have same power. The wavelength of radiation of \(A\) is \(\lambda_{\mathrm{a}}\) and that of \(B\) is \(\lambda_{b}\). The number of photons emitted per second by \(A\) and \(B\) are \(n_{a}\) and \(n_{b}\) respectively, then : (a) \(\lambda_{\mathrm{a}}>\lambda_{b}\) (b) if \(\lambda_{n}>\lambda_{b}, n_{a}\lambda_{b}, n_{a}=n_{b}\)

Work function of a metal is \(10 \mathrm{eV}\). Photons of \(20 \mathrm{eV}\) are bombarded on it. The frequency of photoelectrons will be: (a) \(=\frac{10}{h}\) (b) \(>\frac{10}{h}\) (c) \(<\frac{10}{h}\) (d) \(\geq \frac{10}{h}\)

When the electromagnetic radiations of frequencies \(4 \times 10^{15} \mathrm{~Hz}\) and \(6 \times 10^{15} \mathrm{~Hz}\) fall on the same metal, in different experiments, the ratio of maximum kinetic energy of electrons liberated is \(1: 3 .\) The threshold frequency for the metal is: (a) \(2 \times 10^{15} \mathrm{~Hz}\) (b) \(1 \times 10^{15} \mathrm{~Hz}\) (c) \(3 \times 10^{15} \mathrm{~Hz}\) (d) \(1.67 \times 10^{15} \mathrm{~Hz}\)

Ultraviolet light of wavelength \(66.26 \mathrm{~nm}\) and intensity \(2 \mathrm{~W} / \mathrm{m}^{2}\) falls on potassium surface by which photoelectrons are ejected out. If only \(0.1 \%\) of the incident photons produce photoelectrons, and surface area of metal surface is \(4 \mathrm{~m}^{2}\), how many electrons are emitted per second? (a) \(2.67 \times 10^{15}\) (b) \(3 \times 10^{15}\) (c) \(3.33 \times 10^{17}\) (d) \(4.17 \times 10^{16}\)

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.