/*! 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 9 A \(75-\mathrm{W}\) light source... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A \(75-\mathrm{W}\) light source consumes 75 \(\mathrm{W}\) of electrical power. Assume all this energy goes into emitted light of wavelength 600 \(\mathrm{nm}\) (a) Calculate the frequency of the emitted light. (b) How many photons per second does the source emit?(c) Are the answers to parts (a) and (b) the same? Is the frequency of the light the same thing as the number of photons emitted per second? Explain.

Short Answer

Expert verified
Frequency is \(5.00 \times 10^{14}\ \text{Hz}\); emits \(2.26 \times 10^{20}\) photons/s. Frequency differs from photon emission rate.

Step by step solution

01

Understanding the Problem

We are given a 75-W light source that emits light at a wavelength of \(600 \ \mathrm{nm}\). We need to calculate: (a) the frequency of this light, (b) the number of photons emitted per second, and (c) determine the relationship between frequency and photon emission rate.
02

Calculating the Frequency (Part a)

We know the speed of light \(c = 3 \times 10^8\ \mathrm{m/s}\) and the wavelength \(\lambda = 600\ \mathrm{nm}\). Use the equation \(c = \lambda u\) to find frequency \(u\). Convert the wavelength to meters: \(600 \ \mathrm{nm} = 600 \times 10^{-9} \ \mathrm{m}\). Solve for \(u\): \[ u = \frac{c}{\lambda} = \frac{3 \times 10^8 \ \mathrm{m/s}}{600 \times 10^{-9} \ \mathrm{m}} \approx 5.00 \times 10^{14} \ \mathrm{Hz} \].
03

Calculating Energy per Photon

Using Planck's equation \(E = hu\), where \(h = 6.63 \times 10^{-34} \ \mathrm{J\cdot s}\) is Planck's constant, calculate the energy for a single photon. \[E = (6.63 \times 10^{-34}\ \mathrm{J \cdot s}) \times (5.00 \times 10^{14}\ \mathrm{Hz}) = 3.315 \times 10^{-19} \ \mathrm{J/photon} \].
04

Calculating Number of Photons Emitted per Second (Part b)

The total power is 75 W, meaning 75 J are emitted per second. The number of photons \(n\) can be calculated using \(n = \frac{P}{E_{\text{photon}}}\), where \(P = 75\ \mathrm{J/s}\). \[n = \frac{75 \ \mathrm{J/s}}{3.315 \times 10^{-19} \ \mathrm{J/photon}} \approx 2.26 \times 10^{20}\ \text{photons/second}\].
05

Analysis of Parts (a) and (b) (Part c)

The frequency calculated in part (a) describes the oscillations per second, while part (b) refers to the number of photons emitted per second. These two quantities are not the same; frequency is a property of individual photons, whereas photon emission rate is related to the total number of 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 91Ó°ÊÓ!

Key Concepts

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

Light Frequency
Light frequency is a fundamental characteristic of light, measuring how many wave cycles occur in one second. Frequency, often denoted by \( u \), is measured in hertz (Hz). It tells us how frequently the light's electric and magnetic fields oscillate. The higher the frequency, the more energy the photons carry, and the visible light appears more towards the blue/violet end of the spectrum.

To calculate the frequency of light, one needs to know its wavelength and the speed of light, which is a constant \( c = 3 \times 10^8 \ \text{m/s} \). The formula to find frequency is:
  • \( u = \frac{c}{\lambda} \)
Where \( \lambda \) is the wavelength of the light. In this exercise, we calculated that a 75-W light source emitting light at a wavelength of 600 nm has a frequency of approximately \( 5.00 \times 10^{14} \ \text{Hz} \). This frequency indicates how quickly the light waves are occurring.
Wavelength
Wavelength is the distance between successive peaks of a wave. In the context of light, it is a critical factor in determining the color and type of light within the electromagnetic spectrum.

Wavelength is usually measured in meters (m), but for light studies, it is often expressed in nanometers (nm), where 1 nm equals \( 10^{-9} \) meters. For example, visible light wavelengths range from about 400 nm (violet) to 700 nm (red).
  • Shorter wavelengths equate to higher frequencies and energies.
  • Longer wavelengths equate to lower frequencies and energies.
In this exercise, the given wavelength was 600 nm, corresponding to orange/yellow light in the visible spectrum. This wavelength was used to find the frequency of the emitted light, demonstrating the inverse relationship between wavelength and frequency—shorter wavelengths have higher frequencies, and longer wavelengths have lower frequencies.
Planck's Constant
Planck's constant is a fundamental physical constant that relates a photon's energy to its frequency. It is denoted by the symbol \( h \) and approximately equals \( 6.63 \times 10^{-34} \ \text{J} \cdot \text{s} \). This constant is pivotal in quantum mechanics and was first introduced by Max Planck.

Planck's equation \( E = h u \) allows physicists to compute the energy of a photon when its frequency is known.
  • \( E \) represents the energy of the photon in joules.
  • \( h \) is Planck's constant.
  • \( u \) is the frequency of the light.
In our exercise, Planck's equation was used to determine that the energy of a single photon emitted by the 75-W light source was approximately \( 3.315 \times 10^{-19} \ \text{J/photon} \). This small yet significant quantity shows how energy is quantized at a microscopic scale.
Electromagnetic Spectrum
The electromagnetic spectrum encompasses all types of electromagnetic radiation, which vary by their wavelengths and frequencies. Visible light, which is only a small part of this spectrum, ranges from about 380 to 750 nm in wavelength.

The spectrum includes other forms of electromagnetic radiation, such as:
  • Radio waves with the longest wavelengths and lowest frequencies.
  • Microwaves, which are shorter than radio waves and used in communications and cooking.
  • Infrared waves, experienced as heat, just beyond visible red light.
  • Ultraviolet light, with shorter wavelengths than visible light, known for causing sunburns.
  • X-rays and gamma rays, which have progressively shorter wavelengths and higher frequencies.
This wide range of wavelengths shows how diverse electromagnetic radiation is. In our exercise, we explored a specific point on this spectrum—light with a wavelength of 600 nm, demonstrating the interplay between wavelength, frequency, and the type of electromagnetic radiation observed.

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

A \(2.50-\mathrm{W}\) beam of light of wavelength 124 \(\mathrm{nm}\) falls on a metal surface. You observe that the maximum kinetic energy of the ejected electrons is 4.16 \(\mathrm{eV}\) . Assume that each photon in the beam ejects a photoclectron. (a) What is the work function (in electron volts of this metal? (b) How many photoclectrons are ejected each second from this metal?(c) If the power of the light beam, but not its wavelength, were reduced by half, what would be the answer to part (b)? (d) If the wavelength of the beam, but not its power, were reduced by half, what would be the answer to part (b)?

Removing Vascular Lesions. A pulsed dye laser emits light of wavelength 585 \(\mathrm{nm}\) in \(450-\mu\) s pulses. Because this wave- length is strongly absorbed by the hemoglobin in the blood, the method is especially effective for removing various types of blemishes due to blood, such as port-wine- colored birth-marks. To get a reasonable estimate of the power required for such laser surgery, we can model the blood as having the same specific heat and heat of vaporization as water \(\left(4190 \mathrm{J} / \mathrm{kg} \cdot \mathrm{K}, 2.256 \times 10^{6} \mathrm{J} / \mathrm{kg}\right) .\) Suppose that each pulse must remove 2.0\(\mu g\) of blood by evaporating it, starting at \(33^{\circ} \mathrm{C}\) (a) How much energy must each pulse deliver to the blemish? (b) What must be the power output of this laser? (c) How many photons does each pulse deliver to the blemish?

A photon has momentum of magnitude \(8.24 \times 10^{-28} \mathrm{kg}\) . \(\mathrm{m} / \mathrm{s}\) (a) What is the energy of this photon? Give your answer in joules and in electron volts. (b) What is the wavelength of this photon? In what region of the electromagnetic spectrum does it lie?

(a) An atom initially in an energy level with \(E=-6.52\) eV absorbs a photon that has wavelength 860 \(\mathrm{nm}\) . What is the internal energy of the atom after it absorbs the photon? (b) An atom initially in an energy level with \(E=-2.68 \mathrm{eV}\) emits a photon that has wavelength 420 \(\mathrm{nm}\) . What is the internal energy of the atom after it emits the photon?

(a) What is the minimum potential difference between the filament and the target of an x-ray tube if the tube is to produce x rays with a wavelength of 0.150 \(\mathrm{nm}\) ? (b) What is the shortest wavelength produced in an \(\mathrm{x}\) -ray tube operated at 30.0 \(\mathrm{kV}\) ?

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.