/*! 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 22 To expose photographic film. pho... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

To expose photographic film. photons of light dissociate silver bromide (AgBr) molecules. which requires an energy of \(1.2 \mathrm{eV}\). What limit does this impose on the wavelengths that may be recorded by photographic film?

Short Answer

Expert verified
The maximum wavelength of light that can be recorded by the photographic film is approximately 1031 nm.

Step by step solution

01

Convert Energy

Firstly, convert the energy from electronvolts to Joules using the conversion factor: \(1 \mathrm{eV} = 1.6 \times 10^{-19} \mathrm{J}\). Thus, the energy \(E = 1.2 \mathrm{eV} = 1.2 \times 1.6 \times 10^{-19} \mathrm{J}.\)
02

Apply Planck's Equation to find Wavelength

Planck's equation is given by E = hf = hc/λ. We're solving for λ, so rearranging gives us \(λ = \frac{hc}{E}\). Plug in all the known values: h (Planck's constant) = \(6.63 \times 10^{-34} \mathrm{m^2 kg / s}\), c (speed of light) = \(3 \times 10^8 \mathrm{m/s}\), E = \(1.2 \times 1.6 \times 10^{-19} \mathrm{J}\), and calculate Lambda.
03

Result

Calculate the wavelength λ. The result is the maximum wavelength of light that can be recorded by the photographic film. Please ensure the final answer is in nanometers for the commonly used wavelength unit in light-related problems.

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.

Photon Energy
Photon energy is a fundamental concept when discussing light and its interaction with matter.
Photons are tiny packets of energy that travel at the speed of light. The energy of a photon is directly related to its frequency. The higher the frequency, the more energetic the photon is.
This relationship is crucial for understanding how light affects materials, such as in the case of photographic film where photons have enough energy to cause a chemical change.
  • Photon energy helps determine which wavelengths of light can trigger reactions.
  • Materials only respond to photons with sufficient energy to initiate changes.
Understanding photon energy is essential for applications ranging from capturing images to developing solar cells.
Planck's Equation
Planck's equation is used to calculate the energy of photons and is a vital tool in understanding light phenomena. The equation itself can be written as:
  • \[ E = hf \]
where:
  • \( E \) is the energy of the photon,
  • \( h \) is Planck's constant (\(6.63 \times 10^{-34} \mathrm{m^2 kg / s}\)), and
  • \( f \) is the frequency of the light.
In situations where frequency isn't directly given, you can use the speed of light (\( c \)) and wavelength (\( \lambda \)) to reformulate it as:
  • \[ E = \frac{hc}{\lambda} \]
This version of the equation is extremely helpful for finding either the wavelength or energy of light, which is crucial in materials sciences and quantum physics.
Planck's equation bridges the energy of photons with their measurable properties such as wavelength.
Light Wavelength
The wavelength of light is a key characteristic that determines its color and energy level.
It represents the distance between successive peaks of a light wave and is usually measured in nanometers (nm).
The electromagnetic spectrum covers a wide range of wavelengths, from the very short gamma rays to the long radio waves. Visible light is only a small portion, spanning approximately 400 nm to 700 nm.
  • Shorter wavelengths, like blue and violet light, carry more energy.
  • Longer wavelengths, like red light, have less energy.
When calculating the maximum wavelength that a photon can have to induce reactions, such as dissociating silver bromide in photographic film, understanding light wavelength becomes essential. This knowledge helps in various technological applications including photography and spectroscopy.
Silver Bromide Dissociation
Silver bromide (AgBr) is a chemical compound commonly used in photographic films due to its sensitivity to light.
When exposed to light, silver bromide undergoes a chemical change that is crucial for developing photographs.
Specifically, the energy from the photons dissociates the AgBr molecules, initiating a reaction that records the captured image.
  • The energy threshold for silver bromide dissociation is typically 1.2 electronvolts (eV).
  • Only photons with at least this amount of energy can cause dissociation.
This reaction is what allows light to be effectively recorded as an image on film. Understanding this process is essential for photography and related fields, where precise control over light exposure is crucial for capturing clear and vivid images.

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 the Compton effect, we choose the electron to be at the origin and the initial photon's direction of motion to be in the \(+x\) direction. (a) We may also choose the xy-plane so that it contains the velocities of the outgoing electron and photon. Why? (b) The incoming photon's wavelength \(\lambda\) is assumed to be known. The unknowns after the collision are the outgoing photon's wavelength and direction, \(\Lambda^{\prime}\) and \(\theta,\) and the speed and direction of the electron, \(u_{e}\) and \(\phi .\) With only three equations \(-t w_{0}\) components of momentum conservation and one of energy - we can't find all four. Equation \((3-8)\) gives \(\lambda^{\prime}\) in terms of \(\theta .\) Our lack of knowledge of \(\theta\) after the collision (without an experiment) is directly related to a lack of knowledge of something before the collision. What is it? (lmagine the two objects are hard spheres.) (c) Is it reasonable to suppose that we could know this? Explain.

An electron moving to the lefi at \(0.8 c\) collides with an incoming photon moving to the right. Afuer the collision, the elactron is moving to the right at \(0.6 c\) and an outgoing photon moves to the lefi. What was the wavelength of the incoming photon?

In some implementations of "regular expressions," the notations \(\backslash 1, \backslash 2\), and so on can occur in a search pattern. For example, consider the search pattern \(-([a-z A-Z]) \cdot * \backslash 1 \$$. Here, \)\backslash 1\( represents a recurrence of the same substring that matched \)[\mathrm{a}-\mathrm{z} \mathrm{A}-\mathrm{Z}]$, the part of the pattern between the first pair of parentheses. The entire pattern, therefore, will match a line of text that

Consider two separate objects of unequal temperature. What would you do with them and what would have to happen thereafter to enable them to reach the same common temperature? Use this idea to explain why the electromagnetic radiation enclosed in a cavity has a temperature that is the same as that of the cavity walls.

Suppose we produce X-rays not by smashing elections into targets but by smashing protons, which are far more massive. If the same accelerating potential difference were used for both, how would the cutoff wavelengths of the two X-ray spectra compare? Explain.

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.