Problem 51 - University Physics with Modern Physics

Question

Exposing Photographic Film. The light-sensitive compound on most photographic films is silver bromide, AgBr. A film is "exposed" when the light energy absorbed dissociates this molecule into its atoms. (The actual process is more complex, but the quantitative result does not differ greatly.) The energy of dissociation of AgBr is \(1.00 \times 10^{5} \mathrm{J} / \mathrm{mol}\) . For a photon that is just able to dissociate a molecule of silver bromide, find (a) the photon energy in electron volts; (b) the wavelength of the photon; (c) the frequency of the photon. (d) What is the energy in electron volts of a photon having a frequency of 100 \(\mathrm{MHz}\) (e) Light from a firefly can expose photographic film, but the radiation from an FM station broadcasting \(50,000 \mathrm{W}\) at 100 \(\mathrm{MHz}\) cannot. Explain why this is so.

Step-by-Step Solution

Answer

Photon energy is about 1.038 eV. Wavelength is approximately 119 nm. Frequency is about 2.52 x 10^15 Hz.

Step 1: Convert Molar Dissociation Energy to Per Molecule Energy

First, convert the dissociation energy from joules per mole to joules per molecule. The energy per molecule can be found by dividing the energy per mole by Avogadro's number, \(N_A = 6.022 \times 10^{23} \text{ mol}^{-1}\). Thus, \(E_{molecule} = \frac{1.00 \times 10^{5} \text{ J/mol}}{6.022 \times 10^{23} \text{ mol}^{-1}}\). Calculate this to find \(E_{molecule}\).

Step 2: Calculate Photon Energy in Electron Volts

To convert the energy in joules to electron volts (eV), use the conversion factor \(1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J}\). Multiply the energy per molecule found in Step 1 by this conversion factor.

Step 3: Calculate Wavelength of the Photon

Use the energy-wavelength relationship: \(E = \frac{hc}{\lambda}\), where \(h = 6.626 \times 10^{-34} \, \text{J s}\) is Planck's constant and \(c = 3.00 \times 10^8 \, \text{m/s}\) is the speed of light. Rearrange the formula to find \(\lambda = \frac{hc}{E}\) and calculate \(\lambda\).

Step 4: Calculate Frequency of the Photon

Use the relation between energy and frequency: \(E = h u\), where \(u\) is the frequency. Rearrange to find \(u = \frac{E}{h}\) and calculate \(u\).

Step 5: Energy of a Photon with Given Frequency

To find the energy of a photon with frequency 100 MHz, use: \(E = h u\), with \(u = 100 \times 10^6 \, \text{Hz}\). Convert the obtained energy in joules to eV using the conversion factor from Step 2.

Step 6: Comparison Between Firefly Light and FM Radio Waves

Firefly light can expose the film because it emits photons with sufficient energy to dissociate the AgBr molecules. FM radio waves, such as those at 100 MHz, have a much lower photon energy which is insufficient to dissociate AgBr, even if broadcasted with high power.

Key Concepts

Energy-Wavelength Relationship

Light is an electromagnetic wave characterized by its wavelength and frequency. These two properties are related to the energy of the light through the equation \( E = \frac{hc}{\lambda} \). Here, \( E \) is the energy of the photon, \( h \) is Planck's constant \( 6.626 \times 10^{-34} \, \text{J s} \), \( c \) is the speed of light \( 3.00 \times 10^8 \, \text{m/s} \), and \( \lambda \) is the wavelength.

This relationship demonstrates that energy and wavelength are inversely proportional. That means, as the wavelength increases, the energy decreases and vice versa. For example, ultraviolet light has a shorter wavelength compared to visible light, but since it has higher energy, it can cause chemical reactions like dissociating molecules. Photographic film relies on this principle because the light must have enough energy to break apart the silver bromide molecules responsible for the image capturing process.

Understanding this equation helps us figure out what type of light can cause certain effects, such as affecting photographic film or other image capturing materials.

Molar Dissociation Energy

Molar dissociation energy refers to the amount of energy required to break one mole of a compound into its individual atoms. For silver bromide (AgBr), this energy is given as \(1.00 \times 10^5 \, \text{J/mol}\).
In practical scenarios, it is often necessary to understand energy at the level of individual molecules rather than moles. To find the energy per molecule, we use Avogadro's number, \( N_A = 6.022 \times 10^{23} \text{ mol}^{-1} \). This number represents the number of molecules in one mole of any substance.

To convert molar dissociation energy to the energy required for a single molecule, simply divide the molar energy by Avogadro's number:

• Energy per molecule = \( \frac{1.00 \times 10^5 \, \text{J/mol}}{6.022 \times 10^{23} \, \text{mol}^{-1}} \)

This is a crucial step in calculating how much energy individual photons must provide to dissociate the compound effectively.

Photon Frequency

The frequency of a photon, often symbolized as \( u \), is directly related to its energy through the formula \( E = hu \), where \( h \) is Planck's constant. You can rearrange this formula to solve for frequency: \( u = \frac{E}{h} \).
When calculating the frequency of a photon with a known energy, this formula helps us translate between energy, frequency, and wavelength efficiently.

This concept is critical in understanding why certain frequencies of light, such as visible light, can cause chemical reactions in photographic materials, whereas others, like FM radio waves, cannot. An FM station at 100 MHz has a much lower frequency compared to visible light, resulting in lower photon energy, which is insufficient to initiate chemical processes like those needed to expose silver bromide on film.

Silver Bromide Dissociation

Silver bromide (AgBr) is widely used in photographic film, thanks to its sensitivity to light. The process of exposing film involves breaking down AgBr into its constituent atoms.

The energy needed to do so must be sufficient to overcome the binding energy of the molecule, which is quantified by the molar dissociation energy. Light photons must have enough energy to break these bonds to create what we visualize as a photographic image.
Why is silver bromide a good choice for photographic use?

• It has the right level of sensitivity to visible light, making it efficient for capturing images.
• It uses the energy-wavelength relationship to determine which light frequencies can cause dissociation.

This principle explains why visible light from common sources, like sunlight or camera flashes, can expose film, while radio waves, with their lower photon energies, cannot interact sufficiently with the compound to produce a photo.