Problem 29 - 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 MHz? (e) Light from a firelly 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

The dissociation photon energy is 6.224 eV, wavelength 199 nm, frequency 1.51 x 10¹⁵ Hz. Photon energy at 100 MHz is 4.14 x 10⁻⁷ eV, too low to expose film like firefly light.

Step 1: Convert Dissociation Energy to Electron Volts

The dissociation energy given is \(1.00 \times 10^5 \text{ J/mol}\). First, we need to convert this energy from \(\text{J/mol}\) to \(\text{eV/molecule}\). The conversion factor from Joules to electron volts is \(1 \text{ eV} = 1.602 \times 10^{-19} \text{ J}\). Also, \(1 \text{ mol} = 6.022 \times 10^{23} \text{ molecules}\). Thus:\[E_{\text{eV/molecule}} = \frac{1.00 \times 10^5 \text{ J/mol}}{6.022 \times 10^{23} \text{ molecules/mol}} \times \frac{1 \text{ eV}}{1.602 \times 10^{-19} \text{ J}} \= 6.224 \text{ eV/molecule}\]So, the photon energy needed is approximately 6.224 eV.

Step 2: Calculate the Wavelength of the Photon

To calculate the wavelength, use the energy-wavelength relation:\[ E = \frac{hc}{\lambda} \]where \( h = 6.626 \times 10^{-34} \text{ J·s}\) (Planck's constant) and \( c = 3.00 \times 10^8 \text{ m/s}\) (speed of light). We use the energy calculated in step 1 in Joules, which is\( E = 6.224 \text{ eV} \times 1.602 \times 10^{-19} \text{ J/eV}\).Substituting values, we have:\[\lambda = \frac{6.626 \times 10^{-34} \cdot 3.00 \times 10^8}{6.224 \times 1.602 \times 10^{-19}} \\approx 1.99 \times 10^{-7} \text{ m} \= 199 \text{ nm} \]Thus, the wavelength of the photon is approximately 199 nm.

Step 3: Calculate the Frequency of the Photon

With the wavelength known, the frequency \( u \) can be found using:\[ u = \frac{c}{\lambda} \]Substituting the known values,\[u = \frac{3.00 \times 10^8 }{1.99 \times 10^{-7}} \\approx 1.51 \times 10^{15} \text{ Hz} \]Hence, the frequency of the photon is approximately \(1.51 \times 10^{15} \text{ Hz}.\)

Step 4: Photon Energy of 100 MHz

To find the energy of a photon with frequency \(100 \text{ MHz}\), use \( E = hu \):\[u = 100 \times 10^6 \text{ Hz} \]\[ E = 6.626 \times 10^{-34} \cdot 100 \times 10^6 \]\[ E \approx 6.626 \times 10^{-26} \text{ J} \]Convert the energy to electron volts: \(1 \text{ eV} = 1.602 \times 10^{-19} \text{ J}\).\[ E \approx \frac{6.626 \times 10^{-26} }{1.602 \times 10^{-19}} \text{ eV} \\approx 4.14 \times 10^{-7} \text{ eV} \]The energy in electron volts is approximately \(4.14 \times 10^{-7} \text{ eV}\).

Step 5: Explanation of Firefly vs. FM Station Exposure

Firefly light radiation has energy in the visible range, which can be in the same order as the photon energy required to dissociate AgBr (about 6.224 eV). An FM station at \(100 \text{ MHz}\) only emits much lower energy photons (\(4.14 \times 10^{-7} \text{ eV}\)) which cannot break the bond in AgBr. Hence, FM radio frequencies are insufficient to expose the photographic film.

Key Concepts

Silver Bromide

Silver bromide, represented chemically as AgBr, is a key compound used in photographic film. It's sensitive to light, making it ideal for capturing images. When exposed to light, silver bromide's molecules can absorb this energy and break down into silver and bromine atoms. This chemical change forms the basis of photographic development because it allows the image to be stored on the film. In practical terms, when light photons hit the film, the dissociation of silver bromide allows for a latent image to form, which can be developed into a visible photograph. This process is sensitive enough to capture even subtle light differences, which is why silver bromide has been invaluable in photography for producing high-quality images.

Wavelength Calculation

The wavelength of light is a crucial factor in determining its energy. To calculate the wavelength of photons capable of dissociating silver bromide, we use the energy-wavelength relationship, given by the equation:\[ \lambda = \frac{hc}{E} \]where:

• \(\lambda\) is the wavelength.
• \(h\) is Planck's constant \(6.626 \times 10^{-34} \text{ J·s}\).
• \(c\) is the speed of light \(3.00 \times 10^8 \text{ m/s}\).
• \(E\) is the energy of the photon.

Using this equation, once we have converted the photon's energy into joules, we can determine the wavelength, helping us understand what kind of light can break down silver bromide. In this example, the wavelength was calculated to be approximately 199 nm, which falls in the ultraviolet region of the electromagnetic spectrum.

Frequency Calculation

Frequency refers to the number of waves that pass a point in a given time. The frequency \(u\) can be determined if we know the wavelength \(\lambda\) of the photon, using the equation:\[ u = \frac{c}{\lambda} \]

• Where \(u\) is the frequency.
• \(c\) is the speed of light \(3.00 \times 10^8 \text{ m/s}\).
• \(\lambda\) is the wavelength in meters.

In the case of a photon capable of dissociating silver bromide, the frequency is found to be approximately \(1.51 \times 10^{15} \text{ Hz}\). This high frequency correlates with the high energy required for the dissociation process, showing that only certain light frequencies will effectively expose photographic film.

Photon Dissociation

Photon dissociation refers to the process where a molecule absorbs a photon's energy, causing it to break into smaller parts. In photography, this principle helps to understand how light affects photographic film. Specifically, when a photon hits a silver bromide molecule, the energy absorbed can cause the bond between silver and bromine to break. The dissociation energy for silver bromide was given as \(1.00 \times 10^5 \text{ J/mol}\). Effective dissociation is a measure of the photon's ability to impart enough energy to break these bonds. Understanding this process allows photographers and scientists to determine the type of light needed for photographic development.

Electron Volt Conversion

Conversion between units is essential for understanding the energy involved in photon interactions. The energy given in the exercise needs to be converted from joules per mole (J/mol) to electron volts per molecule (eV/molecule) to quantify photon energy effectively. The conversion process uses the following relationships:

• 1 eV = \(1.602 \times 10^{-19}\) Joules
• 1 mole = \(6.022 \times 10^{23}\) molecules

To convert:\[ E_{\text{eV/molecule}} = \frac{E_{\text{J/mol}}}{6.022 \times 10^{23} \text{ molecules/mol}} \times \frac{1 \text{ eV}}{1.602 \times 10^{-19} \text{ J}} \]This ensures consistency in the measurements and calculations related to photon energy. In this instance, the energy required to dissociate a silver bromide molecule was found to be approximately 6.224 eV, establishing a clear link between the energy required and the photon's effect on the film.