Problem 176
Question
A gas adsorbs a photon of \(355 \mathrm{~nm}\) and emits at two wavelengths. If one of the emissions is at \(680 \mathrm{~nm}\), the other is at: (a) \(743 \mathrm{~nm}\) (b) \(376 \mathrm{~nm}\) (c) \(453 \mathrm{~nm}\) (d) \(581 \mathrm{~nm}\)
Step-by-Step Solution
Verified Answer
The other emission wavelength is (b) 376 nm.
1Step 1: Understanding Energy Conservation
We need to use the principle of energy conservation. The energy absorbed by the gas from the incoming photon must equal the total energy emitted at the two different wavelengths. Let's calculate the absorbed energy first.
2Step 2: Calculate Energy of Absorbed Photon
The energy of a photon is given by the formula: \[ E = \frac{hc}{\lambda} \]where \( h \) 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 in meters. The absorbed wavelength is \( 355 \text{ nm} \), or \( 355 \times 10^{-9} \text{ m} \). Calculate \( E \).
3Step 3: Calculate Energy for Emission at 680 nm
Similarly, calculate the energy of the emitted photon at wavelength \( 680 \text{ nm} \) using the energy formula from Step 2. This gives us the energy emitted for this particular wavelength.
4Step 4: Apply Energy Conservation Principle
Subtract the energy calculated for \( 680 \text{ nm} \) from the absorbed energy calculated in Step 2. This gives us the remaining energy that must be emitted by the second wavelength.
5Step 5: Determine the Wavelength of the Second Emission
Using the remaining energy from Step 4, use the energy formula in Step 2 to calculate the wavelength \( \lambda \) of the second emission. Choose the correct wavelength that matches any of the given options.
Key Concepts
Photon Energy CalculationWavelength EmissionPlanck's Constant
Photon Energy Calculation
When a gas absorbs a photon, its energy is determined by the photon's wavelength using a specific formula. This is essential in photochemistry, where photon absorption leads to chemical reactions. The formula that helps us calculate photon energy is:\[ E = \frac{hc}{\lambda} \]Here, \( E \) represents the photon's energy, \( h \) 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 in meters. By plugging in the wavelength, we calculate the energy of a photon for absorption or emission.
- An understanding of photon energy calculation enables us to explore how energy transformations occur in photochemical reactions.
- This calculation is crucial for determining what happens to absorbed energy within molecules.
Wavelength Emission
In photochemistry, after a molecule absorbs a photon, it often emits light at a different wavelength. This process is thanks to energy conservation, meaning the absorbed energy must equal the total emitted energy. These emitted wavelengths help us analyze the behavior of molecules.
- The calculation involves determining the energy of one wavelength and using that to find any others.
- This provides insights into secondary emissions and how they split energy into different forms.
Planck's Constant
Planck's constant (\( h \)) is a fundamental quantity in physics that links the energy of a photon to its frequency. It's a cornerstone for many calculations involving photon behavior in photochemistry. The value of Planck's constant is \(6.626 \times 10^{-34} \text{ J s}\). Its role is significant, allowing theoretical predictions to match experimental findings.
- Planck's constant helps convert between wavelength and energy.
- It is crucial for understanding quantum mechanics' principles, especially in photochemistry.
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