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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) \(581 \mathrm{~nm}\)

Short Answer

Expert verified
The other emission wavelength is (b) 376 nm.

Step by step solution

01

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.
02

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 \( E \).
03

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.
04

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.
05

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.

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Key Concepts

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

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.
Every time the wavelength changes, the energy changes, which influences how the molecule behaves during and after absorption.
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.
The valuable emissions data, like at 680 nm, enables chemists to assess how energy is distributed during photon emission. Understanding these distributions is the key to material properties analysis and designing devices like lasers and solar cells.
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.
Its consistent value allows scientists to predict how much energy is carried by light at any wavelength. This means that when you perform calculations to find either energy or wavelength, Planck’s constant ensures precision and accuracy in predicting molecular interactions and reactions.

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Most popular questions from this chapter

A 1000 watt radio transmitter operates at a frequen cy of \(880 \mathrm{kc} / \mathrm{sec}\). How many photons per sec does it emit? \(\left[\mathrm{h}=6.626 \times 10^{-34} \mathrm{Js}\right]\) (a) \(2.51 \times 10^{30}\) (b) \(2.27 \times 10^{28}\) (c) \(1.72 \times 10^{30}\) (d) \(1.77 \times 10^{27}\)

Which set has same number of unpaired electrons? (a) \(\mathrm{Fe}^{2+}, \mathrm{Mn}^{+2}\) (b) \(\mathrm{Fe}^{3+}, \mathrm{Mn}^{+2}\) (c) \(\mathrm{Cr}^{+3}, \mathrm{Ni}^{2+}\) (d) \(\mathrm{Zn}^{2+} \mathrm{Cu}^{2+}\)

The value of Planck constant is \(6.63 \times 10^{-34} \mathrm{Js}\). he velocity of light is \(3.0 \times 10^{8} \mathrm{~m} \mathrm{~s}^{-1}\). which value is closest to the wavelength in nanometres of a quantum of light with frequency of \(8 \times 10^{15} \mathrm{~s}^{-1}\) ? (a) \(4 \times 10^{1}\) (b) \(3 \times 10^{7}\) (c) \(2 \times 10^{-25}\) (d) \(5 \times 10^{-18}\)

If \(\mathrm{S}\), be the specific charge \((\mathrm{e} / \mathrm{m})\) of cathode rays and \(\mathrm{S}_{2}\) be that of positive rays then which is true? (a) \(\mathrm{S}_{1}=\mathrm{S}_{2}\) (b) \(\mathrm{S}_{1}<\mathrm{S}_{2}\) (c) \(\mathrm{S}_{1}>\mathrm{S}_{2}\) (d) None of these

For radial probability distribution curves, which of the following is /are correct? (a) the number of spherical or radial nodes is equal to \((\mathrm{n}-l-1)\) (b) the number of maxima in \(2 \mathrm{~s}\) orbital is two (c) the number of angular nodes is ' \(l\). (d) \(3 \mathrm{~d}_{z^{2}}\) has two spherical nodes.

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