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Chapter 3: Problem 33

An atom absorbs a photon of wavelength \(425 \mathrm{nm}\) and immediately emits another photon of wavelength \(643 \mathrm{nm}\). What is the net energy absorbed by the atom in this process?

Short Answer

Expert verified
The net energy absorbed by the atom is \(1.58 \times 10^{-19} \text{J}\).

Step by step solution

01

Convert Wavelengths to Energy

Use the formula for the energy of a photon: \[ E = \frac{hc}{\lambda} \] where \(h\) is Planck's constant \(6.626 \times 10^{-34} \text{Js}\), \(c\) is the speed of light \(3 \times 10^8 \text{m/s}\), and \(\lambda\) is the wavelength.
02

Calculate Energy of Absorbed Photon

Substitute the wavelength of the absorbed photon (\(425 \text{nm} = 425 \times 10^{-9} \text{m}\)) into the energy formula: \[ E_{\text{absorbed}} = \frac{(6.626 \times 10^{-34})(3 \times 10^8)}{425 \times 10^{-9}} = 4.67 \times 10^{-19} \text{J} \]
03

Calculate Energy of Emitted Photon

Substitute the wavelength of the emitted photon (\(643 \text{nm} = 643 \times 10^{-9} \text{m}\)) into the energy formula: \[ E_{\text{emitted}} = \frac{(6.626 \times 10^{-34})(3 \times 10^8)}{643 \times 10^{-9}} = 3.09 \times 10^{-19} \text{J} \]
04

Determine Net Energy Absorbed

Calculate the net energy by subtracting the energy of the emitted photon from the energy of the absorbed photon: \[ E_{\text{net}} = E_{\text{absorbed}} - E_{\text{emitted}} = 4.67 \times 10^{-19} \text{J} - 3.09 \times 10^{-19} \text{J} = 1.58 \times 10^{-19} \text{J} \]

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Key Concepts

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

energy of photons
Photons are elementary particles that carry energy. The energy of a photon is directly related to its wavelength or frequency. Understanding this concept is crucial in fields like quantum mechanics and electronics.
The energy of a photon can be calculated using the formula: \( E = \frac{hc}{\text{wavelength}} \), where:
  • \textbf{E} is the energy.
  • \textbf{h} is Planck's constant.
  • \textbf{c} is the speed of light.
  • \textbf{wavelength} is the photon's wavelength.

This relationship shows that shorter wavelengths correspond to higher energy photons, and longer wavelengths correspond to lower energy photons. This formula is indispensable for calculating the precise amount of energy. Knowing the energy a photon carries helps in understanding how photons interact with other particles and materials.
Planck's constant
Planck's constant is a fundamental constant in physics, denoted as \( h \). It plays a vital role in quantum mechanics.
Its approximate value is \( 6.626 \times 10^{-34} \text{Js} \). Planck's constant links the energy of a photon to its frequency. The relation is given by: \( E = h u \), where:
  • \textbf{E} is energy.
  • \textbf{h} is Planck's constant.
  • \textbf{u} (nu) is the frequency.

This relationship is crucial for understanding the particle nature of light and the quantization of energy levels. Not only is Planck's constant used for calculating photon energy, but it also underpins many principles in quantum physics, such as the Heisenberg Uncertainty Principle.
wavelength to energy conversion
Converting a photon's wavelength to its energy is a fundamental task in physics. As seen in the previous section, the formula used is \( E = \frac{hc}{\text{wavelength}} \). This conversion is done because often, we measure the wavelength of incoming light, especially in experiments and practical applications.
Here's how this works step-by-step:
  • First, identify the wavelength of the photon. Convert it to meters if it's not already.
  • Next, use the values of Planck's constant (h) and the speed of light (c).
  • Substitute these values into the formula and solve for E, the energy.

For example, if a photon has a wavelength of 425 nm (which is \( 425 \times 10^{-9} \text{m} \)), substitute into the formula to get the energy: \[ E = \frac{(6.626 \times 10^{-34})(3 \times 10^8)}{425 \times 10^{-9}} = 4.67 \times 10^{-19} \text{J} \] Practical understanding and ability to perform these conversions are essential in fields like spectroscopy, astrophysics, and photonics.

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Most popular questions from this chapter

A certain gamma-ray detector measures photon energies through the Compton interaction: the photon Compton scatters within the detector material, which then absorbs the kinetic energy of the scattered electron. The absorbed energy of the scattered electron is the response of the Suppose photons of energy \(E\) are incident on this detector. (a) Find an expression for the maximum energy response \(E_{\max }\) of this detector and show that \(E_{\max }\) is less than the original energy of the photon. (b) Evaluate \(E_{\max }\) for an incident photon energy of \(1.5 \mathrm{MeV}\). (c) Occasionally the detector may report events with energy greater than \(E_{\max }\) but less than \(E .\) What other processes might be responsible for such events? \((d)\) What processes might contribute to the detector reporting the full energy \(E\) of the photon?

A hydrogen atom is moving at a speed of \(125.0 \mathrm{m} / \mathrm{s}\). It absorbs a photon of wavelength \(97 \mathrm{nm}\) that is moving in the opposite direction. By how much does the speed of the atom change as a result of absorbing the photon?

What range of photon energies corresponds to radio-wave frequencies of 1 to \(100 \mathrm{MHz} ?\) Our bodies are continuously bombarded by these photons. Why are they not dangerous to us?

The WMAP satellite launched in 2001 studied the cosmic microwave background radiation and was able to chart small fluctuations in the temperature of different regions of the background radiation. These fluctuations in temperature correspond to regions of large and small density in the early universe. The satellite was able to measure differences in temperature of \(2 \times 10^{-5} \mathrm{K}\) at a temperature of \(2.7250 \mathrm{K} .\) At the peak wavelength, what is the difference in the radiation intensity per unit wavelength interval between the "hot" and "cold" regions of the background radiation?

(a) Assuming the human body (skin temperature \(34^{\circ} \mathrm{C}\) ) to behave like an ideal thermal radiator, find the wavelength where the intensity from the body is a maximum. In what region of the electromagnetic spectrum is radiation with this wavelength? (b) Making whatever (reasonable) assumptions you may need, estimate the power radiated by a typical person isolated from the surroundings. (c) Estimate the radiation power absorbed by a person in a room in which the temperature is \(20^{\circ} \mathrm{C}\).
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