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Chapter 2: Problem 148

Photosynthesis uses 660 -nm light to convert \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) into glucose and \(\mathrm{O}_{2}\). Calculate the frequency of this light.

Short Answer

Expert verified
The frequency of the 660-nm light used in photosynthesis is approximately \(4.55 \times 10^{14}\,\mathrm{Hz}\).

Step by step solution

01

Convert the Wavelength to Meters

The given wavelength of light is in nanometers, and we need to convert it into meters to be compatible with the speed of light in meters per second. To do so, use the following conversion factor: \(1\,\mathrm{m} = 10^9\,\mathrm{nm}\). So, the wavelength in meters can be calculated as: \(\lambda_\mathrm{meters} = \cfrac{660\,\mathrm{nm}}{10^9\,\mathrm{nm/m}}\).
02

Calculate the Frequency of Light

Now that we have the wavelength in meters, we can apply the formula mentioned above to determine the frequency of the light used in photosynthesis. \(f = \cfrac{c}{\lambda_\mathrm{meters}}\), Substitute the values: \(f = \cfrac{3.0 \times 10^8\,\mathrm{m/s}}{660\,\mathrm{nm} \times 10^{-9}\,\mathrm{m/nm}}\).
03

Calculate the Final Value

Perform the calculations to obtain the frequency of light: \(f = \cfrac{3.0 \times 10^8}{660 \times 10^{-9}}\), \(f = \cfrac{3.0 \times 10^8}{6.6 \times 10^{-7}}\), \(f = 4.55 \times 10^{14}\,\mathrm{Hz}\). The frequency of the 660-nm light used in photosynthesis is approximately \(4.55 \times 10^{14}\,\mathrm{Hz}\).

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Key Concepts

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

Wavelength and Frequency Calculation
In photosynthesis, light energy plays a crucial role, specifically the energy from 660-nm light. Understanding how to calculate the frequency of light is vital for various scientific analyses. Let's explore how we can do this.

The relationship between wavelength and frequency is expressed by the formula:
  • \( f = \cfrac{c}{\lambda} \)
where:
  • \( f \) is the frequency
  • \( c \) is the speed of light, approximately \( 3.0 \times 10^8 \,\mathrm{m/s} \)
  • \( \lambda \) is the wavelength in meters
To perform this calculation, converting the wavelength from nanometers to meters is necessary, using the conversion factor:
  • \( 1\,\mathrm{m} = 10^9\,\mathrm{nm} \)
With this conversion, substituting the values into the frequency formula yields the frequency from the given wavelength. Calculations result in a frequency of approximately \(4.55 \times 10^{14}\,\mathrm{Hz}\).

This frequency is significant in photosynthesis as it aligns with the energy requirements of the process.
Light Energy in Biological Processes
Photosynthesis is a fundamental biological process where plants convert light energy into chemical energy. This process relies heavily on light that falls within a specific range of wavelengths.

For plants, light in the range around 660 nm contributes to the energy required to transform water and carbon dioxide into glucose. The energy from light facilitates this transformation by exciting electrons, enabling various chemical reactions within the chlorophyll, which acts as a catalyst for energy capture.

This knowledge is not only crucial for understanding plant biology but also provides insights into optimizing agricultural practices. By ensuring adequate exposure to specific wavelengths of light, it can enhance plant growth and photosynthetic efficiency.
Conversion of Units in Scientific Calculations
Conversions are pivotal in scientific computations, especially when working with dimensions like length, mass, or time. Understanding how to convert units ensures accurate calculations and meaningful results.

In our wavelength and frequency calculation, converting the wavelength from nanometers to meters was crucial. Scientific equations, such as the speed of light equation, typically operate in the International System of Units (SI), where the meter is the standard unit for length.

To convert from nanometers to meters, remember that:
  • \(1\,\mathrm{m} = 10^9\,\mathrm{nm} \)
This conversion allows seamless integration into formulas using SI units. Correctly applying conversions maintains consistency and precision in solving scientific questions. It is a fundamental skill that extends beyond physics into many areas of science and engineering.

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Most popular questions from this chapter

An electron is excited from the \(n=1\) ground state to the \(n=\) 3 state in a hydrogen atom. Which of the following statements is/are true? Correct the false statements to make them true. a. It takes more energy to ionize (completely remove) the electron from \(n=3\) than from the ground state. b. The electron is farther from the nucleus on average in the \(n=3\) state than in the \(n=1\) state. c. The wavelength of light emitted if the electron drops from \(n=3\) to \(n=2\) will be shorter than the wavelength of light emitted if the electron falls from \(n=3\) to \(n=1\) d. The wavelcngth of light cmittcd when the clectron returns to the ground state from \(n=3\) will be the same as the wavelength of light absorbed to go from \(n=1\) to \(n=3\) e. For \(n=3,\) the electron is in the first excited state.

Photogray lenses incorporate small amounts of silver chloride in the glass of the lens. When light hits the AgCl particles, the following reaction occurs: $$ \operatorname{AgCl} \stackrel{h v}{\longrightarrow} \mathrm{Ag}+\mathrm{Cl} $$ The silver metal that is formed causes the lenses to darken. The energy change for this reaction is \(3.10 \times 10^{2} \mathrm{kJ} / \mathrm{mol} .\) Assuming all this energy must be supplied by light, what is the maximum wavelength of light that can cause this reaction?

It takes \(208.4 \mathrm{kJ}\) of energy to remove 1 mole of electrons from an atom on the surface of rubidium metal. (1 mol electrons = \(6.022 \times 10^{23}\) electrons. How much energy does it take to remove a single electron from an atom on the surface of solid rubidium? What is the maximum wavelength of light capable of doing this?

Give the maximum number of electrons in an atom that can have these quantum numbers: a. \(n=0, \ell=0, m_{\ell}=0\) b. \(n=2, \ell=1, m_{\ell}=-1, m_{s}=-\frac{1}{2}\) c. \(n=3, m_{s}=+\frac{1}{2}\) d. \(n=2, \ell=2\) e. \(n=1, \ell=0, m_{\ell}=0\)

Ionization energy is the energy required to remove an electron from an atom in the gas phase. The ionization energy of gold is \(890.1 \mathrm{kJ} / \mathrm{mol} .\) Is light with a wavelength of \(225 \mathrm{nm}\) capable of ionizing a gold atom (removing an electron) in the gas phase? ( 1 mol gold \(=6.022 \times 10^{23}\) atoms gold.)
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