RGB Color Television And Computer Displays Use Cathode Ray T ✓ Solved

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Identify the approximate wavelengths of red, green, and blue light in the visible spectrum, and calculate the corresponding frequency and photon energy for each color. Additionally, write suitable net ionic equations demonstrating how each specified species can act as a Brønsted-Lowry base.

Paper For Above Instructions

Introduction

Color displays in television and computer screens utilize red, green, and blue light to create a broad spectrum of colors through the mixing of these primary colors. This process involves understanding the physical properties of light, such as wavelength, frequency, and photon energy, as well as chemical principles relevant to acid-base reactions in aqueous solutions.

Wavelengths of Red, Green, and Blue Light

In the visible spectrum, different colors correspond to specific wavelength ranges. According to standard spectral data:

  • Red light has a wavelength approximately between 620–750 nm.
  • Green light has a wavelength approximately between 495–570 nm.
  • Blue light has a wavelength approximately between 450–495 nm.

For simplicity and approximation, we take:

  • Red: 700 nm
  • Green: 530 nm
  • Blue: 470 nm

Calculating Frequency and Photon Energy

Fundamental Equations

The relationship between wavelength (\(\lambda\)), frequency (\(f\)), and the speed of light (\(c\)) is given by:

f = c / λ

Where:

  • \(c \approx 3.00 \times 10^8\, \text{m/s}\)
  • \(\lambda\) is the wavelength in meters.

The energy of a photon (\(E\)) is given by:

E = h \times f

Where:

  • \(h \approx 6.626 \times 10^{-34}\, \text{J·s}\)

Calculations for Each Color

Red Light

Wavelength: \(\lambda = 700\, \text{nm} = 700 \times 10^{-9}\, \text{m}\)

f = 3.00 \times 10^8\, \text{m/s} / 700 \times 10^{-9}\, \text{m} \approx 4.29 \times 10^{14}\, \text{Hz}
E = 6.626 \times 10^{-34}\, \text{Js} \times 4.29 \times 10^{14}\, \text{Hz} \approx 2.84 \times 10^{-19}\, \text{J}

Green Light

Wavelength: \(\lambda = 530\, \text{nm} = 530 \times 10^{-9}\, \text{m}\)

f = 3.00 \times 10^8\, \text{m/s} / 530 \times 10^{-9}\, \text{m} \approx 5.66 \times 10^{14}\, \text{Hz}
E = 6.626 \times 10^{-34}\, \text{Js} \times 5.66 \times 10^{14}\, \text{Hz} \approx 3.75 \times 10^{-19}\, \text{J}

Blue Light

Wavelength: \(\lambda = 470\, \text{nm} = 470 \times 10^{-9}\, \text{m}\)

f = 3.00 \times 10^8\, \text{m/s} / 470 \times 10^{-9}\, \text{m} \approx 6.38 \times 10^{14}\, \text{Hz}
E = 6.626 \times 10^{-34}\, \text{Js} \times 6.38 \times 10^{14}\, \text{Hz} \approx 4.23 \times 10^{-19}\, \text{J}

Brønsted-Lowry Base Behavior of Species

The Brønsted-Lowry theory defines a base as a species that can accept a proton (\(H^+\)). For each of the given species, we write the net ionic equations demonstrating their ability to act as bases.

(a) H₂O (Water)

Water can accept a proton to form hydronium:

H₂O + H⁺ → H₃O⁺

(b) OH⁻ (Hydroxide Ion)

The hydroxide ion already carries a lone pair and readily accepts a proton:

OH⁻ + H⁺ → H₂O

(c) NH₃ (Ammonia)

Ammonia acts as a base by accepting a proton to form ammonium:

NH₃ + H⁺ → NH₄⁺

(d) CN⁻ (Cyanide Ion)

The cyanide ion can accept a proton to form hydrogen cyanide:

CN⁻ + H⁺ → HCN

(e) S²⁻ (Sulfide Ion)

Sulfide can accept two protons to form hydrogen sulfide:

S²⁻ + 2H⁺ → H₂S

(f) H₂PO₄⁻ (Dihydrogen phosphate ion)

This species can accept a proton to form H₃PO₄:

H₂PO₄⁻ + H⁺ → H₃PO₄

Conclusion

Understanding the physical properties of light used in color displays allows for precise calculations of wavelength, frequency, and energy, which are fundamental to optics and photonics. Simultaneously, knowledge of acid-base chemistry and the Brønsted-Lowry theory provides insight into the behavior of various chemical species as proton acceptors, illustrating their roles as bases in aqueous solutions.

References

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