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(a) A complex absorbs photons with an energy of \(4.51 \times 10^{-19}\). What is the wavelength of these photons? (b) If this is the only place in the visible spectrum where the complex absorbs light, what color would you expect the complex to be?

Short Answer

Expert verified
The wavelength of the photons absorbed by the complex is 441 nm, which falls within the violet range. Since the complex absorbs violet light, the color we perceive is the complementary color to violet, which is yellow. Therefore, the complex would appear yellow in color.

Step by step solution

01

Calculate the wavelength of the photon

To find the wavelength, we can rearrange the formula as follows: \( \lambda = h \cdot c / E \) Now, plug in the values for h, c and E: \( \lambda = (6.626 \times 10^{-34} \text{Js}) (3.00 \times 10^8 \text{m/s}) / (4.51 \times 10^{-19} \text{J}) \) Calculate the value of λ: \( \lambda = 4.41 \times 10^{-7} \text{m} \) Since the answer should be in nanometers (nm), convert the value from meters to nanometers: \( \lambda = 4.41 \times 10^{-7} \text{m} \times \frac{10^9 \text{nm}}{1 \text{m}} \) \( \lambda = 441\,\text{nm} \) The wavelength of the photons absorbed by the complex is 441 nm.
02

Determine the color of the complex

Now that we have the wavelength of the absorbed photons, we can determine the color of the complex. The visible color spectrum is approximately as follows: - Violet: 380 - 450 nm - Blue: 450 - 495 nm - Green: 495 - 570 nm - Yellow: 570 - 590 nm - Orange: 590 - 620 nm - Red: 620 - 750 nm Since the complex absorbs photons with a wavelength of 441 nm, which falls within the violet range, it absorbs violet light. The color we perceive is the color that is not absorbed. In this case, the complex would appear as a complementary color to violet, which is yellow. Therefore, we would expect the complex to be yellow in color.

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Key Concepts

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

Understanding the Photoelectric Effect
The photoelectric effect is a phenomenon in which electrons are emitted from a material when it absorbs light or other electromagnetic radiation. This effect was explained by Albert Einstein and contributed to the foundation of quantum physics. The concept is pivotal in understanding how light interacts with matter on a quantum level.

When light, viewed as a stream of particles called photons, hits a material, each of these photons can transfer energy to an electron if the photon's energy is at least equal to the work function of the material. The work function is the minimum energy required to remove an electron from the surface of the substance. If the photon's energy exceeds this value, the extra energy translates into the kinetic energy of the emitted electron. This directly ties into the exercise where energy and wavelength of photons are related through the equation E = hc/λ, which is the key to calculating wavelength in the given problem.
The Visible Spectrum Demystified
The visible spectrum encompasses the range of wavelengths of electromagnetic radiation that is detectable by the human eye. This range extends approximately from 380 nm (nanometers) to 750 nm. Each wavelength within this spectrum corresponds to a different color that we can observe.

  • Violet light has the shortest wavelength, typically between 380 and 450 nm.
  • Indigo and blue follow, with wavelengths from 450 to 495 nm.
  • Green falls in the middle at 495 to 570 nm.
  • Yellow ranges from 570 to 590 nm.
  • Orange encompasses 590 to 620 nm.
  • Red has the longest wavelength at 620 to 750 nm.
This is important when determining the expected color outcome based on absorbed or emitted light, as demonstrated in the solution to the exercise.
Complementary Colors in Optics
Complementary colors are pairs of colors that, when combined, cancel each other out to produce a white or neutral color. These are located directly opposite each other on a color wheel. Applying this concept in the context of absorbed and emitted light is essential for predicting the resulting color that an object may appear.

For example, when an object absorbs light of a certain color, it typically reflects or transmits the complementary color. In the exercise, since the complex absorbs violet light (441 nm), the complementary color, yellow, would be what the complex appears to be. This understanding of light absorption and perception allows us to predict the expected visual characteristics of an object.
Energy of Photons
The energy of photons, the quantum of light, is directly proportional to its frequency and inversely proportional to its wavelength. This relationship is described by the equation E = hc/λ, where E represents the energy, h is Planck's constant, c is the speed of light, and λ is the wavelength.

In practice, this means that photons with shorter wavelengths (and thus higher frequencies) carry more energy, while those with longer wavelengths have less energy. As seen in the exercise, by rearranging the equation to solve for wavelength, we can determine the characteristics of light absorbed by a substance based on the energy of photons it absorbs.

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Most popular questions from this chapter

The complexes \(\left[\mathrm{V}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}\) and \(\left[\mathrm{VF}_{6}\right]^{3-}\) are both known. (a) Draw the \(d\)-orbital energy-level diagram for V(III) octahedral complexes. (b) What gives rise to the colors of these complexes? (c) Which of the two complexes would you expect to absorb light of higher energy?

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The total concentration of \(\mathrm{Ca}^{2+}\) and \(\mathrm{Mg}^{2+}\) in a sample of hard water was determined by titrating a \(0.100\) - L sample of the water with a solution of EDTA \({ }^{4-}\). The EDTA \({ }^{4-}\) chelates the two cations: $$ \begin{aligned} \mathrm{Mg}^{2+}+[\mathrm{EDTA}]^{4-} & \longrightarrow[\mathrm{Mg}(\mathrm{EDTA})]^{2-} \\ \mathrm{Ca}^{2+}+\left[\mathrm{EDTA}^{4-}\right.& \longrightarrow[\mathrm{Ca}(\mathrm{EDTA})]^{2-} \end{aligned} $$ It requires \(31.5 \mathrm{~mL}\) of \(0.0104 \mathrm{M}[\mathrm{EDTA}]^{4-}\) solution to reach the end point in the titration. A second \(0.100-L\) sample was then treated with sulfate ion to precipitate \(\mathrm{Ca}^{2+}\) as calcium sulfate. The \(\mathrm{Mg}^{2+}\) was then titrated with \(18.7 \mathrm{~mL}\) of \(0.0104 \mathrm{M}\) [EDTA] ]- Calculate the concentrations of \(\mathrm{Mg}^{2+}\) and \(\mathrm{Ca}^{2+}\) in the hard water in mg/I.

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