Problem 48 - Chemistry: The Central Science in SI Units, Global Edition

Question

(a) A complex absorbs photons with an energy of \(4.51 \times 10^{-19} \mathrm{~J}\). 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?

Step-by-Step Solution

Answer

The wavelength of the absorbed photons is \(439\, nm\), which falls in the violet range. Since the complex absorbs violet light, it will appear as the complementary color, which is yellow-green.

Step 1: (a) Calculate the wavelength of absorbed photons

We have the energy of the absorbed photons, which is: \[E = 4.51 \times 10^{-19} J\] Now, we can use the equation: \[E = h \cdot \nu\] where \(h = 6.626 \times 10^{-34} Js\) is the Planck's constant, and \(\nu\) is the frequency. We can rewrite the frequency in terms of the wavelength using the relation: \[\nu = \frac{c}{\lambda}\] where \(c = 2.998 \times 10^8 m/s\) is the speed of light. Let's substitute this equation into the energy equation: \[E = h\cdot \frac{c}{\lambda}\] Now, we can solve for the wavelength \(\lambda\): \[\lambda = \frac{h\cdot c}{E}\] Plugging in the given values for \(E\), \(h\), and \(c\), we get: \[\lambda = \frac{(6.626 \times 10^{-34} Js)(2.998 \times 10^8 m/s)}{4.51 \times 10^{-19} J}\] Calculate the wavelength: \[\lambda = 4.39 \times 10^{-7} m\] Converting to nanometers (1 m = \(10^9\) nm): \[\lambda = 439 \, nm\] So, the wavelength of the absorbed photons is \(439\, nm\).

Step 2: (b) Determine the color of the complex

Now that we know the complex absorbs light at a wavelength of 439 nm, we can determine the color it appears to our eyes. Here's a general range of wavelengths for different colors in the visible light spectrum: - Violet: 380 - 450 nm - Blue: 450 - 495 nm - Green: 495 - 570 nm - Yellow: 570 - 590 nm - Orange: 590 - 620 nm - Red: 620 - 750 nm Given that the complex absorbs light at 439 nm, it falls in the violet range. When a substance absorbs light of a certain color, it usually appears as the complementary color. The complementary color of violet is yellow-green. Thus, we can expect the complex to appear yellow-green.

Key Concepts

Visible Light Spectrum

The visible light spectrum is the portion of the electromagnetic spectrum that is visible to the human eye. This spectrum encompasses wavelengths from approximately 380 nm to 750 nm. Within this range, light varies from violet, which has the shortest wavelength, to red, which has the longest.

Each color in the visible spectrum corresponds to a different wavelength band. Here are some key points to know regarding these colors:

• Violet: 380 - 450 nm
• Blue: 450 - 495 nm
• Green: 495 - 570 nm
• Yellow: 570 - 590 nm
• Orange: 590 - 620 nm
• Red: 620 - 750 nm

Understanding the visible light spectrum helps us determine the perceived color of objects when they absorb specific wavelengths and reflect others. For instance, an object that absorbs light at 439 nm—which is in the violet range—will reflect its complementary color, appearing as yellow-green to our eyes.

Planck's Constant

Planck's Constant is a fundamental constant in physics, denoted by the symbol \(h\). It is a key element in quantum mechanics and relates to the quantization of energy. Planck's constant has a value of \(6.626 \times 10^{-34} \ ext{Js}\).

In the context of photon wavelength calculation, Planck's constant is crucial because it is part of the equation that relates the energy of a photon to its frequency (\(u\)), which is:\[E = h \cdot u\]This implies that the energy of a photon is quantized, meaning it can only take on discrete values rather than any value. This was a groundbreaking discovery in physics, leading to the development of quantum theory.

Planck's Constant allows us to calculate wavelengths from energy, especially useful in determining the characteristics of light absorbed or emitted by different substances. Its application in equations ensures accurate representations of physical phenomena at microscopic levels.

Energy Frequency Relation

The energy frequency relation is a concept that describes the direct relationship between the energy of a photon and its frequency. This relation is given by the equation:\[E = h \cdot u\]where \(E\) is the energy of the photon in joules, \(h\) is Planck's constant, and \(u\) is the frequency in Hertz (Hz).

This equation shows that as the frequency of light increases, so does the energy of the photons. Conversely, lower frequency means lower energy photons. This relation is vital in understanding how electromagnetic radiation interacts with matter.

• Higher frequencies (e.g., ultraviolet light) have higher energy photons, which can cause chemical reactions.
• Lower frequencies (e.g., infrared light) have less energy, often not enough to cause such reactions.

Connecting frequency with wavelength, we use the speed of light \(c\), as in \(c = \lambda \cdot u\), where \(\lambda\) is the wavelength. This allows us to express frequency as:\[u = \frac{c}{\lambda}\]Substituting into the energy equation, we derive:\[E = h \cdot \frac{c}{\lambda}\]This relation is essential for calculating the energy and thus the characteristics of photons across the electromagnetic spectrum.