Problem 48 - Chemistry The Central Science

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 approximately \(441 \mathrm{nm}\), which falls within the violet range of the visible light spectrum. Therefore, we would expect the complex to appear violet in color.

Step 1: Identify the given information

We are given the energy of the absorbed photons as \(E = 4.51 \times 10^{-19} \mathrm{J}\)

Step 2: Write down the energy-wavelength relationship

The formula to relate energy and wavelength is given by: \(E = \dfrac{hc}{\lambda}\) where h is the Planck's constant (\( h \approx 6.63 \times 10^{-34} \mathrm{Js}\)), c is the speed of light (\( c \approx 3 \times 10^{8} \mathrm{m/s}\)), and λ is the wavelength of the photons.

Step 3: Solve for the wavelength λ

We can isolate λ as follows: \[\lambda = \dfrac{hc}{E}\] Now, plug in the given values of E, h, and c: \[\lambda = \dfrac{(6.63 \times 10^{-34} \mathrm{Js})(3 \times 10^{8} \mathrm{m/s})}{4.51 \times 10^{-19} \mathrm{J}}\]

Step 4: Calculate the wavelength λ

After plugging in the values and solving for λ, we get: \[\lambda \approx 4.41 \times 10^{-7} \mathrm{m}\] or \(λ \approx 441 \mathrm{nm}\)

Step 5: Determine the color of the complex

Using the calculated wavelength of \(441 \mathrm{nm}\), we can find the color corresponding to this wavelength. The visible light spectrum ranges from approximately 400 nm to 700 nm, with colors arranged as follows: violet: (380–450 nm), blue: (450-495 nm), green: (495–570 nm), yellow: (570–590 nm), orange: (590-620 nm), and red: (620–750 nm). The calculated wavelength of \(441 \mathrm{nm}\) falls in the violet range. So, we would expect the complex to appear violet in color.

Key Concepts

Planck's Constant

Planck's constant is a fundamental constant in physics, denoted as \( h \), which plays a crucial role in quantum mechanics. Its approximate value is \( 6.63 \times 10^{-34} \) Joule-seconds (Js). Planck's constant is a pivotal component in the formulation of quantum theory.

It relates the energy of a photon to the frequency of its electromagnetic wave. This relationship is expressed in the Planck-Einstein relation:

• \( E = h u \)
• Where \( E \) is the energy of the photon and \( u \) is its frequency.

This fundamental relationship implies that energy is quantized, meaning it exists in discrete "packets" called quanta, or in this case, photons.

Planck's constant is essential in understanding the behavior of particles at atomic and subatomic levels, revolutionizing our comprehension of the microscopic world.

Visible Light Spectrum

The visible light spectrum is a segment of the electromagnetic spectrum that the human eye can perceive. It ranges approximately from 380 nanometers (nm) to 750 nm in wavelength.

Each color within this spectrum corresponds to a specific range of wavelengths:

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

The color we perceive in an object is due to the wavelengths of light that are not absorbed but instead reflected or transmitted by the object. For example, if an object absorbs light primarily within the violet range (such as at 441 nm, like in the given exercise), we perceive it as violet.

Photon Energy Calculation

Photon energy calculation involves determining the energy carried by a photon using its wavelength or frequency. The formula used is:
\[E = \frac{hc}{\lambda}\]Where:

• \( E \) is the energy of the photon.
• \( h \) is Planck's constant (\( 6.63 \times 10^{-34} \) Js).
• \( c \) is the speed of light (\( 3 \times 10^{8} \) meters per second).
• \( \lambda \) is the wavelength of the photon in meters.

This relationship highlights the inverse correlation between wavelength and energy: the shorter the wavelength, the higher the energy.

In practical situations, photon energy calculations are crucial for understanding spectroscopy, especially in identifying elements and compounds based on their light absorption or emission at specific wavelengths.