Problem 25 - Chemistry The Central Science

Question

(a) Calculate the energy of a photon of electromagnetic radiation whose frequency is \(2.94 \times 10^{14} \mathrm{s}^{-1} .\) (b) Calculate the energy of a photon of radiation whose wavelength is 413 nm. (c) What wavelength of radiation has photons of energy \(6.06 \times 10^{-19} \mathrm{J} ?\)

Step-by-Step Solution

Answer

(a) The energy of a photon with frequency \(2.94 \times 10^{14} \mathrm{s}^{-1}\) is \(1.95 \times 10^{-19} \mathrm{J}\). (b) The energy of a photon with a wavelength of 413 nm is \(4.82 \times 10^{-19} \mathrm{J}\). (c) The wavelength of radiation with photons of energy \(6.06 \times 10^{-19} \mathrm{J}\) is 328 nm.

Step 1: (a) Calculate the energy of a photon

We are given the frequency of the electromagnetic radiation: f = 2.94 × 10^14 s^-1. We will use the formula E = h * f to calculate the energy of a photon: E = (6.63 × 10^-34 Js) * (2.94 × 10^14 s^-1) = 1.95 × 10^-19 J The energy of a photon is 1.95 × 10^-19 J.

Step 2: (b) Calculate the energy of a photon with a known wavelength

We are given the wavelength of the radiation, λ = 413 nm. In order to use the formula E = h * (c / λ), we need to convert the wavelength to meters: λ = 413 × 10^-9 m. Now, we can calculate the energy of a photon using the formula: E = (6.63 × 10^-34 Js) * (3 × 10^8 m/s) / (413 × 10^-9 m) = 4.82 × 10^-19 J The energy of a photon with a wavelength of 413 nm is 4.82 × 10^-19 J.

Step 3: (c) Find the wavelength of radiation with a photon energy

We are given the energy of photons: E = 6.06 × 10^-19 J. We will use the formula E = h * (c / λ) to find the wavelength of radiation: 6.06 × 10^-19 J = (6.63 × 10^-34 Js) * (3 × 10^8 m/s) / λ Now, we need to solve for λ: λ = (6.63 × 10^-34 Js) * (3 × 10^8 m/s) / (6.06 × 10^-19 J) = 3.28 × 10^-7 m We can convert the wavelength back to nanometers: λ = 328 nm. The wavelength of radiation with photons of energy 6.06 × 10^-19 J is 328 nm.

Key Concepts

Electromagnetic Radiation Frequency

Electromagnetic radiation is a form of energy that travels through space at the speed of light. It encompasses a wide range of wavelengths and frequencies, including visible light, microwaves, X-rays, and more. Frequency, in this context, refers to how many wave cycles pass a specific point per second. It is measured in Hertz (Hz), or in scientific terms, seconds inverse (s⁻¹). The higher the frequency, the more energy the electromagnetic wave possesses. When solving problems related to electromagnetic radiation, one key formula to remember is the relationship between energy and frequency:

• Energy (E) = Planck’s constant (h) multiplied by frequency (f).

For example, in calculating the energy of a photon with a frequency of \(2.94 \times 10^{14} s^{-1}\), this formula directly applies as seen in the solution provided.Understanding frequency is crucial when analyzing electromagnetic spectra because it helps determine how electromagnetic waves interact with materials, affecting everything from electronic devices to biological systems.

Wavelength-to-Energy Conversion

Wavelength and energy are inversely related in the study of electromagnetic radiation. This means that as the wavelength of a photon (a particle of light) decreases, its energy increases. Wavelength is the distance between consecutive peaks of a wave and is typically measured in meters or nanometers (nm).To convert wavelength into energy, the equation used is:

• E = h \( \times \frac{c}{\lambda} \), where c is the speed of light and \lambda is the wavelength.

Applying this formula allows you to find the energy of a photon when given the wavelength, as seen when solving for a wavelength of 413 nm in the example solution.It’s essential to convert the wavelength into meters before performing calculations, as the speed of light is usually given in meters per second. This ensures that the units are consistent for accurate calculations. Understanding this conversion process is vital for fields like spectroscopy, which relies on analyzing light properties.

Planck's Constant

Planck's constant (h) plays a fundamental role in quantum mechanics and physics. It is a proportionality constant used to describe the sizes of quanta, the smallest possible discrete units of energy. The value of Planck's constant is approximately \(6.63 \times 10^{-34} Js\).This constant is crucial when working with equations involving energy and frequency, as seen in the formula E = h \( \times f \). Planck essentially revolutionized physics by introducing this concept, explaining how energy is quantized and leading to the development of quantum theory.Planck’s constant demonstrates that energy changes occur in discrete steps at the quantum level. This insight is applied in various scientific disciplines, from quantum computing to the study of atomic and molecular structures.Understanding the significance and use of Planck's constant is key to grasping many principles of modern physics. It helps explain phenomena that classical physics could not, and remains a cornerstone in the fields of both theoretical and applied physics.