Problem 25 - Chemistry: The Central Science in SI Units, Global Edition

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 \(\mathrm{nm} .\) (c) What wavelength of radiation has photons of energy \(6.06 \times 10^{-19} \mathrm{~J} ?\)

Step-by-Step Solution

Answer

The energy of a photon with a frequency of \(2.94 \times 10^{14} s^{-1}\) is \(1.95 \times 10^{-19} J\). The energy of a photon with a wavelength of \(413 nm\) is \(4.81 \times 10^{-19} J\). The wavelength of radiation with photon energy of \(6.06 \times 10^{-19} J\) is \(328 nm\).

Step 1: Identify the given parameters

In this case, the given frequency (v) is \(2.94 \times 10^{14} s^{-1}\).

Step 2: Calculate the energy

Use the formula for energy: \(E = h * v\), with Planck's constant (h) equal to \(6.63 \times 10^{-34} J∙s\). Then, the energy is \(E = (6.63 \times 10^{-34} J∙s)(2.94 \times 10^{14} s^{-1}) = 1.95 \times 10^{-19} J\). #b: Calculate the energy of a photon given the wavelength#

Step 1: Identify the given parameters

In this case, the given wavelength (λ) is \(413 nm\). But before using this value, it should be converted to meters: \(λ = 413 × 10^{−9} m\).

Step 2: Calculate the frequency

Use the relationship between speed of light, frequency, and wavelength: \(c = λv\). Rearrange the equation for frequency: \(v = \frac{c}{λ}\). Plug in the speed of light (c) = \(3 \times 10^{8} m/s\) and the wavelength (λ): \(v = \frac{3 \times 10^{8} m/s}{413 × 10^{-9} m} = 7.26 \times 10^{14} s^{-1}\).

Step 3: Calculate the energy

Use the formula for energy: \(E = h \cdot v\). Then, the energy is \(E = (6.63 \times 10^{-34} J∙s)(7.26 × 10^{14} s^{-1}) = 4.81 \times 10^{-19} J\). #c: Calculate the wavelength of radiation given the photon's energy#

Step 1: Identify the given parameters

In this case, the given energy (E) is \(6.06 \times 10^{-19} J\).

Step 2: Calculate the frequency

Use the formula for energy: \(E = h \cdot v\). Rearrange the equation for frequency: \(v = \frac{E}{h}\). Plug in the Planck's constant (h) and the photon's energy (E): \(v = \frac{6.06 \times 10^{-19} J}{6.63 × 10^{-34} J∙s} = 9.14 \times 10^{14} s^{-1}\).

Step 3: Calculate the wavelength

Use the relationship between the speed of light, frequency, and wavelength: \(c = λv\). Rearrange the equation for wavelength: \(λ = \frac{c}{v}\). Plug in the speed of light (c) and the frequency (v): \(λ = \frac{3 \times 10^{8} m/s}{9.14 × 10^{14} s^{-1}} = 3.28 × 10^{-7} m\), or \(328 nm\).

Key Concepts

Electromagnetic Radiation

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, forming what is known as the electromagnetic spectrum.

This includes different types of waves such as radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, and gamma rays. Each type has its typical uses and characteristics. Visible light, for instance, is the light we can see and includes the range of colors from violet to red.

A key point to remember is that electromagnetic waves are composed of oscillating electric and magnetic fields. These waves travel in a vacuum and do not require a medium like sound waves do. The energy carried by electromagnetic radiation can be measured in photons, which are the smallest quantum units of light.

Understanding the nature of electromagnetic radiation is fundamental in different scientific fields, from physics to medicine, as each type interacts with matter in unique ways.

Planck's Constant

Planck's constant is a crucial quantity in quantum mechanics, denoted by the symbol \(h\). It has a very small value of approximately \(6.63 \times 10^{-34} \, \text{J⋅s}\) and fundamentally relates the energy of a photon to the frequency of its associated electromagnetic wave.

The significance of Planck's constant lies in its role in the formula \(E = h \cdot v\), which states that the energy \(E\) of a photon is equal to Planck's constant times the frequency \(v\) of the wave. This formula highlights that energy is quantized and can only exist in discrete amounts in the context of electromagnetic radiation.

Planck's constant is at the heart of quantum theory, where it helps to explain phenomena that classical physics could not, such as the photoelectric effect and atomic spectra. Its introduction was a revolutionary shift in physics, offering insights into the dual nature of light as both a particle and a wave.

Wavelength and Frequency Relationship

The relationship between wavelength and frequency is a fundamental concept in the study of waves, particularly electromagnetic waves. It is given by the equation \(c = \lambda v\), where \(c\) is the speed of light, \(\lambda\) is the wavelength, and \(v\) is the frequency.

This equation illustrates that wavelength and frequency are inversely proportional to each other.

• When the frequency increases, the wavelength decreases, provided the speed of light remains constant.
• Conversely, a decrease in frequency results in an increase in wavelength.

The speed of light \(c\) is a constant, approximately \(3 \times 10^8 \, \text{m/s}\), which means all electromagnetic waves travel at this speed in a vacuum.

Understanding this relationship is crucial for many applications, such as calculating the energy of photons or determining the characteristics of different types of electromagnetic radiation.

These concepts allow scientists to manipulate and harness various wavelengths and frequencies for diverse purposes, ranging from communication technologies to medical imaging.