Problem 53 - Chemistry Matter and Change

Question

A photon has an energy of \(1.10 \times 10^{-13}\) J. What is the photon's wavelength? What type of electromagnetic radiation is it?

Step-by-Step Solution

Answer

The wavelength of the photon with an energy of \(1.10 \times 10^{-13} J\) is \(1.81 \times 10^{-7} m\), and it belongs to the Ultraviolet radiation spectrum.

Step 1: Understanding the Planck's equation

We will use the Planck's equation, which relates the energy (E) of a photon to its wavelength (λ). The equation is given by: \(E = h \cdot f\) where, - E is the energy of a photon - h is the Planck's constant (\(6.63 \times 10^{-34} Js\)) - f is the frequency of the electromagnetic radiation Additionally, we know the speed of light (c) can be expressed as: \(c = \lambda \cdot f\) Our goal is to find the wavelength (λ), so we will combine these two equations, solve for the wavelength, and substitute the given values to find the answer.

Step 2: Combining equations and solving for the wavelength

First, let's express the frequency (f) from the second equation: \(f = \frac{c}{\lambda}\) Now, substitute this in the Planck's equation to get: \(E = h \cdot \frac{c}{\lambda}\) Now, solve for the wavelength (λ): \(\lambda = \frac{h \cdot c}{E}\)

Step 3: Substituting given values and calculating the wavelength

Now, substitute the given energy (E), Planck's constant (h), and speed of light (c) into the equation: \(\lambda = \frac{(6.63 \times 10^{-34} Js)(3.00 \times 10^{8} m/s)}{1.10 \times 10^{-13} J}\) Calculate the value: \(\lambda = 1.81 \times 10^{-7} m\) So, the wavelength of the photon is \(1.81 \times 10^{-7} m\).

Step 4: Identifying the type of electromagnetic radiation

Now that we have the wavelength, we can classify the photon according to the electromagnetic radiation spectrum. Refer to the following classification: - Radio waves: \(> 10^{-1} m\) - Microwaves: \(10^{-1} m\) to \(10^{-3} m\) - Infrared: \(10^{-3} m\) to \(7 \times 10^{-7} m\) - Visible light: \(7 \times 10^{-7} m\) to \(4 \times 10^{-7} m\) - Ultraviolet: \(4 \times 10^{-7} m\) to \(10^{-8} m\) - X-rays: \(10^{-8} m\) to \(10^{-11} m\) - Gamma rays: \< \(10^{-11} m\) The given wavelength falls into the Ultraviolet range (since \(1.81 \times 10^{-7} m\), lies between \(4 \times 10^{-7} m\) and \(10^{-8} m\)) Thus, the photon with an energy of \(1.10 \times 10^{-13} J\) has a wavelength of \(1.81 \times 10^{-7} m\), and it belongs to the Ultraviolet radiation spectrum.

Key Concepts

Exploring Photon Energy

Photon energy is a pivotal concept in physics, especially in the realm of quantum mechanics. When we delve into the microscopic world, we encounter particles such as photons - the fundamental particles of light. The energy carried by a photon is directly linked to the electromagnetic radiation it represents.

The quantization of energy in photons was a groundbreaking discovery by Max Planck, which led to the development of Planck's equation. According to this principle, the energy (\(E\)) of a single photon is directly proportional to its frequency (\(f\)), and this relationship hinges on a fundamental constant of nature - the Planck's constant (\(h\) - roughly 6.63 x 10^-34 J·s). In formulaic terms, this equation is expressed simply as \(E = h \times f\).

Since the frequency of a photon is tied to the color of light in the visible spectrum, the higher the frequency, the more energy a photon carries. For instance, the energy of a blue photon is higher than that of a red photon due to its higher frequency. In the ultraviolet range, photons have even more energy, which is why they can cause sunburn by breaking chemical bonds in our skin. Understanding photon energy is not only fundamental in physics but also in applications such as medical imaging, where high-energy photons are used in X-rays.

Unveiling Electromagnetic Radiation

Electromagnetic radiation is a form of energy that spreads out as it travels through space, and it encompasses a broad spectrum, from gamma rays, with the shortest wavelengths, to radio waves, with the longest wavelengths. This radiation is composed of photons, which are massless particles that travel at the speed of light (\(c\) - approximately 3.00 x 10⁸ m/s) and carry energy.

Each category of electromagnetic radiation is defined by its wavelength (\(λ\)) or frequency (\(f\)), with gamma rays being the most energetic and radio waves the least. The broad spectrum includes, in order of decreasing energy: gamma rays, X-rays, ultraviolet (UV) light, visible light, infrared radiation, microwaves, and radio waves.

Furthermore, electromagnetic radiation is not only fundamental in understanding astronomical phenomena but also in daily technologies. Radio waves are used for broadcasting signals, microwaves in ovens, and infrared in remote controls. Each type of radiation holds unique properties and applications, reflecting the diversity and utility of electromagnetic waves in various fields.

Wavelength Calculation

Wavelength calculation is a critical skill in physics that allows us to understand the properties of electromagnetic waves. Wavelength (\(λ\)) is the distance between two consecutive crests or troughs in a wave. By knowing the energy of a photon, as given in an exercise, we can calculate its wavelength through a rearranged Planck's equation.

The first step involves recognizing the relationship between a photon's energy (\(E\)), its frequency (\(f\)), and the speed of light (\(c\)). We use the fact that \(c = λ \times f\) to express frequency in terms of wavelength. Substituting that into Planck's equation, we get \(E = h \times (c / λ)\), which we rearrange to solve for wavelength as \(λ = h \times c / E\).

In the context of the given exercise, knowing the energy of a photon enables us to perform the calculation to find its wavelength, revealing the type of electromagnetic radiation it belongs to. Through wavelength calculation, we gain insights into the frequency, energy, and practical applications of different forms of radiation, whether it's for communication, medicine, or exploring the universe.