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

Step 1: (a) Calculate the energy of a photon with given frequency

To calculate the energy of the photon, use the Planck's formula: E = hν. Given frequency, ν = 2.94 x 10^14 s⁻¹ Planck's constant, h = 6.626 x 10⁻³⁴ Js So, E = (6.626 x 10^-34 Js) * (2.94 x 10^14 s⁻¹) E ≈ 1.95 x 10^-19 J Thus, the energy of the photon is 1.95 x 10^-19 J.

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

To calculate the energy of the photon, we first need to find its frequency (ν) using the speed of light formula: c = λν. Given wavelength, λ = 413 nm = 413 x 10⁻⁹ m Speed of light, c = 3.0 x 10^8 m/s Now, rearrange the formula for frequency: ν = c / λ ν = (3.0 x 10^8 m/s) / (413 x 10⁻⁹ m) ν ≈ 7.27 x 10^14 s⁻¹ Now we have the frequency, we can find the energy using the Planck's formula: E = hν. E = (6.626 x 10^-34 Js) * (7.27 x 10^14 s⁻¹) E ≈ 4.82 x 10^-19 J Thus, the energy of the photon is 4.82 x 10^-19 J.

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

To find the wavelength of radiation, we first need to find the frequency (ν) of the photons using the Planck's formula: E = hν. Given energy, E = 6.06 x 10^-19 J Planck's constant, h = 6.626 x 10⁻³⁴ Js Now, rearrange the formula for frequency: ν = E / h ν = (6.06 x 10^-19 J) / (6.626 x 10^-34 Js) ν ≈ 9.15 x 10^14 s⁻¹ Now we have the frequency, we can find the wavelength using the speed of light formula: c = λν. Speed of light, c = 3.0 x 10^8 m/s Now, rearrange the formula for wavelength: λ = c / ν λ = (3.0 x 10^8 m/s) / (9.15 x 10^14 s⁻¹) λ ≈ 3.28 x 10^-7 m Convert the wavelength to nm: λ ≈ 328 nm Thus, the wavelength of radiation with photons of energy 6.06 x 10⁻¹⁹ J is 328 nm.

Key Concepts

Planck's formula

Understanding the relationship between energy and frequency in electromagnetic radiation begins with Planck's formula, which is the cornerstone of quantum mechanics. This fundamental equation is expressed as E = h\(u\), where E represents the energy of a photon, h is Planck’s constant (approximately \(6.626 \times 10^{-34} \) Joule-seconds), and \(u\) is the frequency of the radiation.

When observing the behavior of photons, tiny packets of energy, Planck's formula demonstrates that the more frequent the oscillation of electromagnetic waves, the higher the energy carried by the photons. This concept is further highlighted through exercises that prompt students to calculate photon energy, providing a practical understanding of this quantum relationship. Planck's constant provides a conversion factor between the wavelength and the energy, bridging the gap between the macroscopic world we observe and the microscopic realm of quantum particles.

Frequency of electromagnetic radiation

Electromagnetic radiation covers a broad range of frequencies, from radio waves to gamma rays, and it is essentially light, including wavelengths not visible to the human eye. The frequency of this radiation, denoted by \(u\), is the number of times the wave oscillates or cycles per second and is measured in Hertz (Hz), signifying cycles per second or s⁻¹.

Higher frequencies correlate with higher energy photons, as per Planck's formula. For example, gamma rays have frequencies at the high end of the spectrum, and consequently, they possess high photon energies, which is why they can be so penetrative and potentially harmful. When estimating the energy of a photon for a given frequency, as demonstrated in exercise solutions, Planck's formula allows us to directly calculate this energy without needing to consider the photon's wavelength.

Wavelength and frequency relationship

The wavelength (\(\lambda\)) and frequency (\(u\)) of electromagnetic radiation are interconnected through the speed of light with the equation \(c = \lambdau\), where \(c\) stands for the speed of light in a vacuum, which is about \(3.0 \times 10^8 \) meters per second. This inverse relationship means that as the wavelength increases (representing the physical distance between successive crests of the wave), the frequency decreases, and vice versa.

Speed of Light as a Constant

In lessons on electromagnetic radiation, we stress that the speed of light is a constant in a vacuum, making it a pivotal factor in calculations involving wavelengths and frequencies. When the wavelength of a photon is given, like in the provided exercise, this unwavering speed permits the calculation of the unknown frequency or energy of the photon, emphasizing the interconnected nature of these fundamental properties of light.

Speed of light

The speed of light (\(c\)) is a fundamental constant of nature, approximately \(299,792,458 \) meters per second in a vacuum. Its constancy enables precise calculations in various astrophysical and physics problems. In optics, as well as in the provided textbook exercises, the speed of light is instrumental for deriving other quantities such as frequency and wavelength of electromagnetic waves.

One significant implication of the speed of light is that it establishes a speed limit for how fast information, or anything else, can travel in the universe. This speed limit, along with Planck's constant, forms the basis of our ability to understand the scale and energy of various elements in the quantum world, including the photons whose energies students calculate in exercises.