Question

Infrared Radiation How many photons of infrared radiation with a fre-quency of \(4.88 \times 10^{13} \mathrm{Hz}\) are required to provide an energy of 1.00 \(\mathrm{J} ?\)

Short answer

Approximately \(3.09 \times 10^{20}\) photons of infrared radiation with a frequency of \(4.88 \times 10^{13} \mathrm{Hz}\) are required to provide an energy of 1.00 J.

Step-by-step solution

The energy-frequency relationship of a photon is given by the equation: \[E = h \nu\] where E is the energy of the photon, h is the Planck's constant (\(6.63 \times 10^{-34} \mathrm{J \cdot Hz^{-1}}\)), and \(\nu\) is the frequency of the photon. #Step 2: Calculate the energy of one photon with the given frequency#

Using the given frequency (\(4.88 \times 10^{13} \mathrm{Hz}\)), we will calculate the energy of one photon using the energy-frequency equation: \[E = h \nu = (6.63 \times 10^{-34} \mathrm{J \cdot Hz^{-1}})(4.88 \times 10^{13} \mathrm{Hz})\] This calculates to: \[E = 3.236 \times 10^{-20} \mathrm{J}\] #Step 3: Calculate the number of photons required to provide an energy of 1.00 J#

We need the total energy to be 1.00 J and we know the energy of one photon. To calculate the number of photons (n) needed to provide the desired energy, we can use the following equation: \[Total \, Energy = n \times Energy \, per \, Photon\] Replacing the values and re-arranging the equation to solve for the number of photons (n): \[n = \frac{Total \, Energy}{Energy \, per \, Photon} = \frac{1.00 \mathrm{J}}{3.236 \times 10^{-20} \mathrm{J}}\] After calculating, we find: \[n \approx 3.09 \times 10^{20}\] #Step 4: Final answer#

Approximately \(3.09 \times 10^{20}\) photons of infrared radiation with a frequency of \(4.88 \times 10^{13} \mathrm{Hz}\) are required to provide an energy of 1.00 J.

Key concepts

Photon Energy

Every light wave, including infrared radiation, consists of packets of energy known as photons. The energy of a single photon is an essential concept in understanding how light and electromagnetic radiation interact with matter. The energy of a photon is calculated using the equation:\[ E = h u \]In this equation:

• \(E\) stands for the energy of one photon.
• \(h\) is Planck's constant, an important physical constant.
• \(u\) represents the frequency of the light wave.

Photons with higher frequencies carry more energy. That's why gamma rays are much more energetic than radio waves. This relationship between energy and frequency is crucial in fields like quantum mechanics and spectroscopy.

Planck's Constant

Planck's constant, denoted as \(h\), is a fundamental constant in physics. It plays a pivotal role in quantum mechanics, especially in the relationship between the energy and frequency of photons. This constant was introduced by Max Planck in the early 20th century as a part of his quantum theory.The value of Planck's constant is \(6.63 \times 10^{-34} \text{J} \cdot \text{Hz}^{-1}\). This very small value indicates that individual photons carry a minuscule amount of energy unless the frequency is extremely high.Planck's constant is essential for:

• Determining the energy of photons in light and electromagnetic radiation.
• Understanding the behavior of particles at quantum levels.
• Helping bridge classical and quantum physics.

The concept of Planck's constant has paved the way for countless technological and theoretical advances, including the development of lasers and quantum computing.

Frequency Calculation

The frequency of a wave is a key concept when dealing with electromagnetic radiation. Frequency can be defined as the number of wave cycles that pass a point per second, expressed in Hertz (Hz).For photons, a higher frequency means more energy. This is why, in the original exercise, we focus on the frequency \(4.88 \times 10^{13} \text{Hz}\) to calculate the energy of a photon.Calculating frequency can be crucial in various contexts:

• Identifying different types of electromagnetic radiation (e.g., visible light, infrared, ultraviolet).
• Determining the energy photons possess when moving through space.
• Applications in the study of spectral lines in astronomy to find elements in stars.

The link between frequency and energy explains the wide range of behaviors observed for different frequencies of light, from the visible spectrum to infrared and beyond.