Problem 104 - Chemistry Matter and Change

Question

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

Step-by-Step Solution

Answer

Approximately \(3.10 \times 10^{19}\) photons are needed.

Step 1: Understand the Problem

We need to find the number of photons required to produce an energy of 1.00 Joule, given the frequency of infrared radiation is \(4.88 \times 10^{13}\) Hz.

Step 2: Use Planck's Equation

Recall Planck's equation for energy of a photon: \(E= h u\), where \(E\) is the energy of the photon, \(h\) is Planck's constant \(6.626 \times 10^{-34}\) Js, and \(u\) is the frequency of the radiation.

Step 3: Calculate Energy of One Photon

Substitute \(h = 6.626 \times 10^{-34}\) Js and \(u = 4.88 \times 10^{13}\) Hz into the equation: \[ E = (6.626 \times 10^{-34} \text{ Js}) \times (4.88 \times 10^{13} \text{ Hz}) \approx 3.23 \times 10^{-20} \text{ J} \]This is the energy of one photon of the given frequency.

Step 4: Calculate Number of Photons Needed

Divide the total energy required by the energy per photon to find the number of photons:\[ \text{Number of photons} = \frac{1.00 \text{ J}}{3.23 \times 10^{-20} \text{ J/photon}} \approx 3.10 \times 10^{19} \]Thus, approximately \(3.10 \times 10^{19}\) photons are required.

Key Concepts

Planck's Constant

Planck's constant is a fundamental constant in physics that plays a crucial role in the field of quantum mechanics. It is denoted by the symbol \( h \) and has a value of approximately \( 6.626 \times 10^{-34} \text{ Joule seconds (Js)} \). This constant was introduced by Max Planck in the early 20th century when he was studying black-body radiation.

• Planck discovered that energy is quantized, meaning it exists in discrete amounts called photons.
• Using Planck's constant, the energy of each photon can be calculated based on its frequency.

The equation \( E = h u \) epitomizes this relationship, where \( E \) represents the energy of a photon and \( u \) is its frequency. This equation tells us that as the frequency of light increases, so does the energy of its photons. Every step in understanding quantum phenomena relies on comprehending this foundational constant's role.

Photon Energy

Photon energy is the energy carried by each photon, a particle of light. Understanding photon energy is central to solving problems related to radiation and quantum mechanics. The energy of a photon is directly proportional to its frequency and can be determined using Planck's equation \( E = h u \). Here:

• \( E \) stands for the energy of one photon.
• \( h \) is Planck's constant, \( 6.626 \times 10^{-34} \text{ Js} \).
• \( u \) is the frequency in Hertz (Hz), or cycles per second.

In the context of infrared radiation with a frequency of \( 4.88 \times 10^{13} \) Hz, the energy of each photon can be computed. By substituting the given frequency and Planck's constant into the equation, we find that each photon has an energy of approximately \( 3.23 \times 10^{-20} \) Joules. This calculation illustrates how photon energy varies with frequency and underpins tasks like calculating the number of photons required to reach a certain energy threshold.

Frequency and Wavelength Calculations

In physics, frequency and wavelength are key properties of waves, including light waves like infrared radiation. Understanding how these relate helps us perform important calculations related to energy and photon behavior.

• Frequency (\( u \)): It's the number of wave cycles that pass a point per second. Measured in Hertz (Hz), this is crucial for determining photon energy using \( E = h u \).
• Wavelength (\( \lambda \)): This is the distance between consecutive peaks of a wave. Measured in meters (m), it is linked to frequency through the speed of light \( c \) with the equation \( c = \lambda u \).

In exercises involving photon energy, frequency is often given because it directly affects the energy of the photon. However, knowing either frequency or wavelength allows you to find the other, as they are inversely related given a constant speed of light. For example, if the frequency is known, it can be plugged into calculations to determine the energy or the wavelength, using \( c = 3.00 \times 10^8 \text{ m/s} \) for the speed of light in a vacuum.