Problem 32 - Chemistry The Central Science

Question

A stellar object is emitting radiation at 3.55 \(\mathrm{mm}\) . (a) What type of electromagnetic spectrum is this radiation? (b) If a detector is capturing \(3.2 \times 10^{8}\) photons per second at this wavelength, what is the total energy of the photons detected in 1.0 hour?

Step-by-Step Solution

Answer

The radiation emitted by the stellar object falls under the category of microwaves, with a wavelength of 3.55 mm. The total energy of the photons detected in 1 hour is approximately \(6.42\cdot10^{-6}\) Joules.

Step 1: Identify the type of electromagnetic spectrum

Given the wavelength is 3.55 mm, we can use this information to identify the type of electromagnetic spectrum. The electromagnetic spectrum, in increasing order of wavelength, is classified as follows: 1. Gamma rays (< 0.01 nm) 2. X-rays (0.01 nm - 10 nm) 3. Ultraviolet (10 nm - 400 nm) 4. Visible light (400 nm - 700 nm) 5. Infrared (700 nm - 1 mm) 6. Microwaves (1 mm - 1 m) 7. Radio waves (> 1 m) The given wavelength of radiation is 3.55 mm, which falls in the range of microwaves category. So, the type of electromagnetic spectrum emitted by the stellar object is microwaves.

Step 2: Calculate the total energy of the photons detected in 1 hour

We know that the energy of a photon can be calculated using the formula: \(E = hf\) Where, \(E\) = energy of the photon, \(h\) = Planck's constant (Approximately equal to \(6.63\cdot10^{-34} \, \text{J}\cdot\text{s}\)), \(f\) = frequency of the incident light. We have the detector capturing \(3.2\cdot10^{8}\) photons per second. Let's calculate the total number of photons detected in 1 hour: Total number of photons in 1 hour = 3.2\(\cdot10^{8}\) \(\cdot\) 3600 We can first find the frequency (\(f\)) of the given wavelength using the formula: \(f = \frac{c}{\lambda}\) Where, \(c\) = speed of light (approximately equal to \(3\cdot10^{8}\,m/s\)) and \(\lambda\) = wavelength of radiation (3.55 mm). Now, calculate the frequency: \(f = \frac{3\cdot10^{8}\,\text{m/s}}{3.55\cdot10^{-3}\,\text{m}} = 8.45\cdot10^{10} \, \text{Hz}\) Now, calculate the energy of a single photon using the energy formula \(E = hf\): \(E = 6.63\cdot10^{-34} \, \text{J}\cdot\text{s} \cdot 8.45\cdot10^{10} \, \text{Hz} = 5.6\cdot10^{-23} \, \text{J}\) Calculate the total energy of photons detected in 1 hour: Total energy of photons = Total number of photons in 1 hour \(\cdot\) energy per photon Total energy of photons = \(3.2\cdot10^{8} \cdot 3600 \cdot 5.6\cdot10^{-23} \, \text{J}\) Total energy of photons = \(6.42\cdot10^{-6} \, \text{J}\) The total energy of the photons detected in 1 hour is approximately \(6.42\cdot10^{-6}\) Joules.

Key Concepts

Microwaves

Microwaves are a type of electromagnetic radiation with wavelengths ranging from one millimeter to one meter, which places them between infrared radiation and radio waves in the electromagnetic spectrum. Microwaves are commonly known for their use in microwave ovens where they excite water molecules to heat food.

However, they also have numerous applications in telecommunications, for example, they are used for satellite communication and Wi-Fi signals because of their ability to penetrate clouds, rain, and smoke. Additionally, microwaves are utilized in radar systems to detect aircraft and ships due to their relatively shorter wavelengths which allow for the detection of smaller objects.

In the context of our exercise, a stellar object emitting radiation with a wavelength of 3.55 mm falls within the microwave range. This categorization is fundamental in astrophysics for understanding the nature of celestial bodies and their behavior. When observing space, scientists can use the microwave signature to study the cosmic microwave background radiation, which offers insights into the early universe.

Photon Energy Calculation

The energy of a photon is directly proportional to its frequency and inversely proportional to its wavelength. To calculate the photon energy, we can use the fundamental photon energy equation:
\[E = hf\]
where:

• \(E\) is the energy of a single photon,
• \(h\) is Planck's constant,
• \(f\) is the frequency of the photon.

Since frequency \(f\) and wavelength \(\lambda\) are inversely related through the speed of light \(c\), the frequency can be determined using the equation \(f = \frac{c}{\lambda}\).

In practical scenarios like the exercise, where we quantify the total energy from numerous photons, we multiply the single photon's energy by the total number of photons. For instance, if a detector captures photons over a specific period, we can calculate the total energy absorbed or emitted in that duration. This calculation is crucial in various fields such as quantum mechanics, photonics, and the study of astrophysical phenomena.

Planck's Constant

Planck's constant, denoted by \(h\), is a fundamental constant in quantum physics that describes the quantization of energy. It has a value of approximately \(6.62607015 \times 10^{-34} \) joule seconds (J⋅s). This constant plays a pivotal role in the quantum theory of energy and is an essential element in the equation used to calculate the energy of photons.

The introduction of Planck's constant marked a major shift in our understanding of energy, indicating that energy is quantized and can be exchanged only in discrete amounts called quanta. This concept was revolutionary as it challenged the classic view that energy changes were continuous.

Planck's constant is not only crucial for energy calculations but also appears in many other key quantum equations, including the Heisenberg uncertainty principle. Understanding its significance allows students to appreciate the quantum foundation of light and its interactions with matter, which is an integral concept in physics and other scientific disciplines.