Question

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

Short answer

(a) The radiation has a wavelength of 3.55 mm, which corresponds to the microwave part of the electromagnetic spectrum. (b) The total energy of the photons detected in one hour is approximately \(6.44 \times 10^{-10}\) J.

Step-by-step solution

Determining type of Electromagnetic Radiation

The wavelength is given as 3.55 mm, which is equivalent to \(3.55 \times 10^{-3}\) m. The electromagnetic spectrum is divided into several types: Radio > Microwave > Infrared > Visible > Ultraviolet > X-Rays > Gamma rays. The given wavelength of \(3.55 \times 10^{-3}\) m corresponds to the microwave part of the spectrum.

Determining the Energy of One Photon

The energy of a photon is given by the equation \(E = hf\), where \(h\) is Planck's constant (approximately \(6.63 \times 10^{-34}\) Js) and \(f\) is the frequency of the wave. We don't have the frequency, but we can find it using the relation \(c = f \lambda\), where \(c\) is the speed of light (approximately \(3.00 \times 10^8\) m/s) and \(\lambda\) is the wavelength. Firstly, let's find the frequency \(f = \frac{c}{\lambda} = \frac{3.00 \times 10^{8} \text{m/s}}{3.55 \times 10^{-3} \text{m}} \approx 8.45 \times 10^{10} \text{Hz}\). Then, by substituting \(f\) into Einstein's energy equation we find \(E = 6.63 \times 10^{-34} \times 8.45 \times 10^{10} \approx 5.60 \times 10^{-23} \text{J}\). This is the energy of one photon at this wavelength.

Calculating Total Energy in One Hour

Now, let's find out the total energy of the photons detected in one hour. We know that the detector is capturing \(3.2 \times 10^{8}\) photons every second, so in one hour (3600 seconds), the detector captures \(3.2 \times 10^{8} \times 3600 \approx 1.15 \times 10^{13}\) photons. The total energy is then \(1.15 \times 10^{13} \times 5.60 \times 10^{-23} = 6.44 \times 10^{-10} \text{J}\). This is the total energy of the photons detected in one hour.

Key concepts

Microwave Radiation

Microwave radiation is a type of electromagnetic radiation, situated on the electromagnetic spectrum between radio and infrared waves. Its wavelengths range from approximately one millimeter to one meter. In the context of the exercise,
the stellar object emitting radiation at a wavelength of 3.55 mm falls into the microwave category.

Microwaves are used for various applications, most notably in microwave ovens, which heat food through the agitation of water molecules. They're also utilized in satellite communications, as well as in radar technology, which is used for navigation and weather monitoring. When it comes to stellar objects, microwaves can provide information about their structure and composition based on their emission or absorption of microwave radiation.

In practical terms, understanding the properties of microwave radiation allows scientists to harness it for different technologies, ranging from cellular networks to deep space communications. The behavior of microwave radiation also follows fundamental physical laws, and it interacts with matter in predictable ways, for instance, by causing water molecules to vibrate and produce heat, which is the principle behind microwave cooking.

Photon Energy Calculation

Calculating the energy of a photon is a critical exercise in understanding not just stars and galaxies, but also the principles of quantum mechanics and how light interacts with matter. A photon is essentially a packet of light energy, and its energy can be determined using the equation
\( E = hf \) where
\( E \) is the photon's energy,
\( h \) is Planck's constant, and
\( f \) is the frequency of the wave.

To find the energy of a single photon, you must first determine the frequency. Since frequency and wavelength are inversely related, as in the exercise, the frequency can be calculated using the speed of light
\( c \) and the wavelength \( \lambda \) - with the formula
\( f = \frac{c}{\lambda} \) .

Once the frequency is known, multiplying it by Planck's constant gives the energy of a single photon. By then multiplying the photon's energy by the total number of photons, the total energy over a specified time period can be calculated, as showcased in the one-hour total energy calculation in the problem. This calculation applies to any photon, regardless of its position in the electromagnetic spectrum, allowing for a wide array of applications in physics and chemistry.

Planck's Constant

Planck's constant, symbolized
\( h \), is a fundamental quantity in quantum mechanics. It has a value of approximately \( 6.63 \times 10^{-34} \) Joule seconds \( (J\cdot s) \). This small constant is named after Max Planck, the physicist who first introduced it and is critical in the field of quantum mechanics.

Planck's constant is the proportionality factor that relates the energy of a photon to the frequency of its associated electromagnetic wave, expressed in the equation
\( E = hf \), where
\( E \) is energy and
\( f \) is frequency. Because of its role in the energy equation, Planck's constant bridges the gap between the macroscopic classical physics and microscopic quantum world.

Additionally, Planck's constant is key in determining the smallest possible energy transfer at the quantum scale. This explains why energy in the quantum realm is not continuous, but rather comes in discrete amounts called quanta. Planck's constant also finds its use in Heisenberg's uncertainty principle and is fundamental in the quantization of physical quantities such as energy, momentum, and angular momentum.