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 one hour?

Short answer

(a) The radiation is microwave radiation. (b) The total energy of the photons detected in one hour is approximately \(6.44 \times 10^{-11}\) J.

Step-by-step solution

Identify the type of electromagnetic radiation

Using the given wavelength (3.55 mm), we can consult the electromagnetic spectrum toidentify the type of radiation. The common types of electromagnetic radiation, ordered by decreasing wavelength, are: - Radio waves - Microwaves - Infrared - Visible light - Ultraviolet - X-rays - Gamma rays A wavelength of 3.55 mm falls within the microwave range (1 mm to 100 cm). So, the radiation being emitted by the stellar object is microwave radiation.

Calculate the energy of detected photons per second

We are given the detection rate (3.2 x 10^8 photons per second) and the wavelength (3.55 mm). First, let's convert the wavelength to meters: 3.55 mm = 3.55 x 10^-3 m. Now, we'll use the Planck's equation to find the energy of a single photon: \[E = hf\] Where E is the energy, h is the Planck's constant (\(6.63 \times 10^{-34} \mathrm{Js}\)), and f is the frequency. To find the frequency, use the speed of light (c) and the given wavelength (λ): \(c = f\lambda\) Rearranging to get the frequency: \(f = \frac{c}{\lambda}\) Now, plug in the values: \(f = \frac{3 \times 10^8 \mathrm{ms^{-1}}}{3.55 \times 10^{-3}\mathrm{m}} \approx 8.45 \times 10^{10} \mathrm{s^{-1}}\) Now, calculate the energy of a single photon: \(E = (6.63 \times 10^{-34} \mathrm{Js})(8.45 \times 10^{10} \mathrm{s^{-1}}) \approx 5.60 \times 10^{-23} \mathrm{J}\) Next, find the total energy of the detected photons per second by multiplying the energy of a single photon by the detection rate: \(E_\mathrm{total} = (5.60 \times 10^{-23} \mathrm{J})(3.2 \times 10^8 \mathrm{photons/s}) \approx 1.79 \times 10^{-14} \mathrm{J/s}\) The total energy detected per second is approximately \(1.79 \times 10^{-14}\) J/s.

Calculate the total energy detected in one hour

Finally, we'll find the total energy detected in one hour by multiplying the energy per second by the number of seconds in an hour: \(E_\mathrm{1h} = (1.79 \times 10^{-14} \mathrm{J/s})(3600 \mathrm{s/h}) \approx 6.44 \times 10^{-11} \mathrm{J}\) The total energy of the photons detected in one hour is approximately \(6.44 \times 10^{-11}\) J. To summarize the answers: (a) The radiation is microwave radiation. (b) The total energy of the photons detected in one hour is approximately \(6.44 \times 10^{-11}\) J.

Key concepts

Microwave Radiation

Microwave radiation is a type of electromagnetic radiation with wavelengths ranging from 1 millimeter to 1 meter. It fits into the electromagnetic spectrum between radio waves, which have longer wavelengths, and infrared radiation, which has shorter wavelengths.
This range of microwaves allows them to be utilized in various applications, such as cooking food in microwave ovens, transmitting signals in communication devices, and studying astronomical phenomena.

In the context of the electromagnetic spectrum, microwaves have wavelengths typically between 1 mm to 100 cm. Because the wavelength of the radiation absorbed by the stellar object in the exercise is given as 3.55 mm, it clearly lies within this microwave range.

• Microwaves are widely used in modern-day technology.
• They can be found in household appliances and are essential in wireless communication.

Understanding their place in the electromagnetic spectrum helps categorize different kinds of radiation based on their properties.

Photon Energy Calculation

To calculate the energy of photons, it's crucial to consider their wavelength and use relevant physics equations. Photons are packets of electromagnetic energy, and each photon at a given wavelength or frequency carries a specific amount of energy.

In the exercise, the radiation’s wavelength is 3.55 mm, which is first converted to meters (3.55 x 10^{-3} m). To find the energy of a photon, we use the formula:\[E = hf\]where \(E\) is the photon energy, \(h\) is Planck's constant \( (6.63 \times 10^{-34} \, \text{Js}) \), and \(f\) is the frequency. Frequency can be determined by:\[f = \frac{c}{\lambda}\]where \(c\) is the speed of light (\(3 \times 10^8 \, \text{m/s}\)) and \(\lambda\) is the wavelength. After determining the frequency, plug it back into the energy equation to find the energy of a single photon.

This method allows scientists to compute the energy carried by radiation, which is crucial in applications from medical imaging to astrophysics. Understanding photon energy helps decipher how much energy is being transferred in both everyday technologies and cosmic processes.

Planck's Equation

Planck's Equation is a cornerstone in understanding the quantum nature of electromagnetic radiation. It relates the energy of a photon to its frequency and is foundational in quantum mechanics.

The equation is expressed as:\[E = hf\]In this equation, \(E\) represents the energy of the photon, \(h\) is Planck's constant (\(6.63 \times 10^{-34} \, \text{Js}\)), and \(f\) is the frequency of the photon. This simple yet powerful relationship shows that energy is quantized and disproves earlier theories that assumed energy was continuous.

Planck's introduction of this equation revolutionized physics by providing one of the first pieces of evidence for quantization in nature. It explained blackbody radiation and laid the groundwork for later developments in quantum theory.

• Planck's Equation reveals that higher frequency photons have higher energy.
• The constant, \(h\), acts as the proportionality constant between frequency and energy.

Applying this formula helps us calculate the energy of radiation across the electromagnetic spectrum, from microwaves to gamma rays.