Problem 32 - Chemistry The Central Science In Si Units

Question

A stellar object is emitting radiation at \(3.0 \mathrm{~mm} .\) (a) What type of electromagnetic spectrum is this radiation? (b) If a detector is capturing \(3.0 \times 10^{8}\) photons per second at this wavelength, what is the total energy of the photons detected in 1 day?

Step-by-Step Solution

Answer

(a) Microwave; (b) Total energy in 1 day is approximately \(1.72 \times 10^{-9}\) J.

Step 1: Identify the Type of Electromagnetic Spectrum

The wavelength of the radiation is given as \(3.0 \text{ mm}\). The electromagnetic spectrum consists of various types of waves, such as radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, and gamma rays. Millimeter wavelengths (\(1\text{ mm}\) to \(10\text{ mm}\)) fall within the microwave range. Therefore, the radiation is a type of microwave.

Step 2: Calculate the Energy of a Single Photon

To find the total energy of the photons, we first need to determine the energy of a single photon. This is given by the formula \(E = \frac{hc}{\lambda}\), where \(h\) is Planck's constant \( (6.626 \times 10^{-34} \text{ Js}) \), \(c\) is the speed of light \((3 \times 10^8 \text{ m/s})\), and \(\lambda\) is the wavelength \( (3.0 \times 10^{-3} \text{ m}) \). Substituting the values, we get:\[ E = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{3.0 \times 10^{-3}} = 6.626 \times 10^{-23} \text{ J/photon} \]

Step 3: Calculate the Total Energy Detected in 1 Second

The number of photons detected per second is \(3.0 \times 10^8\). The energy detected in 1 second is the energy of one photon multiplied by the number of photons:\[ E_{\text{total, 1s}} = 6.626 \times 10^{-23} \times 3.0 \times 10^{8} \]\[ E_{\text{total, 1s}} = 1.9878 \times 10^{-14} \text{ J/s} \]

Step 4: Calculate Total Energy Detected in 1 Day

To find the total energy detected in 1 day, we multiply the energy per second by the total number of seconds in a day. There are 86400 seconds in a day:\[ E_{\text{total, 1 day}} = 1.9878 \times 10^{-14} \times 86400 \]\[ E_{\text{total, 1 day}} = 1.7176352 \times 10^{-9} \text{ J} \]

Key Concepts

Electromagnetic Spectrum

The electromagnetic spectrum encompasses all types of electromagnetic radiation, which are waves of electric and magnetic energy traveling through space. These waves vary in length from very long radio waves, through microwaves, infrared radiation, visible light, ultraviolet, to very short wavelengths like X-rays and gamma rays. Each segment of the spectrum has its own unique properties and applications. It is important to understand where each type of radiation falls in the spectrum to interpret their interactions and potential uses correctly.

When looking at the electromagnetic spectrum in terms of wavelength, microwaves typically have wavelengths ranging from 1 millimeter to 1 meter. They fall just after radio waves and before infrared radiation. In our problem, the wavelength of the radiation emitted by the stellar object is given as 3.0 millimeters, which places it firmly within the microwave portion of the electromagnetic spectrum.

• Radio Waves:
  • Wavelength: > 1 meter
• Microwaves:
  • Wavelength: 1 millimeter to 1 meter
• Infrared:
  • Wavelength: between 0.75 micrometers and 1 millimeter

Understanding this spectrum helps in identifying how different types of radiation can be used for diverse purposes such as communication, cooking, or even astronomy.

Microwave Radiation

Microwave radiation is a type of electromagnetic wave that falls between radio waves and infrared on the electromagnetic spectrum. With wavelengths ranging from 1 millimeter to 1 meter, microwaves are extensively used in everyday devices and industrial applications.

One of the most well-known uses of microwaves is in microwave ovens, where they heat food by causing molecules to vibrate and generate thermal energy. Microwaves are also essential in telecommunications, particularly in radar technology and satellite communications.

• Application in Communication:
  • Used in radar which measures distances and speeds by reflecting microwaves.
  • Satellites use microwaves to transmit data to receivers on Earth.
• Scientific Contribution:
  • Microwave radiation is used in Astronomy to study cosmic background radiation, giving insights into the early universe.

Microwaves' properties, such as penetrating through clouds and being unaffected by rain, make them ideal for various applications in weather forecasting and navigation systems.

Photon Energy Formula

The energy carried by a photon can be calculated using the photon energy formula: \[ E = \frac{hc}{\lambda} \] where:

• \( E \) is the energy of the photon (in joules).
• \( h \) represents Planck's constant, approximately \( 6.626 \times 10^{-34} \text{Js} \).
• \( c \) is the speed of light, \( 3 \times 10^8 \text{m/s} \).
• \( \lambda \) is the wavelength of the photon.

By inputting the wavelength of the radiation, you can derive the amount of energy each photon carries. For microwaves, as with the given problem, the wavelength is 3.0 millimeters, or \( 3.0 \times 10^{-3} \text{m} \). Using the formula, the energy of one photon emitted by the stellar object can be calculated as \( 6.626 \times 10^{-23} \text{ J/photon} \).

This formula illustrates a direct relationship between the energy and wavelength of a photon. As wavelength increases, the energy of the photon decreases, explaining why photons from gamma rays (shorter wavelengths) carry much more energy compared to those from microwave radiation (longer wavelengths). Understanding how to use this formula and the factors affecting photon energy is essential for interpreting radiation energy from various sources.