Problem 31 - Chemistry The Central Science In Si Units

Question

A diode laser emits at a wavelength of \(987 \mathrm{nm} .\) (a) In what portion of the electromagnetic spectrum is this radiation found? (b) All of its output energy is absorbed in a detector that measures a total energy of \(0.52 \mathrm{~J}\) over a period of \(32 \mathrm{~s}\). How many photons per second are being emitted by the laser?

Step-by-Step Solution

Answer

(a) Infrared portion; (b) \(8.09 \times 10^{16}\) photons per second.

Step 1: Identify the Region of the Electromagnetic Spectrum

First, examine the wavelength of the radiation. Given that the diode laser emits at a wavelength of \(987 \text{ nm}\), we need to identify this against the electromagnetic spectrum. Wavelengths in the range of \(750 \text{ nm}\) to \(1 \text{ mm}\) are classified as infrared (IR). Therefore, the radiation emitted by the laser is in the infrared portion of the spectrum.

Step 2: Calculate the Energy of a Single Photon

Use the formula for energy of a photon: \(E = \frac{hc}{\lambda}\), where \(h = 6.626 \times 10^{-34} \text{ J}\cdot\text{s}\) is Planck's constant and \(c = 3 \times 10^8 \text{ m/s}\) is the speed of light. Convert \(987 \text{ nm}\) to meters by dividing by \(10^9\): \(\lambda = 987 \text{ nm} = 987 \times 10^{-9} \text{ m}\). Now, calculate the energy: \[ E = \frac{6.626 \times 10^{-34} \cdot 3 \times 10^8}{987 \times 10^{-9}} \approx 2.01 \times 10^{-19} \text{ J} \].

Step 3: Determine the Total Number of Photons Emitted

The total energy absorbed by the detector is given as \(0.52 \text{ J}\). Divide this energy by the energy of a single photon to find the total number of photons: \[ N = \frac{0.52}{2.01 \times 10^{-19}} \approx 2.59 \times 10^{18} \text{ photons} \].

Step 4: Calculate the Number of Photons Emitted Per Second

The time period given is \(32 \text{ s}\). To determine the photons emitted per second, divide the total number of photons by the time in seconds: \[ \frac{2.59 \times 10^{18}}{32} \approx 8.09 \times 10^{16} \text{ photons/second} \].

Key Concepts

Infrared Spectrum

The infrared spectrum is part of the electromagnetic spectrum where longer wavelengths exist beyond visible light. Infrared light is invisible to the human eye but can be felt as heat. It's typically found in the range from about 750 nm to 1 mm.

A diode laser operating at a wavelength of 987 nm falls within this infrared region. This is because 987 nm sits comfortably between the border of visible red light and the longer wavelengths characteristic of infrared radiation. In practical terms, infrared is extensively used in everyday technology such as remote controls, night-vision equipment, and in scientific applications like spectroscopy.

Infrared waves are significant in various fields:

• Thermal imaging, where it helps in identifying objects in the dark based on heat emission.
• Weather forecasting, using satellite sensors to monitor thermal patterns on Earth.
• Medical treatments, where it's applied for heating therapies.

Understanding that a diode laser emits in the infrared spectrum allows scientists to utilize it for specific applications, maximizing its benefits in both technological and research areas.

Photon Energy Calculation

Calculating the energy of a photon is essential to understanding how much energy is transferred during emission or absorption processes.

Photon energy, denoted as \(E\), can be calculated using the equation \(E = \frac{hc}{\lambda}\), where:

• \(h\) is Planck's constant \(6.626 \times 10^{-34} \text{ J}\cdot\text{s}\)
• \(c\) is the speed of light \(3 \times 10^8 \text{ m/s}\)
• \(\lambda\) is the wavelength of the photon

In our example, converting the wavelength from nanometers to meters is crucial: \(987 \text{ nm} = 987 \times 10^{-9} \text{ m}\).

Now, solve for the energy: \[E = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{987 \times 10^{-9}} \approx 2.01 \times 10^{-19} \text{ J}.\]

This calculation reveals the energy carried by an individual photon at this wavelength. It's crucial for determining how many such photons contribute to observable energy events, such as those measured by detectors in experiments.

Electromagnetic Spectrum

The electromagnetic spectrum encompasses all types of electromagnetic radiation, ranging from very short gamma rays to long-wavelength radio waves. Each type of radiation defined by its wavelength and frequency offers specific applications and characteristics.

Key areas of the electromagnetic spectrum include:

• Gamma rays: extremely high energy and short wavelength, used in cancer treatment and scanning materials.
• X-rays: capable of penetrating objects, are essential in medical imaging.
• Ultraviolet: shorter than visible light, used in sterilization and fluorescent lights.
• Visible light: the only part perceivable by the human eye, facilitating vision.
• Infrared: longer than visible light, experienced as heat.
• Microwaves: used in communication and cooking appliances.
• Radio waves: with the longest wavelength, primarily used in broadcasting and communications.

Understanding where a particular wavelength fits within this spectrum is vital for determining its potential uses and interactions with matter. Knowing that a 987 nm laser operates in the infrared portion helps in selecting appropriate applications, from thermal imaging to communications.

The orderly nature of the electromagnetic spectrum allows scientists and engineers to tap into each section's unique properties, advancing technology and scientific understanding.