Problem 27 - Chemistry: The Central Science

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 \(0.52\) J over a period of \(32 \mathrm{~s}\). How many photons per second are being emitted by the laser?

Step-by-Step Solution

Answer

The radiation of the given wavelength (987 nm) is found in the Infrared portion of the electromagnetic spectrum. The diode laser emits approximately \( 8.05 \times 10^{16} \) photons per second.

Step 1: Determine the portion of the electromagnetic spectrum corresponding to the given wavelength.

First, let's examine the given wavelength of 987 nm. We need to compare this value to the ranges of the electromagnetic spectrum to determine in which portion it lies: - Infrared: \( \approx 700 \mathrm{~nm} \) to \( 1 \mathrm{~mm} \) - Visible: \( \approx 400 \mathrm{~nm} \) to \( 700 \mathrm{~nm} \) - Ultraviolet: \( \approx 10 \mathrm{~nm} \) to \( 400 \mathrm{~nm} \) Since 987 nm lies in the Infrared range, the radiation is found in the Infrared portion of the electromagnetic spectrum.

Step 2: Calculate the frequency of the radiation.

To calculate the frequency of the radiation, we will use the relationship between the speed of light (c), frequency (ν), and wavelength (λ) given by: \( c = \nu \times \lambda \) We know the wavelength (λ = 987 nm) and the speed of light (c = \(3.0 \times 10^8\) m/s). First, we need to convert the wavelength to meters: \( \lambda = 987 \mathrm{~nm} \times \frac{1 \mathrm{~m}}{10^9 \mathrm{~nm}} \) \( \lambda \approx 9.87 \times 10^{-7} \mathrm{~m} \) Now we can solve for the frequency (ν) using the equation: \( \nu = \cfrac{c}{\lambda} \) \( \nu = \cfrac{3.0 \times 10^8 \mathrm{~m/s}}{9.87 \times 10^{-7} \mathrm{~m}} \) \( \nu \approx 3.04 \times 10^{14} \mathrm{~Hz} \)

Step 3: Calculate the energy of an individual photon.

We can now use the frequency to calculate the energy (E) of an individual photon using the Planck's constant (h = \(6.63 \times 10^{-34}\) J s): \( E = h \times \nu \) \( E = (6.63 \times 10^{-34} \mathrm{~J \cdot s}) \times (3.04 \times 10^{14} \mathrm{~Hz}) \) \( E \approx 2.02 \times 10^{-19} \mathrm{J} \)

Step 4: Calculate the number of photons per second.

Now that we have the energy of an individual photon, we can use the total energy (0.52 J) and the period (32 s) to determine the number of photons per second emitted by the laser. First, let's calculate the energy per second absorbed by the detector: \( \cfrac{0.52 \mathrm{J}}{32 \mathrm{s}} \approx 0.01625 \mathrm{J/s} \) Now divide this value by the energy of an individual photon to find the number of photons per second: \( \cfrac{0.01625 \mathrm{J/s}}{2.02 \times 10^{-19} \mathrm{J}} \approx 8.05 \times 10^{16} \) photons/s The diode laser emits approximately \( 8.05 \times 10^{16} \) photons per second.

Key Concepts

Electromagnetic Spectrum

The electromagnetic spectrum encompasses all types of electromagnetic radiation, which differ in their wavelengths and frequencies. Imagine it like a rainbow of light waves, stretching from the shortest gamma rays to the longest radio waves. Each portion of this spectrum has its own unique properties and uses. For example, the visible portion is where all the colors of light live. But there’s so much more beyond what our eyes can see!
Let's break down the main parts of the spectrum:

• Radio Waves: Longest wavelengths, used for music on the radio and TV shows.
• Microwaves: These heat up your food in microwaves.
• Infrared: Just below visible light, mainly heat.
• Visible Light: The only part visible to human eyes; think ROYGBIV (rainbow colors).
• Ultraviolet (UV): Beyond violet, these can give you sunburns!
• X-Rays: Used by doctors to look inside your body.
• Gamma Rays: Shortest wavelength, but most energy.

When you hear about the electromagnetic spectrum, think of all these differing lights coming together to form a complete picture of how energy travels and interacts with the world.

Infrared Radiation

Infrared radiation is a part of the electromagnetic spectrum that is just beyond the visible red light. It has longer wavelengths than visible light and is often associated with heat. You can sense infrared radiation even if you can't see it, like when you feel the warmth from the sun or a campfire without direct contact.
It's fascinating how infrared works:

• Heat Signature Detection: Infrared cameras can detect the heat emitted by objects, useful in night-vision equipment.
• Applications in Science: Used in astronomy to study celestial objects and atmospheres.
• Remote Controls: Many TV remotes use infrared light to send signals to the television.

This form of radiation is important in both everyday technology and scientific research. Infrared waves are key to heating and cooking technologies as well as in thermal imaging.

Photon Energy

Photon energy is all about how each tiny particle of light, a photon, carries energy that can be calculated. The energy of a photon depends entirely on its frequency, which is linked to its place on the electromagnetic spectrum.
Here's how it breaks down:

• The energy of a photon can be calculated with the formula:
  \( E = h \times u \)
  Where \(E\) is energy, \(h\) is Planck's constant \((6.63 \times 10^{-34} \text{ J·s})\), and \(u\) is the frequency.
• Photons from different parts of the spectrum have different energies. For example, gamma ray photons have much more energy than radio wave photons.
• This energy is what allows photons to interact with matter, such as providing light and warmth when absorbed.

A great example is visible light, where the colors near the blue end have more energy than those near the red end. In technology, photon energy is crucial for devices like lasers, relying on it to perform tasks such as cutting materials or reading data.