Question

In what region of the electromagnetic spectrum would you expect to find radiation having an energy per photon 100 times that associated with 988 nm radiation?

Short answer

The radiations is in the region of \(x\) on the electromagnetic spectrum. (Replace \(x\) with the specific region e.g ultraviolet.)

Step-by-step solution

Calculate the Energy of 988nm radiation

First, the energy of the radiation with a wavelength of 988nm needs to be calculated. Using the formula for energy of a photon, \(E = h * \frac{c}{\lambda}\), where \(h\) is the Planck's constant \(6.63 * 10^{-34} Js\), \(c\) is the speed of light \(3.0 * 10^8 m/s\), and \(\lambda\) is the wavelength which is 988nm or \( 988 * 10^{-9}m \).

Calculate the energy of the given radiation

The problem states that the energy of the given radiation is 100 times that of 988nm radiation. So, multiply the energy obtained in the first step by 100.

Calculate the wavelength of the given radiation

Since energy and wavelength are in a reciprocal relationship, the wavelength of the given radiation can be found after obtaining its energy. Rearrange the energy formula to solve for wavelength, \(\lambda = h * \frac{c}{E}\). Insert the energy from step 2 into the formula to get the wavelength.

Identify the region of the electromagnetic spectrum

With the obtained wavelength of the radiation, its corresponding region on the electromagnetic spectrum can be identified.

Key concepts

Photon Energy

Photon energy is a fundamental concept in understanding electromagnetic radiation. It refers to the energy carried by a single photon, which is the smallest particle of light. The energy of a photon is calculated using the formula:

• \( E = h \cdot \frac{c}{\lambda} \)

In this equation, \(E\) represents the energy of the photon, \(h\) is Planck's constant \(6.63 \times 10^{-34} \text{Js}\), \(c\) is the speed of light \(3.0 \times 10^8 \text{m/s}\), and \(\lambda\) is the wavelength. This formula shows that photon energy is inversely proportional to its wavelength.
This means that shorter wavelengths correspond to higher energy photons. Conversely, longer wavelengths indicate lower energy photons.
Understanding photon energy is crucial for identifying different forms of electromagnetic radiation, as examples include the visible light we see and the ultraviolet radiation from the sun.

Wavelength Calculation

Calculating the wavelength of electromagnetic radiation involves rearranging the photon energy formula to solve for wavelength. This means if you know the energy of a photon, you can find its wavelength using:

• \( \lambda = \frac{h \cdot c}{E} \)

For example, if you have radiation with known energy, inserting it into this formula gives the wavelength. The shorter the wavelength, the higher the energy of the radiation.
This understanding is useful in various scientific fields, from astronomy to medicine. Identifying the exact wavelength helps determine the type of electromagnetic radiation, such as radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, and gamma rays.
Each type of radiation is associated with specific wavelengths and energies, making wavelength calculations essential in identifying and utilizing these rays effectively.

Electromagnetic Radiation

Electromagnetic radiation encompasses all forms of light, including those invisible to the human eye. It travels through space as waves, defined by their frequency, wavelength, and energy. The electromagnetic spectrum categorizes this radiation based on wavelength.
From longest to shortest wavelength, the spectrum includes:

• Radio waves
• Microwaves
• Infrared
• Visible light
• Ultraviolet
• X-rays
• Gamma rays

Each range has different applications and effects. For instance, radio waves are used in communication systems, while ultraviolet rays are responsible for sunburns.
Electromagnetic radiation is both natural, such as sunlight, and man-made, like microwaves. Its study is crucial in technology development, medical imaging, and understanding cosmic phenomena. Additionally, by examining specific wavelengths, scientists can identify new applications and advance our knowledge of the universe.