Problem 24 - Chemistry: The Central Science

Question

The first LEDs were made from GaAs, which has a band gap of \(1.43 \mathrm{eV}\). What wavelength of light would be emitted from an LED made from GaAs? What region of the electromagnetic spectrum does this light correspond to: UV, Visible, or IR?

Step-by-Step Solution

Answer

An LED made from GaAs emits light with a wavelength of approximately \(867 \) nm, which corresponds to the infrared (IR) region of the electromagnetic spectrum.

Step 1: Convert the band gap energy from eV to Joules

Since the formula relating energy, wavelength, and the speed of light uses energy in Joules, we need to first convert the given energy value in eV to Joules by using the conversion factor (1 eV = \(1.60218 \times 10^{-19}\) J): $$ E (J) = E (eV) \times 1.60218 \times 10^{-19} J/eV $$ Plug in the given value for the band gap energy of GaAs (1.43 eV): $$ E = 1.43 \text{ eV} \times 1.60218 \times 10^{-19} \frac{\text{J}}{\text{eV}} \approx 2.29 \times 10^{-19} \text{ J} $$ So, the energy of the emitted photon is approximately \(2.29 \times 10^{-19}\) J.

Step 2: Determine the wavelength of the emitted light

Now that we have the energy in Joules, we can use the formula that relates energy, wavelength and the speed of light: $$ E = \frac{hc}{\lambda} $$ Where: - \(E\) is the energy of the emitted photon (\(2.29 \times 10^{-19}\) J) - \(h\) is the Planck constant (\(6.626 \times 10^{-34} \text{ Js}\)) - \(c\) is the speed of light (\(3.00 \times 10^8 \text{ m/s}\)) - \(\lambda\) is the wavelength, which is what we want to find First, let's rearrange the formula to solve for \(\lambda\): $$ \lambda = \frac{hc}{E} $$ Now, plug in the values of \(h\), \(c\), and \(E\): $$ \lambda = \frac{6.626 \times 10^{-34} \text{ Js} \times 3.00 \times 10^8 \text{ m/s}}{2.29 \times 10^{-19} \text{ J}} \approx 8.67 \times 10^{-7} \text{ m} $$ The wavelength of light emitted by the GaAs LED is approximately \(8.67 \times 10^{-7}\) m, or \(867 \) nm.

Step 3: Determine the region of the electromagnetic spectrum

Now that we have the wavelength of the light emitted by the GaAs LED, let's determine which region of the electromagnetic spectrum it corresponds to. The electromagnetic spectrum is divided into different regions based on the wavelength range: - Ultraviolet (UV): \(10 \text{ nm} < \lambda < 400 \text{ nm}\) - Visible: \(400 \text{ nm} < \lambda < 700 \text{ nm}\) - Infrared (IR): \(700 \text{ nm} < \lambda < 1 \text{ mm}\) Based on these ranges, the GaAs LED's emitted light wavelength (\(867 \) nm) falls into the infrared (IR) region of the electromagnetic spectrum. In conclusion, an LED made from GaAs emits light with a wavelength of approximately \(867 \) nm, which corresponds to the infrared (IR) region of the electromagnetic spectrum.

Key Concepts

Electromagnetic Spectrum

The electromagnetic spectrum is a range of all types of electromagnetic radiation. Radiation is the energy that travels and spreads out as it goes. Visible light that you see is one type of electromagnetic radiation, but it is just a small part of the electromagnetic spectrum.

The electromagnetic spectrum includes, from shortest to longest wavelength:

• Gamma rays
• X-rays
• Ultraviolet (UV)
• Visible light
• Infrared (IR)
• Microwaves
• Radio waves

Wavelengths and frequencies are inversely related. Longer wavelengths correspond to lower frequencies and vice versa. Each region has specific characteristics or applications. For instance, X-rays are used in medical imaging, while radio waves bring us communication signals. Understanding these different regions helps us see how varied electromagnetic energy can be used in technology and science.

Infrared Radiation

Infrared radiation (IR) is a type of electromagnetic radiation with wavelengths longer than visible light but shorter than microwaves. This means infrared falls between visible light and microwave on the electromagnetic spectrum.

Let's dive deeper into infrared characteristics:

• Wavelengths range from about 700 nanometers (nm) to 1 millimeter (mm).
• Infrared waves are generally experienced as heat.
• They can be used in numerous applications such as heat lamps, night vision equipment, and remote controls.

These waves are specially designed to be used for heating purposes as they are more easily absorbed by objects, making them heat up rapidly. The infrared region plays a significant role in various scientific, industrial, and household applications, showing how versatile this kind of radiation can be.

Photon Wavelength Calculation

To calculate the wavelength of a photon, we use a simple relationship between the photon's energy (E), Planck's constant (h), the speed of light (c), and the wavelength (λ). The equation is:\[E = \frac{hc}{\lambda} \]For practical purposes, let's work backwards to find λ when the photon's energy is known:\[ \lambda = \frac{hc}{E} \]It's crucial to first convert the energy from electron volts (eV) to joules (J) because our formula requires it in joules. Remember the conversion: 1 eV = \(1.60218 \times 10^{-19} \) J.

Let's break down the calculation:

• Determine energy in joules using the given formula.
• Plug h (Planck constant = \(6.626 \times 10^{-34}\) Js) and c (speed of light = \(3.00 \times 10^8\) m/s) into the formula.
• Calculate λ to get the wavelength in meters or nanometers.

Understanding this calculation helps solve problems related to the behavior and properties of light. We can figure out what part of the electromagnetic spectrum a particular light belongs to by knowing its wavelength.