Question

The band gap in aluminum phosphide (AIP) is 2.5 electronvolts \(\left(1 \mathrm{eV}=1.6 \times 10^{-19} \mathrm{J}\right) .\) What wavelength of light is emitted by an AlP diode?

Short answer

The wavelength of light emitted by an aluminum phosphide (AlP) diode can be calculated using the Planck's constant formula and converting the given band gap energy of 2.5 eV to Joules. After solving the formula, we find that the wavelength is approximately \(497 \: \text{nm}\).

Step-by-step solution

Convert energy from electronvolts to Joules

We are given that the band gap of AlP is 2.5 eV. Since 1 eV is equal to \(1.6 \times 10^{-19}\) Joules, let's convert the energy to Joules: \(2.5 \text{ eV} \times \frac{1.6 \times 10^{-19}\: \text{J}}{1 \:\text{eV}} = 4 \times 10^{-19}\: \text{J}\) So, the energy in Joules is \(4 \times 10^{-19}\) J.

Apply the Planck's constant formula

The formula that relates energy and wavelength is given by Planck's constant (\(h\)) and the speed of light (\(c\)) as follows: \(E = \frac{hc}{\lambda}\) where, \(E\) = energy in Joules, \(h\) = Planck's constant (\(6.626 \times 10^{-34}\: \text{J s}\)), \(c\) = speed of light (\(3 \times 10^8\: \text{m/s}\)), \(\lambda\) = wavelength in meters. We need to find the wavelength (\(\lambda\)). We can rearrange the formula to solve for \(\lambda\): \(\lambda = \frac{hc}{E}\)

Calculate the wavelength

Now, let's plug the known values into the formula and solve for the wavelength: \(\lambda = \frac{(6.626 \times 10^{-34}\: \text{J s}) \times (3 \times 10^8\: \text{m/s})}{(4 \times 10^{-19}\: \text{J})}\) \(\lambda = \frac{(1.988 \times 10^{-25}\: \text{J m})}{(4 \times 10^{-19}\: \text{J})}\) \(\lambda = 4.97 \times 10^{-7}\: \text{m}\)

Convert the wavelength to nanometers

Finally, we'll convert the wavelength from meters to nanometers for easier interpretation. There are \(10^9\) nanometers in 1 meter, so: \(4.97 \times 10^{-7}\: \text{m} \times \frac{10^9\: \text{nm}}{1\: \text{m}} = 497\: \text{nm}\) The wavelength of light emitted by an aluminum phosphide (AlP) diode is approximately 497 nm.

Key concepts

Electronvolts to Joules Conversion

The conversion from electronvolts to Joules is a fundamental concept in understanding energy at the atomic and subatomic levels. An electronvolt (eV) is a unit of energy equal to the amount of kinetic energy gained by a single electron when it accelerates through an electric potential difference of one volt. In terms of Joules, the SI unit of energy, 1 eV is defined as exactly 1.6 x 10^-19 Joules.

When working with photon energies and the band gaps of materials in physics, it's common to use eV because the values are more manageable. For instance, the band gap in aluminum phosphide (AlP) is given as 2.5 eV. To convert this to Joules, simply multiply by the conversion factor:\[2.5 \text{ eV} \times \dfrac{1.6 \times 10^{-19}\: \text{J}}{1 \:\text{eV}} = 4 \times 10^{-19}\: \text{J}\].

It's important to remember this conversion factor, as it links the intuitive concept of energy in electronvolts to the more formally defined energy in Joules, making it easier to apply universal physics equations such as Planck's relation or calculations involving the speed of light.

Planck's Constant Formula

Planck's constant is a physical constant that is essential to the field of quantum mechanics. Represented by the symbol \(h\), it relates the energy \(E\) of a photon to its frequency \(u\) through the equation:\[E = hu\].

However, since the frequency of light is related to its wavelength \(\lambda\) by the speed of light \(c\), (where \(u = \frac{c}{\lambda}\)), Planck's constant can also be used to connect energy and wavelength:\[E = \frac{hc}{\lambda}\].

Planck's constant has a value of approximately 6.626 x 10^-34 Joules seconds (J s). This formula is fundamental when we talk about the energy of photons emitted or absorbed by atoms – for example, in LEDs or lasers. When we solve for the wavelength, we're effectively using Planck's constant to translate between the microscopic world of atoms and the macroscopic world of observable light. This relationship illuminates why certain materials emit specific colors of light, which is defined by the material's band gap energy.

Wavelength Calculation

The wavelength of light is directly tied to its energy, and we can determine one if we know the other. The formula derived from Planck's constant involves the formula:\[\lambda = \frac{hc}{E}\],

where \(\lambda\) is the wavelength, \(h\) is Planck's constant, \(c\) is the speed of light, and \(E\) is the energy of the photon. To understand how this works, consider a diode emitting light with a known band gap energy. For aluminum phosphide (AlP), with a band gap of 2.5 eV, once we've converted this energy into Joules, we can plug the value into the above expression to find the emitted light's wavelength.For instance, the calculated energy for AlP in Joules was \(4 \times 10^{-19}\) J. By substituting this into the formula along with Planck's constant and the speed of light, we can find the respective wavelength.After performing the calculation, we might convert the wavelength into more useful units like nanometers (nm), given that 1 meter equals 1 billion nanometers (10^9 nm). This makes the result much more relatable since the visible spectrum ranges from approximately 380 to 750 nm.