The semiconductor CdSe has a band gap of 1.74 eV. What wavelength of light would be emitted from an LED made from CdSe? What region of the electromagnetic spectrum does this correspond to?

Verified step by step guidance

1. Step 1: Understand the relationship between energy and wavelength. The energy of a photon is related to its wavelength by the equation: \( E = \frac{hc}{\lambda} \), where \( E \) is the energy in electron volts (eV), \( h \) is Planck's constant (\( 4.1357 \times 10^{-15} \) eV·s), \( c \) is the speed of light (\( 3.00 \times 10^8 \) m/s), and \( \lambda \) is the wavelength in meters.
2. Step 2: Rearrange the equation to solve for wavelength \( \lambda \). The equation becomes: \( \lambda = \frac{hc}{E} \).
3. Step 3: Substitute the given values into the equation. Use \( E = 1.74 \) eV, \( h = 4.1357 \times 10^{-15} \) eV·s, and \( c = 3.00 \times 10^8 \) m/s.
4. Step 4: Calculate the wavelength \( \lambda \) in meters. After substituting the values, perform the calculation to find \( \lambda \).
5. Step 5: Convert the wavelength from meters to nanometers (1 m = 10^9 nm) and determine the region of the electromagnetic spectrum it corresponds to. Typically, visible light ranges from about 400 nm to 700 nm.