Question

A GaP–GaAs semiconductor LED display has a band gap of 1.9 eV. The wavelength of emitted light in \( \mu \)m is (rounded off to two decimal places) .........

Hint:

To find the wavelength of emitted light from the energy of the photon, use the equation \( \lambda = \frac{hc}{E} \) and ensure to convert units correctly. Always remember to convert from eV to joules if necessary.

Answer 1

The energy of a photon is related to its wavelength by the equation:
\[ E = \frac{hc}{\lambda} \] Where:
- \( E \) is the energy of the photon (in joules),
- \( h \) is Planck's constant (\( 6.63 \times 10^{-34} \, {J s} \)),
- \( c \) is the speed of light (\( 3 \times 10^8 \, {m/s} \)),
- \( \lambda \) is the wavelength of the emitted light (in meters).
Given that the band gap \( E = 1.9 \, {eV} \), and using the conversion factor \( 1 \, {eV} = 1.6 \times 10^{-19} \, {J} \), the energy of the photon in joules is:
\[ E = 1.9 \times 1.6 \times 10^{-19} = 3.04 \times 10^{-19} \, {J} \] Now, using the photon energy-wavelength relation, we can solve for \( \lambda \):
\[ \lambda = \frac{hc}{E} = \frac{(6.63 \times 10^{-34} \, {J s})(3 \times 10^8 \, {m/s})}{3.04 \times 10^{-19} \, {J}} = 6.55 \times 10^{-7} \, {m} \] To convert this to micrometers (\( \mu m \)), we multiply by \( 10^6 \):
\[ \lambda = 6.55 \times 10^{-7} \times 10^6 = 0.655 \, \mu m \] Thus, the wavelength of the emitted light is approximately \( 0.655 \, \mu m \), which lies between 0.60 and 0.70 micrometers. Therefore, the correct answer is between 0.60 and 0.70.