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:

Remember the shortcut: \(\lambda (\text{nm}) \approx \frac{1240}{E(eV)}\). For \(E=1.9\), \(\lambda \approx 653 \, nm\), which matches our detailed calculation.

Answer 1

Correct Answer: 0.6

Step 1: Relation between energy and wavelength.
The photon energy is: \[ E = \frac{hc}{\lambda} \] where \(h\) = Planck’s constant, \(c\) = speed of light, \(\lambda\) = wavelength.

Step 2: Convert band gap energy to joules.
\[ E = 1.9 \, eV = 1.9 \times (1.6 \times 10^{-19}) \, J \] \[ E = 3.04 \times 10^{-19} \, J \]

Step 3: Solve for wavelength.
\[ \lambda = \frac{hc}{E} \] Substitute values: \[ \lambda = \frac{(6.63 \times 10^{-34})(3 \times 10^8)}{3.04 \times 10^{-19}} \] \[ \lambda = \frac{1.989 \times 10^{-25}}{3.04 \times 10^{-19}} \] \[ \lambda = 6.54 \times 10^{-7} \, m \]

Step 4: Convert to micrometers.
\[ 6.54 \times 10^{-7} \, m = 0.654 \, \mu m \]



Step 5: Round off.
\[ \lambda \approx 0.65 \, \mu m \] Final Answer:
\[ \boxed{0.65 \, \mu m} \]