Question

An LED is constructed from a p-n junction diode using GaAsP. The energy gap is 1.9 eV. The wavelength of the light emitted will be equal to

• A. \( 10.4 \times 10^{-6} \, \text{m} \)
• B. 654 nm
• C. 654 \( \text{Å} \)
• D. \( 654 \times 10^{-11} \, \text{m} \)

Hint:

To calculate the wavelength of light emitted from an LED, use the relation \( E_g = \frac{hc}{\lambda} \) and convert the energy gap from eV to joules before calculating.

Answer 1

The Correct Option is B

To find the wavelength of light emitted by an LED made from a GaAsP p-n junction diode with an energy gap of 1.9 eV, we use the relation between energy (E) and wavelength (λ):

\[ E = \frac{hc}{\lambda} \]

Where:

• \( E \) is the energy of the photon (1.9 eV),
• \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{J s} \)),
• \( c \) is the speed of light (\( 3 \times 10^8 \, \text{m/s} \)),
• \( \lambda \) is the wavelength.

Since 1 eV = \( 1.602 \times 10^{-19} \, \text{J} \), convert 1.9 eV to joules:

\( E = 1.9 \times 1.602 \times 10^{-19} = 3.044 \times 10^{-19} \, \text{J} \)

Rearrange the energy-wavelength equation to solve for λ:

\[ \lambda = \frac{hc}{E} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{3.044 \times 10^{-19}} \]

Calculate λ:

\[ \lambda = \frac{1.9878 \times 10^{-25}}{3.044 \times 10^{-19}} \approx 6.54 \times 10^{-7} \, \text{m} \]

Convert meters to nanometers (1 meter = \( 10^9 \) nanometers):

\( \lambda \approx 654 \, \text{nm} \)

Thus, the wavelength of light emitted is 654 nm.

Answer 2

The energy gap \( E_g \) of the LED is given as 1.9 eV. The wavelength \( \lambda \) of the emitted light is related to the energy gap \( E_g \) by the equation: \[ E_g = \frac{hc}{\lambda} \] where: 
- \( h \) is Planck’s constant (\( 6.626 \times 10^{-34} \, \text{J·s} \)), 
- \( c \) is the speed of light (\( 3 \times 10^8 \, \text{m/s} \)), 
- \( \lambda \) is the wavelength of the emitted light. We also need to convert the energy gap from eV to joules. 
Using the conversion factor \( 1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J} \), the energy gap in joules is: \[ E_g = 1.9 \, \text{eV} = 1.9 \times 1.602 \times 10^{-19} \, \text{J} = 3.05 \times 10^{-19} \, \text{J} \] Now, using the formula for the wavelength: \[ \lambda = \frac{hc}{E_g} \] Substitute the known values: \[ \lambda = \frac{(6.626 \times 10^{-34}) \times (3 \times 10^8)}{3.05 \times 10^{-19}} = 6.54 \times 10^{-7} \, \text{m} = 654 \, \text{nm} \] 
Thus, the wavelength of the emitted light is 654 nm. Therefore, the correct answer is: \[ \text{(B) } 654 \, \text{nm} \]