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An LED is constructed from a p-n junction diode using GaAsP. The energy gap is 1.9 eV1.9 \text{ eV}. The wavelength of the light emitted will be equal to:

A

10.4×10−26 m10.4 \times 10^{-26} \text{ m}

B

654 nm654 \text{ nm}

C

654 m654 \text{ m}

D

654×10−11 m654 \times 10^{-11} \text{ m}

Step-by-Step Solution

Given: Energy gap, Eg=1.9 eVE_g = 1.9 \text{ eV} We know that the wavelength λ\lambda of the emitted light is related to the energy gap by the equation: Eg=hcλE_g = \frac{hc}{\lambda} λ=hcEg\lambda = \frac{hc}{E_g} Substituting the standard values (hc≈1240 eV⋅nmhc \approx 1240 \text{ eV}\cdot\text{nm} or more precisely 1242 eV⋅nm1242 \text{ eV}\cdot\text{nm}): λ=1242 eV⋅nm1.9 eV≈653.68 nm≈654 nm\lambda = \frac{1242 \text{ eV}\cdot\text{nm}}{1.9 \text{ eV}} \approx 653.68 \text{ nm} \approx 654 \text{ nm}

Alternatively, using standard SI units: h=6.63×10−34 J sh = 6.63 \times 10^{-34} \text{ J s} c=3×108 m/sc = 3 \times 10^8 \text{ m/s} Eg=1.9 eV=1.9×1.6×10−19 J=3.04×10−19 JE_g = 1.9 \text{ eV} = 1.9 \times 1.6 \times 10^{-19} \text{ J} = 3.04 \times 10^{-19} \text{ J} λ=6.63×10−34×3×1083.04×10−19\lambda = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{3.04 \times 10^{-19}} λ=19.89×10−263.04×10−19 m=6.54×10−7 m\lambda = \frac{19.89 \times 10^{-26}}{3.04 \times 10^{-19}} \text{ m} = 6.54 \times 10^{-7} \text{ m} λ=654×10−9 m=654 nm\lambda = 654 \times 10^{-9} \text{ m} = 654 \text{ nm}

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