Question

Question: The photons emitted by an LED arise from the energy given up in electron-hole recombinations across the energy gap. How large should the energy gap be to give photons at the red end of the visible spectrum (700nm) ?

Short answer

Answer

The energy gap to give photons at the red end of the visible spectrum is 1.8 eV .

Step-by-step solution

Given data 

Wavelength of emitted photons, $\lambda =700\text{nm}$.

Concept of photon energy 

Energy corresponding to wavelength of photon is,$\mathit{E}\mathbf{=}\frac{\mathbf{h}\mathbf{c}}{\mathbf{\lambda }}$ .

Where, h is Planck's constant $\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{64}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{34}}\mathbf{\text{J}}\mathbf{·}\mathbf{\text{s}}$.

C Is velocity of light$\mathbf{=}\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{8}}\mathbf{\text{m}}\mathbf{/}\mathbf{\text{s}}$ .

Step 3:Determine the energy gap to give photons at the red end of the visible spectrum

Substitute numerical values in photon energy equation.

$$\begin{gathered} E=\frac{6.64\times {10}^{-34}\text{J}·\text{s}\times 3\times {10}^8\text{m}/\text{s}}{700\times {10}^{-9}\text{m}} \\ =0.0284\times {10}^{-17}\text{J} \\ =1.8\text{eV} \end{gathered}$$

The energy gap to give photons at the red end of the visible spectrum is 1.8eV

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