Chapter 29, Problem 22P

To determine

The wavelength at which LED radiate.

Answer to Problem 22P

  $0.89\mu m$ .

Explanation of Solution

Given:

Energy gap is $1.4eV$

Formula used:

It is known that $\lambda =\frac{c}{f}$

Where,

  $\lambda$ is wavelength

  $c$ is speed of light, $3.00\times {10}^{8}m/s$

f is frequency

The gap energy equals, $E_g=hf=\frac{hc}{\lambda }$

Where,

  $E_g$ is energy gap

f is frequency.

  $\lambda$ is wavelength

  $hf$ is emitted photon

  $h$ is Planck’s constant, $6.63\times {10}^{-34}J\cdot s$

  $c$ is speed of light, $3.00\times {10}^{8}m/s$

Calculation:

The photon will have an energy equal to the energy gap:

  $\lambda =\frac{c}{f}=\frac{hc}{hf}=\frac{hc}{E_g}$

  $=\frac{\left(6.63\times {10}^{-34}J\cdot s\right)\left(3.00\times {10}^{8}m/s\right)}{\left(1.4eV\right)}$

  $=\frac{\left(1.24\times {10}^{3}eV\cdot nm\right)}{\left(1.4eV\right)}$

  $=8.9\times {10}^{2}nm=0.89\mu m$

Conclusion: The required value is $0.89\mu m$ .