Question

If an LED has to emit 662nm wavelength of light, then what should be the band gap energy of its semiconductor? h=6.62×10⁻³⁴Js

Answer 1

The band gap energy of a semiconductor that emits light at a wavelength of 662 nm is approximately 1.87 electronvolts (eV), calculated using the relationship between energy, Planck's constant, and wavelength.

Step-by-step explanation:

To determine the band gap energy of a semiconductor that emits light at a wavelength of 662nm, we can use the equation E = hc/λ, where E is the energy in joules, h is Planck's constant (6.626 x 10-34 J·s), c is the speed of light in a vacuum (3.00 x 108 m/s), and λ is the wavelength in meters.

Converting the wavelength from nanometers to meters (662nm = 662 x 10-9 m), and substituting the values into the equation, we get:

E = (6.626 x 10-34 J·s) x (3.00 x 108 m/s) / (662 x 10-9 m) ≈ 3.0 x 10-19 J

To convert this energy from joules to electronvolts (eV), where 1 eV = 1.602 x 10-19 J, we do the following:

E (eV) = 3.0 x 10-19 J / (1.602 x 10-19 J/eV) ≈ 1.87 eV

Therefore, the band gap energy of the semiconductor should be approximately 1.87 eV for it to emit light at a wavelength of 662 nm.