Question

If an LED has to emit 662 nm wavelength of light Lhen what should be the band gap energy of its semiconductor? [h= 6.62*10⁻³⁴ J S] [Ans: 1.875 eV]

Answer 1

Dear Student,

◆ Answer -

BGE = 1.875 eV

◆ Explaination -

Band gap energy of LED is semiconductor is -

BGE = hc/λ

In electron volts, this is represented as -

BGE = hc/eλ

BGE = (6.62×10^-34 × 3×10^8) / (1.6×10^-19 × 662×10^-9)

BGE = 1.875 eV

Thanks for asking. Hope this helps you...