A photodiode is made froma semiconductor In_(0.53) Gu_(0.47)A_(s), with E_(g)=0.73eV.What is the maximum wavelength which it can detect ?h=6.63xx10^(-34)J_(s)

A photodiode is made froma semiconductor In_(0.53) Gu_(0.47)A_(s), wit

1.	A photodiode is made froma semiconductor In_(0.53) Gu_(0.47)A_(s), with E_(g)=0.73eV.What is the maximum wavelength which it can detect ?h=6.63xx10^(-34)J_(s)
Answer» Solution :The maximum wavelength`(LAMBDA)`, which a photodiode , which a photodiode can detect corresponds to the energy `E_(g).So h_(c)/lambda=E_(g)`,or`lambda=h_(c)/E_(g)=(6.63xx10^(-34)XX (3xx10^(8)))/(0.73xx1.6xx10^(-19))`=`1703xx 10^(-9)m=1703nm`

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A photodiode is made froma semiconductor In_(0.53) Gu_(0.47)A_(s), with E_(g)=0.73eV.What is the maximum wavelength which it can detect ?h=6.63xx10^(-34)J_(s)

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