1.

If the incidentwavelength on a photodiode is 1700 nm, what is its energy gap (E_(g)) ?

Answer»

0.073 eV
1.20 eV
0.73 eV
1.16 eV

Solution :0.73 eV
`E_(G)=(hc)/(LAMBDA)`
`=(6.62xx10^(-34)xx3xx10^(8))/(17xx10^(-7)xx1.6xx10^(-19))=0.73014~~0.73eV`


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