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If the incidentwavelength on a photodiode is 1700 nm, what is its energy gap (E_(g)) ? |
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Answer» 0.073 eV `E_(G)=(hc)/(LAMBDA)` `=(6.62xx10^(-34)xx3xx10^(8))/(17xx10^(-7)xx1.6xx10^(-19))=0.73014~~0.73eV` |
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