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A `p-n` photodiode is fabricated from a semiconductor with a band gap of `2.5 eV`. It can detect a signal of wavelengthA. 6000 ÅB. 4000 nmC. 6000 nmD. 4960 Å |
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Answer» Correct Answer - D Only signal having wavelength less than threshold wavelength will be detercted. Energy `E=hv=h(c)/(lamda)implieslamda=(hc)/(E)` Substituting the value of h, c and E in the above equation `lamda=(6.6xx10^(-34)xx3xx10^(8))/(2.5xx1.6xx10^(-19))=5000Å` As `4000Ålt5000Å` Signal of wavelength `4000Å` can be detected by the photodiode |
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