A p–n photodiode is made of a material with a band gap of 2.0 eV. The minimum frequency of the radiation that can be absorbed by the material is nearly :\[1\times 10_{14} Hz\]\[20\times 10_{14} Hz\]\[10\times 10_{14} Hz\]\[5\times 10_{14} Hz\]Solution:Energy of EM wave E = hυ ⇒ 2 eV = 6.626 × 10^-34×υ ⇒ \[v= 5\times 10_{14} Hz\]
Leave a Reply