Advertisements
Advertisements
Question
Answer the following question.
Three photodiodes D1, D2, and D3 are made of semiconductors having band gaps of 2.5 eV, 2 eV, and 3 eV respectively. Which of them will not be able to detect light of wavelength 600 nm?
Solution
The energy of the incident light
`E = (hC)/lambda`
= `((6.6 xx 10^-34) xx (3 xx 10^8))/((600 xx 10^-9)(1.6 xx 10^-19))`
E = 2.06 eV
The incident radiations can be detected by a photodiode if the energy of incident radiation photon is greater than the band gap. This is true only for D2 (2 eV). Hence, only D2 will detect the light of 600 nm wavelength.
RELATED QUESTIONS
In an n-type silicon, which of the following statement is true:
In the case of metals, the valence and conduction bands ______.
In the case of semiconductors, the valence and conduction bands ______.
Doping of semiconductor is the process of ______.
The energy band gap is maximum in
ASSERTION (A): The electrical conductivity of a semiconductor increases on doping.
REASON (R): Doping always increases the number of electrons in the semiconductor.
The energy band gap of semiconducting material to produce violet (wavelength = 4000 Å).
LED is ______ eV. (Round off to the nearest integer)
When a diode is forward biased, it has a voltage drop of 0.5 V. The safe limit of current through the diode is 10 mA. If a battery of emf 1.5 V is used in the circuit, the value of minimum resistance to be connected in series with the diode so that the current does not exceed the safe limit is ______.
Draw an energy band diagram for a p-type semiconductor at T > 0 K.
Draw an energy band diagram for an n-type semiconductor at T > 0 K.
