Three photodiodes D_1 , D_2 and D_3 are made of semiconductors having band gaps of 2.5 eV , 2 eV and 3 eV , respectively . Which one will be able to detect light of wavelength 6000 Ã ?
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The photo diode D3 will be able to detect light of wave length 6000 Ã.
- Let us go through the given parameters in problem
Band gap of D1 = 2.5 eV
Band gap of D2 = 2 eV
Band gap of D3 = 3 eV
- The maximum WAVE LENGTH of diode is given by = hc/E
where h = 6.6 x 10⁻³⁴
c = 3 x 10⁸
- Maximum wavelngth of D1 = hc/E
=6.6 x 10⁻³⁴ x 3 x 10⁸/ 2.5 x 1.6 x 10⁻¹⁹
=4.95x 10⁻⁷m
- Maximum wavelngth of D2 = hc/E
=6.6 x 10⁻³⁴ x 3 x 10⁸/ 2 x 1.6 x 10⁻¹⁹
=6.18 x 10⁻⁷ m
- Maximum wavelngth of D3 =hc/E
=6.6 x 10⁻³⁴ x 3 x 10⁸/ 3 x 1.6 x 10⁻¹⁹
= 4.125 x 10⁻⁷m
- For detection of optical signals wavelength of incident ray must be greater which is possible in case of diode D3.So it will be able to detect light
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