An optical pulse containing 6 - 106 photons is incident on a photodiode and 4-5 - 106 electron-hole pairs are created- The maximum possible quantum efficiency in - of the photodiode is

Number of incident photons 'N' = 6 × 10^6 Number of electron hole pairs generated n = 4.5 × 10^6 Quantum efficiency of photo diode = nN× 100 = 4.5 × 10^66 × ...

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Standard XII

Physics

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An optical pu...

Question

An optical pulse containing $6 \times 10^{6}$ photons is incident on a photodiode and $4.5 \times 10^{6}$ electron-hole pairs are created. The maximum possible quantum efficiency (in $%$ ) of the photodiode is
75

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Solution

The correct option is A 75
Number of incident photons $^{'} N^{'} = 6 \times 10^{6}$

Number of electron-hole pairs generated

$n = 4.5 \times 10^{6}$

Quantum efficiency of photo-diode
$= \frac{n}{N} \times 100 = \frac{4.5 \times 10^{6}}{6 \times 10^{6}} = 75$

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An optical pulse containing 6×106 photons is incident on a photodiode and 4.5×106 electron-hole pairs are created. The maximum possible quantum efficiency (in %) of the photodiode is75

The correct option is A 75Number of incident photons ′N′=6×106 Number of electron-hole pairs generated n=4.5×106 Quantum efficiency of photo-diode =nN×100=4.5×1066×106=75

An optical pulse containing $6 \times 10^{6}$ photons is incident on a photodiode and $4.5 \times 10^{6}$ electron-hole pairs are created. The maximum possible quantum efficiency (in $%$ ) of the photodiode is
75

The correct option is A 75
Number of incident photons $^{'} N^{'} = 6 \times 10^{6}$

Number of electron-hole pairs generated

$n = 4.5 \times 10^{6}$

Quantum efficiency of photo-diode
$= \frac{n}{N} \times 100 = \frac{4.5 \times 10^{6}}{6 \times 10^{6}} = 75$