Ncert Exercises & Solutions

Three photodiodes D₁, D₂ and D₃ are made of semiconductors having band gaps of 2.5eV, 2eV and 3eV, respectively. Which ones will be able to detect light of wavelength 6000 Å?

Hint: To detect radiation, its energy should be more than that of the energy bandgap.

Step 2: Find the energy of the radiation.

Given, the wavelength of light

λ = 6000 Å = 6000 \times 10^{- 10} m

The energy of the light photon:

E = \frac{hc}{λ} = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{6000 \times 10^{- 10} \times 1.6 \times 10^{- 19}} eV = 2.06 eV

Step 2: Compare the radiation energy with the energy of the bandgap.

The incident radiation which is detected by the photodiode having energy should be greater than the band-gap. So, it is only valid for diode D₂. Then, diode D₂ will detect this radiation.

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