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Physics Question on Semiconductor electronics: materials, devices and simple circuits

Conductivity of a photodiode starts changing only if the wavelength of incident light is less than 660 nm. The band gap of the photodiode is found to be X8 eV.\frac{X}{8} \, \text{eV}.The value of XX is:\text{(Given, h=6.6×10−34 Jsh = 6.6 \times 10^{-34} \, \text{Js}, e=1.6×10−19 Ce = 1.6 \times 10^{-19} \, \text{C})}

A

15

B

11

C

13

D

21

Answer

15

Explanation

Solution

Given: - Wavelength of light, λ=660 nm=660×10−9 m\lambda = 660 \, \text{nm} = 660 \times 10^{-9} \, \text{m} - Planck's constant, h=6.6×10−34 Jsh = 6.6 \times 10^{-34} \, \text{Js} - Charge of an electron, e=1.6×10−19 Ce = 1.6 \times 10^{-19} \, \text{C}

Step 1: Calculating the Energy of the Incident Photon

The energy EE of a photon is given by:

E=hcλE = \frac{hc}{\lambda}

where cc is the speed of light, c=3×108 m/sc = 3 \times 10^8 \, \text{m/s}. Substituting the given values:

E=6.6×10−34×3×108660×10−9 JE = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{660 \times 10^{-9}} \, \text{J}

Simplifying:

E=19.8×10−26660×10−9 JE = \frac{19.8 \times 10^{-26}}{660 \times 10^{-9}} \, \text{J} E=3×10−19 JE = 3 \times 10^{-19} \, \text{J}

Step 2: Converting Energy to Electron Volts

To convert the energy from joules to electron volts (eV), we use:

1 eV=1.6×10−19 J1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}

Thus:

E=3×10−191.6×10−19 eVE = \frac{3 \times 10^{-19}}{1.6 \times 10^{-19}} \, \text{eV} E=1.875 eVE = 1.875 \, \text{eV}

Step 3: Finding the Value of XX

Given that the band gap of the photodiode is X8 eV\frac{X}{8} \, \text{eV}:

X8=1.875\frac{X}{8} = 1.875

Solving for XX:

X=1.875×8X = 1.875 \times 8 X=15X = 15

Conclusion:

The value of XX is 15.