Question

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 \[\frac{X}{8} \, \text{eV}.\]The value of $X$ is:\text{(Given, $h = 6.6 \times 10^{-34} \, \text{Js}$, $e = 1.6 \times 10^{-19} \, \text{C}$)}

• A. 15
• B. 11
• C. 13
• D. 21

Answer 1

The Correct Option is A

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

Step 1: Calculating the Energy of the Incident Photon

The energy \( E \) of a photon is given by:

\[ E = \frac{hc}{\lambda} \]

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

\[ E = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{660 \times 10^{-9}} \, \text{J} \]

Simplifying:

\[ E = \frac{19.8 \times 10^{-26}}{660 \times 10^{-9}} \, \text{J} \] \[ E = 3 \times 10^{-19} \, \text{J} \]

Step 2: Converting Energy to Electron Volts

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

\[ 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \]

Thus:

\[ E = \frac{3 \times 10^{-19}}{1.6 \times 10^{-19}} \, \text{eV} \] \[ E = 1.875 \, \text{eV} \]

Step 3: Finding the Value of \( X \)

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

\[ \frac{X}{8} = 1.875 \]

Solving for \( X \):

\[ X = 1.875 \times 8 \] \[ X = 15 \]

Conclusion:

The value of \( X \) is 15.