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

The band gap energy, represented by \(X\), of the photodiode is determined by the wavelength of light at which its conductivity begins to change, specifically when illuminated with light having a wavelength less than 660 nm.

The relationship between photon energy \(E\) and its wavelength \(\lambda\) is defined by the equation:

\(E = \frac{hc}{\lambda}\)

The constants and values used are:

• Planck's constant, \(h = 6.6 \times 10^{-34} \, \text{Js}\).
• The speed of light in a vacuum, \(c = 3 \times 10^8 \, \text{m/s}\).
• The incident light wavelength, \(\lambda = 660 \times 10^{-9} \, \text{m}\).

Substituting these values into the formula yields:

\(E = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{660 \times 10^{-9}}\)

The calculation proceeds as follows:

\(E = \frac{6.6 \times 3}{660} \times 10^{-34 + 8 + 9} \, \text{eV}\)

Simplification results in:

\(E = \frac{6.6 \times 3}{660} \times 10^{-17} \, \text{eV}\)

\(E = \frac{19.8}{660} \times 10^{-17} \, \text{eV}\)

\(E \approx 3.0 \times 10^{-19} \, \text{J}\)

To convert the energy from joules to electron volts, the conversion factor \(1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}\) is applied:

\(E = \frac{3.0 \times 10^{-19}}{1.6 \times 10^{-19}} \, \text{eV}\)

\(E \approx 1.875 \, \text{eV}\)

Given that the band gap \(E_g\) is expressed as \(\frac{X}{8} \, \text{eV}\), we set the calculated energy equal to this expression to solve for \(X\):

\(\frac{X}{8} = 1.875\)

Solving for \(X\):

\(X = 1.875 \times 8 = 15\)

The value of \(X\) is determined to be 15. The correct answer is 15.