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, $h = 6.6 \times 10^{-34} \, \text{Js}$, $e = 1.6 \times 10^{-19} \, \text{C}$)}
- 15
- 11
- 13
- 21
The Correct Option is A
Approach Solution - 1
To determine the value of \(X\), we need to calculate the band gap energy of the photodiode, given that it starts changing its conductivity when illuminated with light whose wavelength is less than 660 nm.
The energy \(E\) of a photon is related to its wavelength \(\lambda\) by the formula:
\(E = \frac{hc}{\lambda}\)
where:
- \(h = 6.6 \times 10^{-34} \, \text{Js}\) is Planck's constant.
- \(c = 3 \times 10^8 \, \text{m/s}\) is the speed of light in a vacuum.
- \(\lambda = 660 \times 10^{-9} \, \text{m}\) is the wavelength of the incident light.
Substituting these values into the formula, we get:
\(E = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{660 \times 10^{-9}}\)
Calculating the above expression gives:
\(E = \frac{6.6 \times 3}{660} \times 10^{-34 + 8 + 9} \, \text{eV}\)
Simplifying gives:
\(E = \frac{6.6 \times 3}{660} \times 10^{-17} \, \text{eV}\)
\(E = \frac{19.8}{660} \times 10^{-17} \, \text{eV}\)
\(E \approx 3.0 \times 10^{-19} \, \text{J}\)
To convert energy from joules to electron volts, use the conversion factor \(1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}\):
\(E = \frac{3.0 \times 10^{-19}}{1.6 \times 10^{-19}} \, \text{eV}\)
\(E \approx 1.875 \, \text{eV}\)
The band gap \(E_g\) is given as \(\frac{X}{8} \, \text{eV}\), so we equate and solve for \(X\):
\(\frac{X}{8} = 1.875\)
Therefore,
\(X = 1.875 \times 8 = 15\)
Hence, the value of \(X\) is 15. Thus, the correct answer is 15.
Approach Solution -2
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.
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