Question

Three phtodiodes D₁, D₂, D₃ are made of semiconductors having band gaps of 2.5 eV, 2 eV and 3 eV respectively. Which one will be able to detect light of wavelength 600 nm?

• A. D₁ only
• B. Both D₁ and D₃
• C. D₂ only
• D. All the three diodes

Answer 1

The Correct Option is C

To determine which photodiode can detect light of wavelength 600 nm, we need to compare the energy of the photon with the band gap energy of each photodiode.

1. Calculate the energy of a photon with a wavelength of 600 nm:

We can use the formula: $E = \frac{hc}{\lambda}$

where:

$E$ is the energy of the photon
$h$ is Planck's constant ($6.626 \times 10^{-34} \, \text{Js}$)
$c$ is the speed of light ($3 \times 10^8 \, \text{m/s}$)
$\lambda$ is the wavelength ($600 \, \text{nm} = 600 \times 10^{-9} \, \text{m}$)

$E = \frac{(6.626 \times 10^{-34} \, \text{Js}) \times (3 \times 10^8 \, \text{m/s})}{(600 \times 10^{-9} \, \text{m})}$
$E \approx 3.313 \times 10^{-19} \, \text{J}$

Now, convert Joules to electron volts (eV):
$1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J}$
$E \, (\text{in eV}) = \frac{(3.313 \times 10^{-19} \, \text{J})}{(1.602 \times 10^{-19} \, \text{J/eV})}$
$E \approx 2.07 \, \text{eV}$

2. Compare the photon energy with the band gap energies of the photodiodes:

$D_1$: Band gap = $2.5 \, \text{eV}$
$D_2$: Band gap = $2.0 \, \text{eV}$
$D_3$: Band gap = $3.0 \, \text{eV}$

For a photodiode to detect the light, the photon energy must be greater than or equal to the band gap energy of the semiconductor material. If the photon energy is less than the band gap, the photon will not have enough energy to excite an electron across the band gap and create an electron-hole pair, which is the basis of photodetection.

$D_1$: Photon energy ($2.07 \, \text{eV}$) < Band gap ($2.5 \, \text{eV}$) --> Cannot detect
$D_2$: Photon energy ($2.07 \, \text{eV}$) > Band gap ($2.0 \, \text{eV}$) --> Can detect
$D_3$: Photon energy ($2.07 \, \text{eV}$) < Band gap ($3.0 \, \text{eV}$) --> Cannot detect

3. Conclusion:

Only photodiode $D_2$ has a band gap energy ($2.0 \, \text{eV}$) less than the photon energy ($2.07 \, \text{eV}$), so only $D_2$ will be able to detect the light with a wavelength of 600 nm.

Therefore, the answer is: $D_2$ only

Answer 2

Explanation: To determine which photodiode can detect light of wavelength 600 nm:

1. Calculate photon energy (E): 
   \( E = \frac{hc}{\lambda} \), where \( hc = 1240 \, eV·nm \), \( \lambda = 600 \, \text{nm} \). 
   \( E = \frac{1240}{600} \approx 2.07 \, \text{eV} \).
2. Compare with band gaps:
   • D₁ (2.5 eV): Photon energy (2.07 eV) < band gap → No detection.
   • D₂ (2 eV): 2.07 eV > band gap → Detection possible.
   • D₃ (3 eV): 2.07 eV < band gap → No detection.

Only D₂ (2 eV band gap) can detect 600 nm light.