Question

A p-n junction diode is manufactured from a semiconductor with band gap of 3.1 eV. Which of the following wavelengths can be detected by it?
$
  {\text{A}}{\text{. 4000}}\mathop {\text{A}}\limits^0 \\
  {\text{B}}{\text{. 3900}}\mathop {\text{A}}\limits^0 \\
  {\text{C}}{\text{. 4200}}\mathop {\text{A}}\limits^0 \\
  {\text{D}}{\text{. Both A & B}} \\
$

Answer 1

Hint: The band gap of a semiconductor represents the minimum amount of energy required to jump an electron from valence band to the conduction band. The energy of a photon is given as the product of the Planck’s constant and the frequency of the incident light and the frequency is equal to the ratio of speed of light and the wavelength of light. Using this formula, we can obtain the required answer.

Formula used:
The band gap energy is given as
${E_g} = h\nu = \dfrac{{hc}}{\lambda }$
The values of constants are
$
  h = 6.626 \times {10^{23}}Js \\
  c = 3 \times {10^8}m/s \\
$

Complete answer:
The band gap between the valence band and the conduction band of a semiconductor is known as the band gap of the semiconductor. The band gap is very large in the case of insulators; for semiconductors, it lies between the range of insulators while in the case of conductors the band gap is almost or completely absent as the valence and conduction bands overlap each other.
Now, to solve the above question, we will require to use the very basic formula for the band gap energy in a semiconductor which is
${E_g} = h\nu = \dfrac{{hc}}{\lambda }$
Here h is known as the Planck’s constant, $\nu $ is the minimum frequency and $\lambda $ is the maximum wavelength at which the excitation from valence band to the conduction band occurs and the diode can detect these frequencies or wavelengths while c is known as the velocity of light.
We are given the value of the band gap of a p-n junction diode. Its value is given as
${E_g} = 3.1eV$
Now, we can insert this value and values of various constants in our formula. Doing so, we get
$
  3.1 \times 1.6 \times {10^{ - 19}} = \dfrac{{6.626 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{\lambda } \\
  \lambda = \dfrac{{6.626 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{3.1 \times 1.6 \times {{10}^{ - 19}}}} \\
  \lambda = 4.007 \times {10^{ - 7}}m \\
  \lambda = 4007\mathop {\text{A}}\limits^0 \\
$
So, the maximum wavelength that can be detected by the semiconductor is $4007\mathop {\text{A}}\limits^0 $.
Therefore, we can conclude that the semiconductor detect both $4000\mathop {\text{A}}\limits^0 $ and $3900\mathop {\text{A}}\limits^0 $ but not $4200\mathop {\text{A}}\limits^0 $.

So, the correct answer is “Option D”.

Note:
It should be noted that the frequency corresponding to the given energy will be the minimum frequency that can be detected by the diode. This means all the frequencies above this threshold and all the wavelengths below this threshold can be detected by the diode. This behaviour arises due to the inverse relationship between frequency and wavelength.