
An electron hole pair is formed when light of maximum wavelength \[6000\,A^\circ \] is incident on the semiconductor. What is the band gap energy of the semiconductor? \[\left( {h = 6.62 \times {{10}^{ - 34}}\,J - s} \right)\]
A. \[3.31 \times {10^{ - 19}}\,J\]
B. \[3.07 \times {10^{ - 19}}\,J\]
C. \[2.07 \times {10^{ - 19}}\,J\]
D. \[2.07\,J\]
Answer
505.2k+ views
Hint:In solid-state physics, a band gap, also known as an energy gap, is an energy spectrum in which no electronic states can exist. The band gap is an important factor in deciding a solid's electrical conductivity.
Complete step by step answer:
A semiconductor is a material with a non-zero band gap that acts as an insulator at absolute zero but enables thermal excitation of electrons into its conduction band below its melting point. A substance with a large band gap, on the other hand, is an insulator. The valence and conduction bands in conductors may overlap, so there is no band difference.
The band gap has a major impact on the conductivity of intrinsic semiconductors. The electrons with enough thermal energy to be excited through the band gap and the electron holes that are left off when such an excitation occurs are the only charge carriers required for conduction.
In the question it is given that, \[\lambda = 6000A^\circ = 6000 \times {10^{ - 10}}\,m\] and \[h = 6.62 \times {10^{ - 34}}J - s\] kg/s.
We are finding the bond energy gap using the formula \[E = \dfrac{{hc}}{\lambda }\] .
We know that \[c = 3 \times {10^8}\] m/s. That is,
\[E = \dfrac{{6.62 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{6000 \times {{10}^{ - 10}}}}\]
\[ \Rightarrow E = \dfrac{{19.86}}{{6 \times {{10}^{ - 7}}}} \times {10^{ - 19}}\]
\[ \therefore E = 3.31 \times {10^{ - 19}}\,J\]
So, the answer is option A.
Note:Remember the formula \[E = \dfrac{{hc}}{\lambda }\]. With increasing temperature, the band-gap energy of semiconductors tends to decrease. It also depends on the amount of pressure.The method of regulating or altering a material's band gap by controlling the composition of some semiconductor alloys, such as GaAlAs, InAlAs, and InGaAs, is known as band-gap engineering. Techniques like molecular-beam epitaxy can also be used to create layered materials with alternating compositions. These techniques are used to create heterojunction bipolar transistors, solar cells and laser diodes.
Complete step by step answer:
A semiconductor is a material with a non-zero band gap that acts as an insulator at absolute zero but enables thermal excitation of electrons into its conduction band below its melting point. A substance with a large band gap, on the other hand, is an insulator. The valence and conduction bands in conductors may overlap, so there is no band difference.
The band gap has a major impact on the conductivity of intrinsic semiconductors. The electrons with enough thermal energy to be excited through the band gap and the electron holes that are left off when such an excitation occurs are the only charge carriers required for conduction.
In the question it is given that, \[\lambda = 6000A^\circ = 6000 \times {10^{ - 10}}\,m\] and \[h = 6.62 \times {10^{ - 34}}J - s\] kg/s.
We are finding the bond energy gap using the formula \[E = \dfrac{{hc}}{\lambda }\] .
We know that \[c = 3 \times {10^8}\] m/s. That is,
\[E = \dfrac{{6.62 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{6000 \times {{10}^{ - 10}}}}\]
\[ \Rightarrow E = \dfrac{{19.86}}{{6 \times {{10}^{ - 7}}}} \times {10^{ - 19}}\]
\[ \therefore E = 3.31 \times {10^{ - 19}}\,J\]
So, the answer is option A.
Note:Remember the formula \[E = \dfrac{{hc}}{\lambda }\]. With increasing temperature, the band-gap energy of semiconductors tends to decrease. It also depends on the amount of pressure.The method of regulating or altering a material's band gap by controlling the composition of some semiconductor alloys, such as GaAlAs, InAlAs, and InGaAs, is known as band-gap engineering. Techniques like molecular-beam epitaxy can also be used to create layered materials with alternating compositions. These techniques are used to create heterojunction bipolar transistors, solar cells and laser diodes.
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