The energy gap of compound semiconductor and the colour emitted by it in forward bias, is
A. 1.9 eV, green
B. 1.9 eV, red
C. 2.2 eV, green
D. 2.2 eV, red
Answer
608.1k+ views
Hint:The above problem is resolved by using the mathematical and conceptual relation for the energy bandgap. The energy band gap is that visible range of the energy in the atomic spectrum, which the energy difference for the highest energy level and the lower energy level is taken into consideration. Moreover, for specific semiconductors, the value of the energy gap is also specified.
Complete step by step answer:
As we know that energy band gap lies between 0.5 eV to 2 eV.
As we are given with the semiconductor \[GaAs\]. So, its standard value for the energy band gap is,
\[E = 1.9\;{\rm{eV}}\]
Now, the expression for the wavelength is given as,
\[\lambda = \dfrac{{hc}}{E}\]
Here, h is the Planck's constant and its value is \[4.136 \times {10^{ - 15}}\;{\rm{eV/s}}\] and c is speed of light and its value \[3 \times {10^{17}}\;{\rm{nm/s}}\].
Solve by substituting the values as,
\[\begin{array}{l}
\lambda = \dfrac{{4.136 \times {{10}^{ - 15}}\;{\rm{eV/s}} \times 3 \times {{10}^{17}}\;{\rm{nm/s}}}}{{1.9\;{\rm{eV}}}}\\
\lambda = \dfrac{{1240.8}}{{1.9}}\;{\rm{nm}}\\
\lambda = 653.05\;{\rm{nm}}
\end{array}\]h
So, this wavelength \[653.05\;{\rm{nm}}\] is between the longest visible range$\left( {625\;{\rm{nm}} - 740\;{\rm{nm}}} \right)$ and it correspond to the red colour.
Therefore, the colour emitted by it in forward bias, is red and energy band is of 1.9 eV and option (B) is correct.
Note: To resolve the given problem, one must understand the concept of the energy band in the atomic spectra. When an electron receives some energy, then it excites from the lower energy level to the higher energy level. The numerical difference for the energy at both the state is then calculated to obtain the desired energy bandgap.
Complete step by step answer:
As we know that energy band gap lies between 0.5 eV to 2 eV.
As we are given with the semiconductor \[GaAs\]. So, its standard value for the energy band gap is,
\[E = 1.9\;{\rm{eV}}\]
Now, the expression for the wavelength is given as,
\[\lambda = \dfrac{{hc}}{E}\]
Here, h is the Planck's constant and its value is \[4.136 \times {10^{ - 15}}\;{\rm{eV/s}}\] and c is speed of light and its value \[3 \times {10^{17}}\;{\rm{nm/s}}\].
Solve by substituting the values as,
\[\begin{array}{l}
\lambda = \dfrac{{4.136 \times {{10}^{ - 15}}\;{\rm{eV/s}} \times 3 \times {{10}^{17}}\;{\rm{nm/s}}}}{{1.9\;{\rm{eV}}}}\\
\lambda = \dfrac{{1240.8}}{{1.9}}\;{\rm{nm}}\\
\lambda = 653.05\;{\rm{nm}}
\end{array}\]h
So, this wavelength \[653.05\;{\rm{nm}}\] is between the longest visible range$\left( {625\;{\rm{nm}} - 740\;{\rm{nm}}} \right)$ and it correspond to the red colour.
Therefore, the colour emitted by it in forward bias, is red and energy band is of 1.9 eV and option (B) is correct.
Note: To resolve the given problem, one must understand the concept of the energy band in the atomic spectra. When an electron receives some energy, then it excites from the lower energy level to the higher energy level. The numerical difference for the energy at both the state is then calculated to obtain the desired energy bandgap.
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