Chapter 18, Problem 18.33QA

Interpretation Introduction

To find:

The energies corresponding to radiation in the visible region of the spectrum for the given nitride ceramics.

Answer to Problem 18.33QA

Solution:

$InN$

Explanation of Solution

1) Concept:

To find the wavelength corresponding to the visible regions, we need to use the $E_g$ absorbed by the semiconductor. We need to convert the given $E_g$ into Joule by using an appropriate unit conversion factor. We can calculate the wavelengthby using the $E_g$ band gap and the formula.  The wavelength of radiation in visible region is $400 nm \mathrm{t}\mathrm{o} 800 nm$.

2) Formula:

$E_g=h\times \frac{ C}{\lambda }$             

3) Given:

i) $E_{g\left(AlN\right)}=580.6 \frac{kJ}{mol}$

ii) $E_{g\left(GaN\right)}=322.1 \frac{kJ}{mol}$

iii) $E_{g\left(InN\right)}=192.9 \frac{kJ}{mol}$

iv) $h=6.626 \times {10}^{-34} J.s$

v) $c= 3 \times {10}^8 \frac{m}{s}$

vi) Avogadro’s number, $1 mol=6.022 \times {10}^{23} photons$

4) Calculations:

a) The energy gap is the minimum energy needed to excite the electrons from the ground state for the conduction. From this energy, we can calculate the light energy needed to excite the electrons from their ground state to conduction band.

First we will calculate the energy, J/photons, by using Avogadro’s number.

$580.6 \frac{kJ}{mol}\times \frac{1000 J}{1 kJ} \times \frac{1 mol }{6.022 \times {10}^{23} photons}=9.6413 \times {10}^{-19} \frac{J}{photons}$

$E_g per photon= 9.6413 \times {10}^{-19} J$

Wavelength of $AlN$ is calculated as follows:

$E_g=h\times \frac{ c}{\lambda }$

$\lambda = \frac{h\times c}{E_g}$

$\lambda = \frac{\left( 6.626 \times {10}^{-34} J.s\right) \times \left(3 \times {10}^8 \frac{m}{s}\right)}{9.6413 \times {10}^{-19} J} = 2.06 \times {10}^{-7} m$

$\lambda =2.06 \times {10}^{-7} m \times \frac{1 \times {10}^{9}nm}{1 m} =206 nm$

$\lambda =206 nm$

Since the wavelength of $AlN$ is less than $400 nm,$ the energy of $AlN$ does not correspond to the radiation in the visible region of the spectrum.

b) Wavelength of $GaN$.

First we will calculate the energy, J/photons, by using Avogadro’s number.

$322.1 \frac{kJ}{mol}\times \frac{1000 J}{1 kJ} \times \frac{1 mol }{6.022 \times {10}^{23} photons}=5.349 \times {10}^{-19} \frac{J}{photons}$

$E_g per photon= 5.349 \times {10}^{-19} J$

Wavelength of $AlN$ is calculated as follows:

$E_g=h\times \frac{ c}{\lambda }$

$\lambda = \frac{h\times c}{E_g}$

$\lambda = \frac{\left( 6.626 \times {10}^{-34} J.s\right) \times \left(3 \times {10}^8 \frac{m}{s}\right)}{5.349 \times {10}^{-19} J} = 3.72 \times {10}^{-7} m$

$\lambda =3.72 \times {10}^{-7} m \times \frac{1 \times {10}^{9}nm}{1 m} =372nm$

$\lambda =372 nm$

Since the wavelength of $GaN$ is less than $400 nm,$ the energy of $GaN$ does not correspond to the radiation in the visible region of the spectrum.

c) Wavelength of $InN$ is calculated as follows,

First we will calculate the energy, J/photons, by using Avogadro’s number.

$192.9 \frac{kJ}{mol}\times \frac{1000 J}{1 kJ} \times \frac{1 mol }{6.022 \times {10}^{23} photons}=3.203 \times {10}^{-19} \frac{J}{photons}$

$E_g per photon= 3.203 \times {10}^{-19} J$

Wavelength of $AlN$ is calculated as follows:

$E_g=h\times \frac{ c}{\lambda }$

$\lambda = \frac{h\times c}{E_g}$

$\lambda = \frac{\left( 6.626 \times {10}^{-34} J.s\right) \times \left(3 \times {10}^8 \frac{m}{s}\right)}{3.203 \times {10}^{-19} J} = 6.21 \times {10}^{-7} m$

$\lambda =6.21 \times {10}^{-7} m \times \frac{1 \times {10}^{9}nm}{1 m} =621 nm$

$\lambda =621 nm$

Since the wavelength of $InN$ is between $400 nm and 800 nm$, the energy of $InN$ corresponds to the radiation in the visible region of the spectrum.

Conclusion:

The wavelength is calculated from the energy gaps present in the semiconductors and Avogadro’s value.