Respuesta :
Answer:
The intrinsic carrier density of diamond is [tex]1.330\times10^{-26}[/tex].
Explanation:
Given that,
Energy band gape = 5.4 eV
Temperature = 300 K
We need to calculate the intrinsic carrier density of diamond
Using formula of density
[tex]n_{1}^2=N_{c}N_{v}e^{-\dfrac{E_{g}}{kT}}[/tex]....(I)
We need to calculate [tex]N_{c}[/tex]
[tex]N_{c}=2(\dfrac{2\pi m_{n}kT}{h^2})^{\frac{3}{2}}[/tex]
Put the value into the formula
[tex]N_{c}=2(\dfrac{2\pi\times9.1\times10^{-28}\times300\times1.3807\times10^{-16}}{(6.63\times10^{-27})})^{\frac{3}{2}}[/tex]
[tex]N_{c}=2.501\times10^{19}\ cm^{-3}[/tex]
We need to calculate [tex]N_{v}[/tex]
[tex]N_{v}=2(\dfrac{2\pi m_{n}kT}{h^2})^{\frac{3}{2}}[/tex]
[tex]N_{v}=2(\dfrac{2\pi\times9.1\times10^{-28}\times300\times1.3807\times10^{-16}}{(6.63\times10^{-27})})^{\frac{3}{2}}[/tex]
[tex]N_{v}=2.501\times10^{19}\ cm^{-3}[/tex]
So. [tex]N_{c}=N_{v}[/tex]
Now, Put the value of [tex]N_{c}[/tex] and [tex]N_{v}[/tex] in equation (I)
[tex]n_{i}^2=(2.501\times10^{19})^2 e^{\dfrac{-5.4}{0.0259}}[/tex]
[tex]n_{i}^2=1.7715080315\times10^{-52}[/tex]
[tex]n_{i}=1.330\times10^{-26}[/tex]
Hence, The intrinsic carrier density of diamond is [tex]1.330\times10^{-26}[/tex].
