Respuesta :
Explanation :
It is given that,
Length of silicon bar, [tex]l=10\mu m= 0.001\ cm[/tex]
Free electron density, [tex]n_e=104\ cm^3[/tex]
Hole density, [tex]n_h=1016\ cm^3[/tex]
[tex]\mu_n=1200\ cm^2/Vs[/tex]
[tex]\mu_p=500\ cm^2/Vs[/tex]
The total current flowing in bar is the sum of drift current due to hole and the electrons.
[tex]J=J_e+J_h[/tex]
[tex]J=nqE\mu_n+pqE\mu_p[/tex]
where, n and p are electron and hole densities.
[tex]J=Eq[n\mu_n+p\mu_p][/tex]
we know that [tex]E=\dfrac{V}{l}[/tex]
So, [tex]J=\dfrac{V}{l}q[n\mu_n+p\mu_p][/tex]
[tex]J=\dfrac{1.6\times 10^{-19}\ C}{0.001\ V}[104\ cm^{-3}\times 1200\ cm^2/V\ s+1016\ cm^{-3}\times 500\ cm^2/V\ s][/tex]
[tex]J=1012480\times 10^{-16}\ A/m^2[/tex]
or
[tex]J=1.01\times 10^{-9}\ A/m^2[/tex]
Current, [tex]I=JA[/tex]
A is the area of bar, [tex]A= 20\mu m=0.002\ cm[/tex]
So, [tex]I=1.01\times 10^{-9}\ A/m^2\times 0.002\ m^2[/tex]
Current, [tex]I=2.02\times 10^{-12}\ A[/tex]
So, the current flowing in silicon bar is [tex]I=2.02\times 10^{-12}\ A[/tex].
