Answer:
[tex]I =38.46KA[/tex]
Explanation:
Given parameters include:
V = 100mV = 100 × 10⁻³ V
mean free path(W) = 150nm = 150 × 10⁻⁹ m
Length = 250 nm
From Ohm's Law ;
V = IR
[tex]I = \frac{V}{R}[/tex] ---------- equation (1)
Also;
[tex]R = \rho__0\frac{L}{A}[/tex]
where:
[tex]\rho __o[/tex] = resistivity of InSb = 4 × 10⁻¹³ Ω-m
L = length
A = area of cross section
Replacing our values in the above equation; we have:
[tex]R = 4*10^{-13}*\frac{250nm}{150nm*250nm}[/tex]
[tex]R = 4*10^{-13}*\frac{1}{150*10^{-9}m}[/tex]
[tex]R = 2.6*10^{-6}[/tex] Ω
[tex]R = 2.6 \mu[/tex]Ω
From equation (1):
[tex]I = \frac{V}{R}[/tex]
Therefore;
[tex]I = \frac{100*10^{-3}}{2.6*10^{-6}}[/tex]
[tex]I =38461.54 V/[/tex]Ω
Since 1 V/Ω = 0.001 kilo ampere (KA)
Then;
[tex]I =(38461.54 *0.001)KA[/tex]
[tex]I =38.46KA[/tex]
