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An n-type semiconductors has 4 xx 10^(-3)m width, 25 xx 10^(-5)m thickness and 6 xx 10^(-2) m length. 4.8 mA current is flowing through it. Here voltage is applied parallel to the length of the semiconductor. Calculate the current density. The density of the free electron is equal to 10^(22) m^(-3) .What will be the time taken by the electron across the length of the semiconductor ? |
Answer» Solution :l = 6 `xx 10^(-2) m , b = 4 xx 10^(-3 ) m`  h = 25 ` xx 10^(-5) ` m I = `4.8 xx 10^(-3) ` A n= `10^(22) (1)/(m^(3))` J = ? , t = ? Voltage is applied parallel to length, so area is obtained by taking product of breaths and thickness. Area of cross section of conductor, `thereforeA = b xx h ` ` = 4 xx 10^(-3) xx 25 xx 10^(-5)` ` A = 10^(-6) m^(2) ` Current density j = `(I)/(A) = (4.8 xx 10^(-3))/(10^(-6))` `therefore J = 4.8 xx 10^(3) (A)/(m^(2))` Now, I = nAve.... (1) But, suppose time TAKEN by electron to cover distance l is t s., `therefore v = (I)/(t) "" `.... (2) ` therefore I = nA [ (l)/(t) ] e""` [ from equ. (1) and (2) ] ` therefore t = ("nAle")/(l)` ` therefore t = (10^(22) xx 10^(-6) xx 6 xx 10^(-2)xx 1.6 xx 10^(-19))/( 4.8 xx 10^(-3))` `therefore t = 2 xx 10^(-2) ` s |
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