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 ?

Explanation

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,
`therefore A = 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