If the current density is 1.55 and the drift velocity is , how many charge carriers are present? The charge on electron is .(Enter your answer to three significant figures.)

Question

asked Jun 22, 2021 199k views

1 Answer

Cyrus Loree · Answer 1 · 2021-06-28T05:14:04+0000

Answer:

There are
9.053* 10^(14) charge carriers.

Step-by-step explanation:

Given that,

Current density (current per unit time),
$(I)/(A)=1.55\ A/cm^2=0.000155\ A/m^2$

The drift velocity of the electron,
$v_d=107\ cm/s=1.07\ m/s$

We need to find the no charge carriers present. The formula for drift velocity is given by :

$v_d=(I)/(neA)\\\\\text{n is no of charge carriers}\\\\n=(I)/(eAv_d)\\\\n=(0.000155)/(1.6* 10^(-19)* 1.07)\\\\n=9.053* 10^(14)$

So, there are
9.053* 10^(14) charge carriers.