Question

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.)

Answer 1

Answer:

There are $9.053\times 10^{14}$ charge carriers.

Explanation:

Given that,

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

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 :

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

So, there are $9.053\times 10^{14}$ charge carriers.