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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.)

User RyanZim
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1 Answer

2 votes

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.

User Cyrus Loree
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